# 5: Trigonometric Functions

The trigonometric functions are functions of an angle. They are important in the study of triangles and modeling periodic phenomena, among many other applications.

• 5.0: Prelude to Trigonometric Functions
A function that repeats its values in regular intervals is known as a periodic function. The graphs of such functions show a general shape reflective of a pattern that keeps repeating. This means the graph of the function has the same output at exactly the same place in every cycle. And this translates to all the cycles of the function having exactly the same length.
• 5.1: Angles
An angle is formed from the union of two rays, by keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise.
• 5.2: Unit Circle - Sine and Cosine Functions
In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
• 5.3: The Other Trigonometric Functions
Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.
• 5.4: Right Triangle Trigonometry
We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle. In this section, we will see another way to define trigonometric functions using properties of right triangles.
• 5.E: Trigonometric Functions (Exercises)
• 5.R: Trigonometric Functions (Review)
We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle. In this section, we will see another way to define trigonometric functions using properties of right triangles.

## 5: Trigonometric Functions

We have

sec 4 θ – sec 2 θ = tan 4 θ + tan 2 θ

Taking LHS

= sec 4 θ – sec 2 θ

= sec 2 θ(sec 2 θ – 1)

Using sec 2 θ = tan 2 θ + 1, we get

= (1 + tan 2 θ)tan 2 θ

= tan 2 θ + tan 4 θ

Hence, LHS = RHS (Proved)

### Question 2. sin 6 θ + cos 6 θ = 1 – 3sin 2 θcos 2 θ

We have

sin 6 θ + cos 6 θ = 1 – 3sin 2 θcos 2 θ

Taking LHS

= sin 6 θ + cos 6 θ

= (sin 2 θ) 3 + (cos 2 θ) 3

Using a 3 + b 3 = (a + b)(a 2 + b 2 – ab), we get

= (sin 2 θ + cos 2 θ)(sin 4 θ + cos 4 θ – sin 2 θcos 2 θ)

Using a 2 + b 2 = (a + b) 2 – 2ab and sin 2 θ + cos 2 θ = 1, we get

= (1)[(sin 2 θ + cos 2 θ) 2 – 2sin 2 θcos 2 θ – sin 2 θcos 2 θ]

= (1)[(1) 2 – 3sin 2 θcos 2 θ]

= 1 – 3sin 2 θcos 2 θ

Hence, LHS = RHS (Proved)

### Question 3. (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

We have

(cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

Taking LHS

= (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)

Using cosecθ = 1/sinθ and secθ = 1/cosθ

=

=

=

= 1

### Question 4. cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

We have

cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Taking LHS

=

=

=

=

=

= />

= />

Hence, LHS = RHS(Proved)

### Question 5.

We have

Taking LHS

=

Using a 2 – b 2 = (a + b)(a – b) and a 3 + b 3 = (a + b)(a 2 + b 2 ab), we get

=

=

=

=

= sinA

Hence, LHS = RHS(Proved)

### Question 6.

We have

Taking LHS

=

Using tanA = sinA/cosA and cotA = cosA/sinA, we get

=

=

=

=

Using a 3 – b 3 = (a – b)(a 2 + b 2 + ab), we get

=

=

=

Using cosecA = 1/sinA and secA = 1/cosA, we get

= secAcosecA + 1

Hence, LHS = RHS(Proved)

### Question 7.

We have

Taking LHS

=

Using a 3 ± b 3 = (a ± b)(a 2 + b 2 ± ab), we get

=

Using sin 2 θ + cos 2 θ = 1, we get

= 1 – sinAcosA + 1 + sinAcosA

= 2

Hence, LHS = RHS(Proved)

### Question 8. (secAsecB + tanAtanB) 2 – (secAtanB + tanAsecB) 2 = 1

We have

(secAsecB + tanAtanB) 2 – (secAtanB + tanAsecB) 2 = 1

Taking LHS

= (secAsecB + tanAtanB) 2 – (secAtanB + tanAsecB) 2

Expanding the above equation using the formula

(a + b) 2 = a 2 + b 2 + 2ab

= (secAsecB) 2 + (tanAtanB) 2 + 2(secAsecB)(tanAtanB) –

(secAtanB) 2 – (tanAsecB) 2 – 2(secAtanB)(tanAsecB)

= sec 2 Asec 2 B + tan 2 Atan 2 B – sec 2 Atan 2 B – tan 2 Asec 2 B

= sec 2 A(sec 2 B – tan 2 B) – tan 2 A(sec 2 B – tan 2 B)

= sec 2 A – tan 2 A -(Using sec 2 θ – tan 2 θ = 1)

= 1

Hence, LHS = RHS(Proved)

### Question 9.

We have

Taking RHS

=

=

= ×

=

=

=

=

=

=

=

=

=

= ×

=

=

=

=

Hence, RHS = LHS(Proved)

### Question 10.

We have

Taking LHS

=

Using 1 + tan 2 x = sec 2 x and 1 + cot 2 x = cosec 2 x, we get

=

=

=

=

=

Using a 2 + b 2 = (a + b) 2 – 2ab, we get

=

=

=

Hence, LHS = RHS (Proved)

### Question 11.

We have

Taking LHS

=

By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get

=

=

=

Using a 3 +b 3 = (a + b)(a 2 + b 2 – ab), we get

=

=

= 1 – (sin 2 θ + cos 2 θ) + sinθcosθ

= 1 – 1 + sinθcosθ

= sinθcosθ

Hence, LHS = RHS (Proved)

### Question 12.

We have

=

Taking LHS

=

=

=

=

=

=

=

=

=

Hence, LHS = RHS(Proved)

### Question 13. (1 + tanαtanβ) 2 + (tanα – tanβ) 2 = sec 2 αsec 2 β

We have

(1 + tanαtanβ) 2 + (tanα – tanβ) 2 = sec^2αsec 2 β

Taking LHS

= (1 + tanαtanβ) 2 + (tanα – tanβ) 2

= (1 + tan 2 αtan 2 β + 2tanαtanβ) + (tan 2 α + tan 2 β – 2tanαtanβ)

= 1 + tan 2 αtan 2 β + tan 2 α + tan 2 β

= (1 + tan 2 β) + tan 2 α(1 + tan 2 β)

= (1 + tan 2 β)(1 + tan 2 α)

= sec 2 αsec 2 β

Hence, LHS = RHS (Proved)

## Subsection 0.5.1 Measuring Angles with Radians

Sines, cosines, and tangents are very useful when studying triangles. The input into each of these functions is an angle, and the output tells us the ratio of the lengths of the sides of the triangle. There are two commonly used units for measuring angles, degrees and radians, and so there are two commonly used versions of the trigonometric functions. There's (sin x) where (x) is in degrees, and there's (sin x) where (x) is in radians. Calculus is a lot easier if we measure angles in radians, so that's what we'll use throughout this course. If you ever have trouble getting the right numbers from your calculator, you may want to double check that your calculator is in radian mode.

A radian is a measure of angle which is defined so that if we have an angle with a size of one radian on a unit circle (with a radius (r= 1)), then the length of the arc along the circumference of the circle is also equal to one, as we see in Figure 2. Because the circumference of a circle is (2 pi r ext<,>) this means that for one complete circle,

Similarly half a circle is

egin 57.3^circ approx frac<180^circ> = 1 ext < radian >. end Figure 0.5.2 Common radian measures.

Because we define the radian in this way, this means that the arc length (s) along the circumference of a circle with radius (r) over angle ( heta) can be calculated as (s = r heta) as long as the angle ( heta) is measured in radians.

Figure 0.5.3 Arc length, angle, and radius on a circle.

## 5.6: Phase Shift of Sinusoidal Functions

A periodic function that does not start at the sinusoidal axis or at a maximum or a minimum has been shifted horizontally. This horizontal movement allows for different starting points since a sine wave does not have a beginning or an end.

What are five other ways of writing the function (f(x)=2 cdot sin x ?)

### Phase Shift of Sinusoidal Functions

The general sinusoidal function is:

The constant (c) controls the phase shift. Phase shift is the horizontal shift left or right for periodic functions. If (c=frac<2>) then the sine wave is shifted left by (frac<2>). If (c=-3) then the sine wave is shifted right by (3 .) This is the opposite direction than you might expect, but it is consistent with the rules of transformations for all functions.

To graph a function such as (f(x)=3 cdot cos left(x-frac<2> ight)+1,) first find the start and end of one period. Then sketch only that portion of the sinusoidal axis. Finally, plot the 5 important points for a cosine graph while keeping the amplitude in mind. The graph is shown below.

Generally (b) is always written to be positive. If you run into a situation where (b) is negative, use your knowledge of even and odd functions to rewrite the function.

### Examples

Earlier, you were asked to write (f(x)=2 cdot sin x) in five different ways. The function (f(x)=2 cdot sin x) can be rewritten an infinite number of ways.

(
2 cdot sin x=-2 cdot cos left(x+frac<2> ight)=2 cdot cos left(x-frac<2> ight)=-2 cdot sin (x-pi)=2 cdot sin (x-8 pi)
)

It all depends on where you choose start and whether you see a positive or negative sine or cosine graph.

Given the following graph, identify equivalent sine and cosine algebraic models.

Either this is a sine function shifted right by (frac<4>) or a cosine graph shifted left (frac<5 pi><4>).

At (t=5) minutes William steps up 2 feet to sit at the lowest point of the Ferris wheel that has a diameter of 80 feet. A full hour later he finally is let off the wheel after making only a single revolution. During that hour he wondered how to model his height over time in a graph and equation.

Since the period is 60 which works extremely well with the (360^) in a circle, this problem will be shown in degrees.

William chooses to see a negative cosine in the graph. He identifies the amplitude to be 40 feet. The vertical shift of the sinusoidal axis is 42 feet. The horizontal shift is 5 minutes to the right

The period is 60 (not 65 ) minutes which implies (b=6) when graphed in degrees.

Thus one equation would be:

Tide tables report the times and depths of low and high tides. Here is part of tide report from Salem, Massachusetts dated September 19, 2006.

Find an equation that predicts the height based on the time. Choose when (t=0) carefully.

There are two logical places to set (t=0). The first is at midnight the night before and the second is at 10: 15 AM. The first option illustrates a phase shift that is the focus of this concept, but the second option produces a simpler equation. Set (t=0) to be at midnight and choose units to be in minutes.

These numbers seem to indicate a positive cosine curve. The amplitude is 4 and the vertical shift is 5. The horizontal shift is 615 and the period is 720.

Use the equation from Example 4 to find out when the tide will be at exactly (8 mathrm) on September (19^).

This problem gives you the (y) and asks you to find the (x). Later you will learn how to solve this algebraically, but for now use the power of the intersect button on your calculator to intersect the function with the line (y=8). Remember to find all the (x) values between 0 and 1440 to account for the entire 24 hours.

There are four times within the 24 hours when the height is exactly 8 feet. You can convert these times to hours and minutes if you prefer.

(t approx 532.18) (8:52), 697.82 (11:34), 1252.18 (20:52), 1417.82 (23:38)

Graph each of the following functions.

Give one possible sine equation for each of the graphs below.

Give one possible cosine function for each of the graphs below.

The temperature over a certain 24 hour period can be modeled with a sinusoidal function. At 3: 00 , the temperature for the period reaches a low of (22^ mathrm). At (15: mathrm), the temperature for the period reaches a high of (40^ F)

12. Find an equation that predicts the temperature based on the time in minutes. Choose (t=0) to be midnight.

13. Use the equation from #12 to predict the temperature at (4: 00 mathrm).

14. Use the equation from #12 to predict the temperature at 8: 00 AM.

15. Use the equation from #12 to predict the time(s) it will be (32^ mathrm).

## 5: Trigonometric Functions

Class Notes

The McGraw-Hill Ryerson PreCalculus 12 Text is used as the Main Resource.

Assignments in the Powerpoint Lesson Plans refer to pages and questions in the PreCalculus 12 text.

5.1 Graphing Sine and Cosine Functions

5.1 Formative Assessment Amplitude and Period

Digital Resources

5.1 Sin and Cos curves 2

Graphing Point on Circle

Unit Circle Graphing Sine and Cosine

Pedagogical Shifts: TRANSFORM, Moving from Traditional to Student-Centered

Shifting from Student as Knowledge Recipient to Student as Inquirer and Creator

Shifting from Memorization to Higher-level Thinking

Desmos pre-created interactive graphs for trig functions are available online. Follow this link.

https://www.desmos.com/calculator Click on the bars in the upper left corner to view all pre-created interactive graphs.

Building a Sine Curve and Making Connections to the Unit Circle.

This is a nice idea to make connections between the unit circle and the graph of the sine function. Students can trace the heights of the curve to build half a period of a sine curve. Possible follow up questions include "What would the other half of the graph look like?" Connections could also be made to characteristics of a sine function graph such as amplitude and period.

## 5: Trigonometric Functions

With this section we’re going to start looking at the derivatives of functions other than polynomials or roots of polynomials. We’ll start this process off by taking a look at the derivatives of the six trig functions. Two of the derivatives will be derived. The remaining four are left to you and will follow similar proofs for the two given here.

Before we actually get into the derivatives of the trig functions we need to give a couple of limits that will show up in the derivation of two of the derivatives.

See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits.

Before proceeding a quick note. Students often ask why we always use radians in a Calculus class. This is the reason why! The proof of the formula involving sine above requires the angles to be in radians. If the angles are in degrees the limit involving sine is not 1 and so the formulas we will derive below would also change. The formulas below would pick up an extra constant that would just get in the way of our work and so we use radians to avoid that. So, remember to always use radians in a Calculus class!

Before we start differentiating trig functions let’s work a quick set of limit problems that this fact now allows us to do.

1. (displaystyle mathop limits_ < heta o 0>frac<><<6 heta >>)
2. (displaystyle mathop limits_ frac < ight)>>)
3. (displaystyle mathop limits_ frac< ight)>>)
4. (displaystyle mathop limits_ frac < ight)>>< ight)>>)
5. (displaystyle mathop limits_ frac < ight)>><>)
6. (displaystyle mathop limits_ frac < ight) - 1>>)

There really isn’t a whole lot to this limit. In fact, it’s only here to contrast with the next example so you can see the difference in how these work. In this case since there is only a 6 in the denominator we’ll just factor this out and then use the fact.

Now, in this case we can’t factor the 6 out of the sine so we’re stuck with it there and we’ll need to figure out a way to deal with it. To do this problem we need to notice that in the fact the argument of the sine is the same as the denominator (i.e. both ( heta )’s). So we need to get both of the argument of the sine and the denominator to be the same. We can do this by multiplying the numerator and the denominator by 6 as follows.

Note that we factored the 6 in the numerator out of the limit. At this point, while it may not look like it, we can use the fact above to finish the limit.

To see that we can use the fact on this limit let’s do a change of variables. A change of variables is really just a renaming of portions of the problem to make something look more like something we know how to deal with. They can’t always be done, but sometimes, such as this case, they can simplify the problem. The change of variables here is to let ( heta = 6x) and then notice that as (x o 0) we also have ( heta o 6left( 0 ight) = 0). When doing a change of variables in a limit we need to change all the (x)’s into ( heta )’s and that includes the one in the limit.

Doing the change of variables on this limit gives,

And there we are. Note that we didn’t really need to do a change of variables here. All we really need to notice is that the argument of the sine is the same as the denominator and then we can use the fact. A change of variables, in this case, is really only needed to make it clear that the fact does work.

In this case we appear to have a small problem in that the function we’re taking the limit of here is upside down compared to that in the fact. This is not the problem it appears to be once we notice that,

and then all we need to do is recall a nice property of limits that allows us to do ,

With a little rewriting we can see that we do in fact end up needing to do a limit like the one we did in the previous part. So, let’s do the limit here and this time we won’t bother with a change of variable to help us out. All we need to do is multiply the numerator and denominator of the fraction in the denominator by 7 to get things set up to use the fact. Here is the work for this limit.

This limit looks nothing like the limit in the fact, however it can be thought of as a combination of the previous two parts by doing a little rewriting. First, we’ll split the fraction up as follows,

Now, the fact wants a (t) in the denominator of the first and in the numerator of the second. This is easy enough to do if we multiply the whole thing by (< extstyle>) (which is just one after all and so won’t change the problem) and then do a little rearranging as follows,

At this point we can see that this really is two limits that we’ve seen before. Here is the work for each of these and notice on the second limit that we’re going to work it a little differently than we did in the previous part. This time we’re going to notice that it doesn’t really matter whether the sine is in the numerator or the denominator as long as the argument of the sine is the same as what’s in the numerator the limit is still one.

Here is the work for this limit.

This limit almost looks the same as that in the fact in the sense that the argument of the sine is the same as what is in the denominator. However, notice that, in the limit, (x) is going to 4 and not 0 as the fact requires. However, with a change of variables we can see that this limit is in fact set to use the fact above regardless.

So, let ( heta = x - 4) and then notice that as (x o 4) we have ( heta o 0). Therefore, after doing the change of variable the limit becomes,

The previous parts of this example all used the sine portion of the fact. However, we could just have easily used the cosine portion so here is a quick example using the cosine portion to illustrate this. We’ll not put in much explanation here as this really does work in the same manner as the sine portion.

All that is required to use the fact is that the argument of the cosine is the same as the denominator.

Okay, now that we’ve gotten this set of limit examples out of the way let’s get back to the main point of this section, differentiating trig functions.

We’ll start with finding the derivative of the sine function. To do this we will need to use the definition of the derivative. It’s been a while since we’ve had to use this, but sometimes there just isn’t anything we can do about it. Here is the definition of the derivative for the sine function.

Since we can’t just plug in (h = 0) to evaluate the limit we will need to use the following trig formula on the first sine in the numerator.

[sin left( ight) = sin left( x ight)cos left( h ight) + cos left( x ight)sin left( h ight)]

As you can see upon using the trig formula we can combine the first and third term and then factor a sine out of that. We can then break up the fraction into two pieces, both of which can be dealt with separately.

Now, both of the limits here are limits as (h) approaches zero. In the first limit we have a (sin left( x ight)) and in the second limit we have a (cos left( x ight)). Both of these are only functions of (x) only and as (h) moves in towards zero this has no effect on the value of (x). Therefore, as far as the limits are concerned, these two functions are constants and can be factored out of their respective limits. Doing this gives,

At this point all we need to do is use the limits in the fact above to finish out this problem.

[frac<>left( ight) = sin left( x ight)left( 0 ight) + cos left( x ight)left( 1 ight) = cos left( x ight)]

Differentiating cosine is done in a similar fashion. It will require a different trig formula, but other than that is an almost identical proof. The details will be left to you. When done with the proof you should get,

With these two out of the way the remaining four are fairly simple to get. All the remaining four trig functions can be defined in terms of sine and cosine and these definitions, along with appropriate derivative rules, can be used to get their derivatives.

Let’s take a look at tangent. Tangent is defined as,

Now that we have the derivatives of sine and cosine all that we need to do is use the quotient rule on this. Let’s do that.

Now, recall that (left( x ight) + left( x ight) = 1) and if we also recall the definition of secant in terms of cosine we arrive at,

The remaining three trig functions are also quotients involving sine and/or cosine and so can be differentiated in a similar manner. We’ll leave the details to you. Here are the derivatives of all six of the trig functions.

#### Derivatives of the six trig functions

At this point we should work some examples.

1. (gleft( x ight) = 3sec left( x ight) - 10cot left( x ight))
2. (hleft( w ight) = 3> - an left( w ight))
3. (y = 5sin left( x ight)cos left( x ight) + 4csc left( x ight))
4. (displaystyle Pleft( t ight) = frac<><<3 - 2cos left( t ight)>>)

There really isn’t a whole lot to this problem. We’ll just differentiate each term using the formulas from above.

In this part we will need to use the product rule on the second term and note that we really will need the product rule here. There is no other way to do this derivative unlike what we saw when we first looked at the product rule. When we first looked at the product rule the only functions we knew how to differentiate were polynomials and in those cases all we really needed to do was multiply them out and we could take the derivative without the product rule. We are now getting into the point where we will be forced to do the product rule at times regardless of whether or not we want to.

We will also need to be careful with the minus sign in front of the second term and make sure that it gets dealt with properly. There are two ways to deal with this. One way it to make sure that you use a set of parentheses as follows,

Because the second term is being subtracted off of the first term then the whole derivative of the second term must also be subtracted off of the derivative of the first term. The parenthesis make this idea clear.

A potentially easier way to do this is to think of the minus sign as part of the first function in the product. Or, in other words the two functions in the product, using this idea, are ( - ) and ( an left( w ight)). Doing this gives,

[h'left( w ight) = - 12> - 2w an left( w ight) - left( w ight)]

So, regardless of how you approach this problem you will get the same derivative.

As with the previous part we’ll need to use the product rule on the first term. We will also think of the 5 as part of the first function in the product to make sure we deal with it correctly. Alternatively, you could make use of a set of parentheses to make sure the 5 gets dealt with properly. Either way will work, but we’ll stick with thinking of the 5 as part of the first term in the product. Here’s the derivative of this function.

In this part we’ll need to use the quotient rule to take the derivative.

Be careful with the signs when differentiating the denominator. The negative sign we get from differentiating the cosine will cancel against the negative sign that is already there.

This appears to be done, but there is actually a fair amount of simplification that can yet be done. To do this we need to factor out a “-2” from the last two terms in the numerator and the make use of the fact that (left( heta ight) + left( heta ight) = 1).

As a final problem here let’s not forget that we still have our standard interpretations to derivatives.

where (t) is in years. During the first 10 years in which the account is open when is the amount of money in the account increasing?

To determine when the amount of money is increasing we need to determine when the rate of change is positive. Since we know that the rate of change is given by the derivative that is the first thing that we need to find.

[P'left( t ight) = - 100sin left( t ight) - 150cos left( t ight)]

Now, we need to determine where in the first 10 years this will be positive. This is equivalent to asking where in the interval (left[ <0,10> ight]) is the derivative positive. Recall that both sine and cosine are continuous functions and so the derivative is also a continuous function. The Intermediate Value Theorem then tells us that the derivative can only change sign if it first goes through zero.

So, we need to solve the following equation.

[egin - 100sin left( t ight) - 150cos left( t ight) & = 0 100sin left( t ight) & = - 150cos left( t ight) frac<><> & = - 1.5 an left( t ight) & = - 1.5end]

The solution to this equation is,

[egint = 2.1588 + 2pi n,&hspace<0.25in>n = 0, pm 1, pm 2, ldots t = 5.3004 + 2pi n, & hspace<0.25in>n = 0, pm 1, pm 2, ldots end]

If you don’t recall how to solve trig equations go back and take a look at the sections on solving trig equations in the Review chapter.

We are only interested in those solutions that fall in the range (left[ <0,10> ight]). Plugging in values of (n) into the solutions above we see that the values we need are,

So, much like solving polynomial inequalities all that we need to do is sketch in a number line and add in these points. These points will divide the number line into regions in which the derivative must always be the same sign. All that we need to do then is choose a test point from each region to determine the sign of the derivative in that region.

Here is the number line with all the information on it.

So, it looks like the amount of money in the bank account will be increasing during the following intervals.

[2.1588 < t < 5.3004hspace<0.5in>8.4420 < t < 10]

Note that we can’t say anything about what is happening after (t = 10) since we haven’t done any work for (t)’s after that point.

In this section we saw how to differentiate trig functions. We also saw in the last example that our interpretations of the derivative are still valid so we can’t forget those.

Also, it is important that we be able to solve trig equations as this is something that will arise off and on in this course. It is also important that we can do the kinds of number lines that we used in the last example to determine where a function is positive and where a function is negative. This is something that we will be doing on occasion in both this chapter and the next.

## A right triangle has acute angles A and B. If and , what are and ?

Since A and B are the acute angles in a right triangle, they are complementary angles.

Substitute for B. Use the identity (the cofunctions are equal). Substitute the given value.

Substitute for A. The cofunctions of any pair of complementary angles are equal. Substitute the given value.

What are the values of and ?

Incorrect. You probably used the acute angle W, and found . Remember that you get different ratios for the two acute angles, so pay careful attention to which angle you are using. The correct answer is C.

Incorrect. You may have used the acute angle W and also switched cosine and cosecant. Remember that you get different ratios for the two acute angles, so pay careful attention to which angle you are using. The correct answer is C.

Correct. Using the definition of cosine, . Using the definition of cosecant, .

Incorrect. It looks like you switched the values of cosine and cosecant. The names are very similar, so be careful to use the right definition. The correct answer is C.

Relationships Among the Trigonometric Functions

The six ratios or functions are usually thought of as two groups of three functions. The first group is:

One way to remember these three definitions is with a memory device that uses the first letter of each word. The definition of sine is represented by soh (sine equals opposite over hypotenuse). Likewise, the definition of cosine is represented by cah (cosine equals adjacent over hypotenuse), and the definition of tangent is represented by toa (tangent equals opposite over adjacent). Putting these together gives you sohcahtoa.

If you compare these three ratios to the three above them, you’ll see that these three fractions are the reciprocals of the three fractions above them. That is, cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. Writing this gives three more identities:

If you remember sohcahtoa plus these three identities, you can find the values of any trigonometric functions, as seen in the following example.

For acute angle A, and . Find the values of the other four trigonometric ratios for angle A.

The definition of sine tells you that . A triangle with and will have this ratio.

You also know that . You are given , so .

Now you have all three sides of the triangle and you can use the definition of tangent.

Next, use the three reciprocal identities to obtain the other three ratios.

The value of any trigonometric function is a ratio, or a fraction. Remember that fractions can be reduced.

For acute angle A, and . Find the values of and .

You want a right triangle where the ratio of the side adjacent to angle A over the hypotenuse is . A triangle with sides and would have this ratio.

You can use the definition of tangent to find the opposite side. Substitute the value you are given for tangent and then solve the equation.

Now you have all three sides. Use the definition of sine to find its value.

Now using the reciprocal identity, the csc can be found by taking the reciprocal of the sin.

Remember that the sides of a right triangle satisfy the Pythagorean Theorem. So if a and b are the lengths of the legs, and c is the hypotenuse, you must have . In the last example, the lengths of the legs were 2 and 3, and the hypotenuse was , and it is true that .

Which of the following could be the values of the trigonometric functions of the same angle?

Incorrect. You can have these values of sine and tangent for the same angle. However, the values of sine and cosecant of the same angle are reciprocals. If , then , not . The correct answer is D.

Incorrect. The values of cosine and secant are reciprocals, as they should be. However, you cannot have the given values of sine and cosine for the same angle. If , you can draw a right triangle with the leg opposite angle X having length 4 and the hypotenuse having length 5. If you have , then the adjacent leg length is 2. However, the lengths 2, 4, and 5 do not satisfy the Pythagorean Theorem. The correct answer is D.

Incorrect. You can draw a right triangle with the side opposite angle Y having length 12, the adjacent side having length 5, and the hypotenuse having length 13. This will give you . Using the definition of tangent, , you would then have , not . The correct answer is D.

Correct. If , then , because they are reciprocals. You can draw a right triangle with both legs having length 1, and the hypotenuse will have length because of the Pythagorean Theorem. Using the definition of cosine, , you will then have .

Using a Calculator to Find the Values of Trigonometric Functions

You know that if you draw similar triangles with angle measures 35°, 55°, and 90°, the ratio of the side opposite 35° to the hypotenuse will be the same for all those triangles. This is . The easiest way to find what this ratio actually equals is with a scientific or graphing calculator.

Looking at a calculator, you will find a key that says SIN on it. You can use this to find the value of . Keep this in mind: you need to know that there are different units for measuring angles. For our purposes, make sure that your calculator is set in the “degree mode.” (The following instructions are generalized, but you may need to refer to your calculator’s instruction manual for how to perform these calculations on your particular calculator.)

If you use a scientific calculator, look in the display and see if it says DEG in small letters above the 0 (as opposed to RAD or GRAD). If it does not, press the DRG key until the display says DEG. Now enter 35, and then press the SIN key. The result is :

If you have a graphing calculator, press the MODE key. The third line of the display will say RADIAN DEGREE. Use the arrows to select DEGREE, then press ENTER, 2ND, QUIT. Now the calculator is in degree mode. On a graphing calculator, you enter things the same way as you would write them. So press the keys to give you sin(35) on the display and then press ENTER. You should now see the value on the next line of the display.

Because sine is a function, given an angle measure X (the input), your calculator will give you the value of (the output). All the right triangles with acute angle measure X will be similar, so the ratio of the opposite side to the hypotenuse will be the same for all of those triangles. Therefore, the ratio depends only on the value of X it does not depend on the triangle.

Likewise, the other five trigonometric ratios are functions. You can use your calculator to find the value of those functions. You will notice that next to the SIN key there are COS and TAN keys, which can be used to find the values of cosine and tangent.

Use your calculator to find the values of and to the nearest thousandth.

On a scientific calculator, enter 35, then press COS. Do this in the reverse order for a graphing calculator.

Remember to look at the ten thousandths place to help you round to the nearest thousandth.

Use the same procedure for tangent.

You may have noticed that your calculator has no keys for csc, sec, or cot. You can still use it to find the values of these functions. You can do this by using the calculator in combination with the reciprocal identities. You must first find the value of sin, cos, or tan, and then find the reciprocal, as this next example shows.

Use your calculator to find the values of and to the nearest thousandth.

First use your calculator to find the value of . Do not round this value until you are writing the final answer.

Press the key that says or . This will give you the value of cosecant.

Now round your final answer to the nearest thousandth.

Find the value of . Then find the reciprocal and round off.

Find the value of . Then find the reciprocal and round off.

What is the value of to the nearest thousandth?

Correct. First use the calculator to find . Find the reciprocal of this: .

Incorrect. You found the value of . Remember that cosecant is the reciprocal of sine (not cosine). The correct answer is 3.420.

Incorrect. You found the value of . You must first find , then find the reciprocal. The correct answer is 3.420.

Incorrect. Your calculator was not set to degrees. Before doing any calculations, make sure that you first have it set on degrees. The correct answer is 3.420.

Using a Calculator to Find Angle Measures

So far you have learned the definitions of the six trigonometric functions. Remember that a function has an input and an output. For each of these functions, the input is the angle measure and the output equals a certain ratio of sides. Your calculator can be used to find the values of these functions. For example, if the angle measures 60°, the cosine of the angle is 0.5. This can be represented as .

Now what if the situation were reversed? What if you knew the value of the ratio and wanted to know the angle that produced it? That is, what if you knew the output of a trigonometric function, and wanted to know the input? For example, you might know that the cosine of some angle is 0.5 and want to find out what the angle is. You can use your calculator to find these values, too.

In general, when you reverse the input and the output of a function, what you get is called an inverse function. Your calculator can find the inverses of sine, cosine, and tangent. In the example above, on a scientific calculator you would enter 0.5, press the 2ND key, then press COS. The display would show 60. (Make sure that your calculator is set on degrees!) This tells you that the angle is 60°. On a graphing calculator, you would press 2ND, then COS, then 0.5, and finally ENTER. (Keep in mind that you may need to refer to your calculator’s instruction manual for how to perform these calculations on your particular calculator.)

Above the SIN, COS, and TAN keys you will see . These are the inverse trigonometric functions, and the way to read them out loud is: arcsine, arccosine, and arctangent. The result mentioned above can be written as or .

If you were given the value of the sine (or tangent) function and wanted to know what angle produced it, you would follow a procedure similar to that described above. So on a scientific calculator, you would enter the value, press the 2ND key, then press SIN (or TAN).

Use your calculator to find the angle, to the nearest degree, whose tangent value is 0.75.

On a scientific calculator, enter 0.75, then press the 2ND key and TAN. Do this in the reverse order for a graphing calculator.

Look at the tenths place to help you round to the nearest degree.

If you are given the expression , for example, you can interpret this as saying, “Find the angle whose cosine equals 0.24.”

Determine to the nearest tenth of a degree.

On a scientific calculator, enter 0.24, then press the 2ND key and COS. Do this in the reverse order for a graphing calculator.

Look at the hundredths place to help you round to the nearest tenth.

Here is a real-world example using an inverse function.

A skateboard ramp is 7 feet long with one end on the ground and the other end 2 feet above the ground. What is the angle of elevation to the nearest tenth of a degree?

The angle of elevation is angle A. Because you know the opposite side and the hypotenuse, you can use the sine function.

Use the definition of sine. The unknown is the input.

You can rewrite this equation using arcsine. You need to reverse the input and the output.

On a scientific calculator, divide 2 by 7, then press the 2ND key and SIN. Do this in the reverse order for a graphing calculator.

Remember that the sine or cosine function cannot have an output greater than 1. With arcsine and arccosine, you are reversing inputs and outputs. Consequently, the input of these functions cannot be a number bigger than 1. If you try to compute with your calculator, for example, you will get an error message.

If , what is x to the nearest hundredth of a degree?

Incorrect. Instead of finding , you found . The correct answer is 19.47°.

Incorrect. You did not have your calculator set on degrees. Before doing any calculations, make sure that you first have it set on degrees. The correct answer is 19.47°.

Correct. The solution to the equation is given by computing .

Incorrect. Instead of finding , you found . The correct answer is 19.47°.

The six trigonometric functions are defined as ratios of sides in a right triangle. Their values depend only on the angle and not on any particular right triangle. A good way to remember the definitions of sine, cosine, and tangent is with the memory device sohcahtoa. The other three functions—cosecant, secant, and cotangent—are reciprocals of the first three.

You can use a calculator to find the values of these functions or ratios. You can also use a calculator to find the values of the inverse trigonometric functions. That is, given the ratio, you can find the angle that produced it.

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## Unit circle radians

If you have your number line marked with radians, this is how it would look:

First, you have a usual unit circle. In one quarter of a circle is $frac<2>$, in one half is $pi$, in three quarters is $frac<3 pi><2>$, and one whole is $2 pi$.

Now what when you start another lap?

You’re again in zero, but now with 2π of the line around the circle. If you add another $frac<2>$ that will lead you into the point where the ‘old’ $frac<2>$ is, but now that value will be $2 pi + frac <2>= frac<5 pi><2>$. If you continue and add another $frac<2>$ you’ll find yourself in a point where the ‘old’ π lies. Now that point will be $frac<5 pi> <2>+ frac <2>= 3 pi$. And you continue like that.

In degrees you can conclude that $frac <2>= 90^$, $pi = 180^$, $frac<3 pi> <2>= 270^$, $2 pi = 360^$, $frac<5 pi> <2>= 450^$ and so on.

By dividing radians into smaller and smaller parts we can determine measure of every angle.

Angles that are mostly used are 0, $frac<6>$, $frac<3>$, $frac<2>$ and so on.

Can you see a pattern here? If you only observe first and second quadrant you’ll notice that lines that are perpendicular on y-axis that go through $frac<3>$ and $frac<2 pi><3>$ cut off equal parts of y-axis. The same applies with $frac<3 pi><4>$ and $frac<4>$, and also with $frac<5 pi><6>$ and $frac<6>$.

If you take a look at first and fourth quadrant you’ll notice that lines that are perpendicular of x-axis $frac<6>$ and $frac<11 pi><6>$ cut off equal length of x – axis, and so on with other angles. This can help you in drawing them. For example, if you get a task to draw $frac<5 pi><6>$, you can simply draw $frac<6>$ and translate it to second quadrant. Using this way you’ll only need to remember angles in first quadrant and translate them.

Example 1: Find following angles on the unit circle

When you have a fraction whose value is greater than two, that means that you’re starting another “lap” around the circle. When you are dealing with these kinds of values, you have to apply a process of finding the right measure of an angle. That means that you have to find an angle that suits given angle but in your first lap. You do that by subtracting with a multiple of 2π.

For example, let’s say you have $frac<5 pi><2>$. $frac<5 pi><2>$ is greater than $2 pi$ by $frac<2>$. That means that you’ll finish first lap and end up in $frac<2>$.

Example 2: Find following angles on the unit circle

Since we have a minus in front of our values, we start looking from zero but in an opposite way. Whole circle is equal to $2 pi$, which means that $-frac<4>$ will have the same value as $2 pi – frac <4>= frac<7 pi><4>$, $- pi$ as $pi$, and $-2 pi$ as 0.

## Computing Trigonometric Functions

This is a completely optional page. It is not necessary to know how to compute the trig functions and their inverses in order to use them. Nonetheless, many people are interested in how values of these functions were computed before and after the invention of calculators and computers. If you&rsquore interested, then read on. Otherwise, go on to the next section on oblique triangles.

#### Before computers: tables

Rather than repeating what he did for chords, let&rsquos look at how to create tables for sines and cosines using his methods. First, based on the Pythagorean theorem and similar triangles, the sines and cosines of certain angles can be computed directly. In particular, you can directly find the sines and cosines for the angles 30°, 45°, and 60° as described in the section on cosines. Ptolemy knew two other angles that could be constructed, namely 36° and 72°. These angles were constructed by Euclid in Proposition IV.10 of his Elements. Like Ptolemy, we can use that construction to compute the trig functions for those angles. At this point we could compute the trig functions for the angles 30°, 36°, 45°, 60°, and 72°, and, of course we know the values for 0° and 90°, too.

Keep in mind that if you know the sine of an angle &theta, then you know the cosine of the complementary angle 90° &ndash &theta likewise, if you know then cosine of an angle &theta then you know the sine of the complementary angle 90° &ndash &theta:

So you have the trig functions for 18° and 54°, too.

Next, you can use the half-angle formulas for sines and cosines to compute the values for half of an angle if you know the values for the angle. If &theta is an angle between 0° and 180°, then

Using these, from the values for 18°, 30°, and 54°, you can find the values for 27°, 15°, and 9°, and, therefore, their complements 63°, 75°, and 81°.

With the help of the sum and difference formulas

you can find the sine and cosine for 3° (from 30° and 27°) and then fill in the tables for sine and cosine for angles from 0° though 90° in increments of 3°.

Again, using half-angle formulas, you could produce a table with increments of 1.5° (that is, 1° 30'), then 0.75° (which is 45'), or even of 0.375° (which is 22' 30"). But how do you get a table with 1° increments? Ptolemy recognized that there was no Euclidean construction to trisect an angle of 3° to get an angle of 1°, but since the sine function is almost linear for small angles, you could approximate sin 1° just by interpolating a third of the way beteen the values of sin 0.75° and sin 1.5°. With that step, we can construct trig tables for trig functions with increments of 1°.

Better trig tables have been created throughout the centuries. For instance, Ulugh Beg (15th century) constructed sine and tangent tables for every minute of arc to about nine digits of accuracy!

Incidentally, if you have a table of sines, you can read it in reverse to compute arcsine, so only one table is needed for both.

#### After computers: power series

In the late 17th century, Newton and other mathematicians developed power series. A power series is like a polynomial of unbounded degree. For the various trig functions, these mathematicians found power series. Here are the power series for sine and cosine (where x is an angle measured in radians):

The three dots . mean that the expression is to go on forever, adding another term, then subtracting a term, etc. The exclamation point ! is to be read &ldquofactorial&rdquo, and it means you multiply together the whole numbers from 1 up through the given number. For example, 5!, &ldquofive factorial&rdquo, equals 1 times 2 times 3 times 4 times 5, which is 120, and so, 6! = 720.

These power series have infinitely many terms, but they get small so very fast that only the first few terms contribute much.

0.78539816 — 0.78539816 3 /3! + 0.78539816 5 — 0.78539816 7 /7! +.
0.78539816 = 0.78539816
0.70465265 = 0.78539816 — 0.78539816 3 /3!
0.70714304 = 0.78539816 — 0.78539816 3 /3! + 0.78539816 5 /5!
0.70710647 = 0.78539816 — 0.78539816 3 /3! + 0.78539816 5 /5! — 0.78539816 7 /7!
0.70710678 = 0.78539816 — 0.78539816 3 /3! + 0.78539816 5 /5! — 0.78539816 7 /7! + 0.78539816 9 /9!

A little bit of analysis is needed to determine how many terms of the power series are needed to achieve the desired accuracy. Also, certain other tricks can be used to speed up the computations. In any case, the essential idea is to use the first few terms of a power series to compute trig functions.

The power series for the rest of the trig functions and the power series for the inverse trig functions can be found in most books on calculus that discuss power series.