]> Chapter 06 Notes (Sullivan & Sullivan 3e)

Chapter 06: Analytic Trigonometry

§6.1: Inverse Trigonometric Functions

The Problem

As you recall from our earlier work, a function can only have an inverse function if it is one-to-one. Are any of our trigonometric functions one-to-one? Alas, no…which is a problem! There are times when we would really like to be able to find an angle, given the value of, say, the sine ratio…

The Solution

There is a way—we even did it afew times before—to force a function to have an inverse: restrict thedomain of the function.

Consider the sine function:

An Image

Can you see a domain restriction (set of x-values) that would make the function one-to-one? (there are several)

Restrictions

Here are the standard restrictions:

Function	Restricted Domain
Sine	$[- \frac{π}{2}, \frac{π}{2}]$
Cosine	$[0, π]$
Tangent	$(- \frac{π}{2}, \frac{π}{2})$

Remember that the domain and range switch places once we actually find the inverse function. Also, note that these inverse functions are only defined for radian angle measures!

Thus, we can summarize the basic inverse trigonometric functions:

Function	Domain	Range
$y = \sin^{- 1} (x)$	$x \in [- 1, 1]$	$y \in [- \frac{π}{2}, \frac{π}{2}]$
$y = \cos^{- 1} (x)$	$x \in [- 1, 1]$	$y \in [0, π]$
$y = \tan^{- 1} (x)$	$x \in ℝ$	$y \in (- \frac{π}{2}, \frac{π}{2})$

Using Your Calculator

These basic inverse trigonometric functions are built into your calculator, so no special techniques are needed to use them.

Examples

[1.] Find $\sin^{- 1} (\frac{1}{2})$ .

Since $\sin (\frac{π}{6}) = \frac{1}{2}$ , $\sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$ . There are an infinite number of angles whose sine value is one half, but this angle is the only one that is in the range of the inverse sine function.

[2.] Find $\cos^{- 1} (- \frac{1}{\sqrt{2}})$ .

Since $\cos (\frac{3 π}{4}) = - \frac{1}{\sqrt{2}}$ , $\cos^{- 1} (- \frac{1}{\sqrt{2}}) = \frac{3 π}{4}$ . The same comment above applies here, too.

§6.2: More Inverse Trigonometric Functions

Mixing Regular and Inverse Functions

Note that the inverse functions take in a number and spit out an angle; thus, an expression like $\cos (\sin^{- 1} (\frac{2}{3}))$ is really just cosine of some angle. One way to deal with expressions like this is to draw a picture of that mystery angle, and then find the trig function value of that angle.

The situation is more difficult if you have the inverse function on the outside—like, say $\sin^{- 1} (\cos (π))$ . In cases like this, the inner trig function must produce a value which is one of your memorized trig values if you’re to have any hope of working out the value without a calculator.

Of course, situations like $\sin^{- 1} (\sin (\frac{7 π}{6}))$ are usually easier…but be careful! $\sin^{- 1} (\sin (\frac{7 π}{6})) \neq \frac{7 π}{6}$ ! Remember the restrictions on the inverse functions…

The Other Inverse Functions

There are three other trigonometric functions whose inverse we haven’t looked at yet…let’s have at it, then.

Function	Domain	Range
$y = \sec^{- 1} (x)$	$x \in (- \infty, - 1] \cup [1, \infty)$	$y \in [0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
$y = \csc^{- 1} (x)$	$x \in (- \infty, - 1] \cup [1, \infty)$	$y \in [- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$
$y = \cot^{- 1} (x)$	$x \in ℝ$	$y \in (0, π)$

There is some disagreement about these restrictions…in fact, there are many who don’t even bother to use these inverse functions at all, instead relying on just the basic three.

Using Your Calculator

The domain restrictions are one problem; calculator issues are the other issue. Very few, if any calculators have an inverse secant function…thus, the problem must be reworked so that one of the basic trig functions is being used, and then the problem is solved using the basic inverse trig functions.

Examples

[3.] Find $\cos (\sin^{- 1} (\frac{2}{3}))$ .

Let’s just say that $\sin^{- 1} (\frac{2}{3}) = θ$ . I can draw a triangle that uses that angle:

An Image

Note that the other leg must have length $\sqrt{5}$ . So, I can now find $\cos (θ)$ …it is $\frac{\sqrt{5}}{3}$ . Thus, $\cos (\sin^{- 1} (\frac{2}{3})) = \frac{\sqrt{5}}{3}$ . I know that the answer must be positive since inverse sine of a positive number must return an angle in the first quadrant, and cosine of that type of angle is positive.

[4.] Find $\sec^{- 1} (\frac{2}{\sqrt{3}})$ .

Let’s say $\sec^{- 1} (\frac{2}{\sqrt{3}}) = a$ ; thus, $\sec (a) = \frac{2}{\sqrt{3}} \Rightarrow \cos (a) = \frac{\sqrt{3}}{2} \Rightarrow a = \cos^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ . This is how you’d need to approach the problem with a calculator, also.

§6.3: Trigonometric Identities

The Process

We’ve encountered proofs before! The main things to remember are [1] you can’t work both sides, [2] pick the more complicated side to work on, and [3] don’t panic!

Working these proofs requires that you see work a lot on your own, and that you see and recognize patterns in the proofs. There are plenty of problems to practice in the textbook!

Examples

[5.] Show that $(1 - \sin (θ)) (1 + \sin (θ)) = \cos^{2} (θ)$ .

Well…I’m certainly going to work the left side of that!

$(1 - \sin (θ)) (1 + \sin (θ)) = 1 - \sin^{2} (θ)$ , which looks a lot like the Pythagorean identity: $\sin^{2} (θ) + \cos^{2} (θ) = 1 \Rightarrow \cos^{2} (θ) = 1 - \sin^{2} (θ)$ . Thus, I’ve done it. That was easy!

[6.] Show that $\tan (x) + \cot (x) = \sec (x) \csc (x)$ .

Again, I’ll work the left side of that.

$\tan (x) + \cot (x) = \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\sin (x)} = \frac{\sin (x)}{\sin (x)} \cdot \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} \cdot \frac{\cos (x)}{\sin (x)} = \frac{\sin^{2} (x)}{\sin (x) \cos (x)} + \frac{\cos^{2} (x)}{\sin (x) \cos (x)} = \frac{\sin^{2} (x) + \cos^{2} (x)}{\sin (x) \cos (x)} = \frac{1}{\sin (x) \cos (x)} = \frac{1}{\sin (x)} \cdot \frac{1}{\cos (x)} = \sec (x) \csc (x)$ .

[7.] Show that $\frac{1 + \sin (α)}{\cos (α)} = \frac{\cos (α)}{1 - \sin (α)}$ .

Hmmm…neither side looks more complicated…but I like the denominator on the right less, so I’ll work that side. Note the trick! $\frac{\cos (α)}{1 - \sin (α)} \cdot \frac{1 + \sin (α)}{1 + \sin (α)} = \frac{(\cos (α)) (1 + \sin (α))}{1 - \sin^{2} (α)} = \frac{(\cos (α)) (1 + \sin (α))}{\cos^{2} (α)} = \frac{1 + \sin (α)}{\cos (α)}$ .

§6.4: Sum and Difference Formulas

Deriving the Formulas

Let’s see where these come from! I’ll use the unit circle approach.

Cosine

Let’s draw a picture of the angle $α - β$ .

An Image

The red line and coordinates are there for a reason!

Now, the angle $α - β$ is not in standard position…so let’s rotate those two points on the unit circle until the angle is in standard position.

An Image

Note that the red line must have the same length in this picture as it did in the first—and that’s the key. Distance formula, anyone? Distance from the top picture first:

$\sqrt{{(\cos (α) - \cos (β))}^{2} + {(\sin (α) - \sin (β))}^{2}}$

$\sqrt{\cos^{2} (α) - 2 \cos (α) \cos (β) + \cos^{2} (β) + \sin^{2} (α) - 2 \sin (α) \sin (β) + \sin^{2} (β)}$

$\sqrt{2 - 2 \cos (α) \cos (β) - 2 \sin (α) \sin (β)}$

Now let’s work on the distance in the second picture:

$\sqrt{{(\cos (α - β) - 1)}^{2} + {(\sin (α - β) - 0)}^{2}}$

$\sqrt{\cos^{2} (α - β) - 2 \cos (α - β) + 1 + \sin^{2} (α - β)}$

$\sqrt{2 - 2 \cos (α - β)}$

Let’s finally set those equal to each other:

$\sqrt{2 - 2 \cos (α) \cos (β) - 2 \sin (α) \sin (β)} = \sqrt{2 - 2 \cos (α - β)}$

$2 - 2 \cos (α) \cos (β) - 2 \sin (α) \sin (β) = 2 - 2 \cos (α - β)$

$- 2 \cos (α) \cos (β) - 2 \sin (α) \sin (β) = - 2 \cos (α - β)$

$\cos (α) \cos (β) + \sin (α) \sin (β) = \cos (α - β)$

…and there it is.

Replace $β$ with $- β$ and you’ll get the formula for the sum of two angles.

In summary:

$\cos (α \pm β) = \cos (α) \cos (β) \mp \sin (α) \sin (β)$

Sine

Here are two important identities that will establish the formulas for sine:

$\sin (x) = \cos (\frac{π}{2} - x)$ ; $\cos (x) = \sin (\frac{π}{2} - x)$

So—if we want to find the formula for the sine of a sum, we’ll do this:

$\sin (α + β) = \cos (\frac{π}{2} - (α + β)) = \cos ((\frac{π}{2} - α) - β)$

Now apply the cosine formula that we just developed!

$\cos ((\frac{π}{2} - α) - β) = \cos (\frac{π}{2} - α) \cos (β) + \sin (\frac{π}{2} - α) \sin (β)$

Finally, use those two identities again:

$\cos (\frac{π}{2} - α) \cos (β) + \sin (\frac{π}{2} - α) \sin (β) = \sin (α) \cos (β) + \cos (α) \sin (β)$

Replace $β$ with $- β$ and you’ll get the formula for the difference of two angles.

In summary:

$\sin (α \pm β) = \sin (α) \cos (β) \pm \cos (α) \sin (β)$

Tangent

Tangent is defined as sine over cosine…use our already developed formulas to create one for tangents!

$\tan (α + β) = \frac{\sin (α + β)}{\cos (α + β)} = \frac{\sin (α) \cos (β) + \cos (α) \sin (β)}{\cos (α) \cos (β) - \sin (α) \sin (β)}$

Looks nasty…because there’s one more step: divide both the numerator and denominator by $\cos (α) \cos (β)$ .

$\frac{\frac{\sin (α) \cos (β) + \cos (α) \sin (β)}{\cos (α) \cos (β)}}{\frac{\cos (α) \cos (β) - \sin (α) \sin (β)}{\cos (α) \cos (β)}} = \frac{\frac{\sin (α)}{\cos (α)} + \frac{\sin (β)}{\cos (β)}}{1 - \frac{\sin (α)}{\cos (α)} \cdot \frac{\sin (β)}{\cos (β)}} = \frac{\tan (α) + \tan (β)}{1 - \tan (α) \tan (β)}$

Replace $β$ with $- β$ and you’ll get the formula for the difference of two angles.

In summary:

$\tan (α \pm β) = \frac{\tan (α) \pm \tan (β)}{1 \mp \tan (α) \tan (β)}$

Examples

[8.] Evaluate $\cos (\frac{π}{12})$ .

Note that $\frac{π}{12} = \frac{4 π}{12} - \frac{3 π}{12} = \frac{π}{3} - \frac{π}{4}$ . Thus, $\cos (\frac{π}{12}) = \cos (\frac{π}{3} - \frac{π}{4}) = \cos (\frac{π}{3}) \cos (\frac{π}{4}) + \sin (\frac{π}{3}) \sin (\frac{π}{4}) =$ $\frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{3}}{2 \sqrt{2}}$ .

[9.] Evaluate $\sin (\cos^{- 1} (\frac{1}{3}) + \cos^{- 1} (\frac{2}{5}))$ .

This requires a bit from this chapter and a bit from two chapters ago! $\sin (\cos^{- 1} (\frac{1}{3}) + \cos^{- 1} (\frac{2}{5})) = \sin (\cos^{- 1} (\frac{1}{3})) \cos (\cos^{- 1} (\frac{2}{5})) + \cos (\cos^{- 1} (\frac{1}{3})) \sin (\cos^{- 1} (\frac{2}{5}))$

Draw some triangles to find $\sin (\cos^{- 1} (\frac{1}{3}))$ and $\sin (\cos^{- 1} (\frac{2}{5}))$ .

The result is $\frac{\sqrt{8}}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{\sqrt{21}}{5} = \frac{2 \sqrt{8} + \sqrt{21}}{15}$ .

[10.] Show that $\tan (x - \frac{π}{4}) = \frac{\tan (x) - 1}{\tan (x) + 1}$ .

Naturally, I’ll work the left side: $\tan (x - \frac{π}{4}) = \frac{\tan (x) - \tan (\frac{π}{4})}{1 + \tan (x) \tan (\frac{π}{4})} = \frac{\tan (x) - 1}{1 + \tan (x)} = \frac{\tan (x) - 1}{\tan (x) + 1}$ . QED.

§6.5: Double and Half Angle Formulas

The Double Angle Formulas

Simply replace $β$ with $α$ in each formula!

$\sin (2 x) = 2 \sin (x) \cos (x)$

$\cos (2 x) = \cos^{2} (x) - \sin^{2} (x)$

$\tan (2 x) = \frac{2 \tan (x)}{1 - \tan^{2} (x)}$

The Pythagorean Identity allows you to make some changes to the cosine formula, resulting in either $\cos (2 x) = 2 \cos^{2} (x) - 1$ or $\cos (2 x) = 1 - 2 \sin^{2} (x)$ .

The Power Reducing Formulas

Take those last two formulas and solve them for the squared term. These identities are really useful in Calculus!

$\cos^{2} (x) = \frac{1 + \cos (2 x)}{2}$

$\sin^{2} (x) = \frac{1 - \cos (2 x)}{2}$

If you put those two together, you get a power reducing identity for tangent:

$\tan^{2} (x) = \frac{1 - \cos (2 x)}{1 + \cos (2 x)}$

The Half Angle Formulas

Take those identities, square root them and then replace $x$ with $\frac{x}{2}$ to obtain the half angle identities!

$\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$

$\sin (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{2}}$

$\tan (\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos (x)}{1 + \cos (x)}}$

Rearranging things a bit can make that last tangent identity look like this:

$\tan (\frac{x}{2}) = \frac{1 - \cos (x)}{\sin (x)} = \frac{\sin (x)}{1 + \cos (x)}$

Examples

[11.] Find $\sin (2 x)$ if $\sin (x) = \frac{3}{4}$ and $\frac{π}{2} < x < π$ .

Draw a triangle to see that this makes $\cos (x) = - \frac{\sqrt{7}}{4}$ . Thus, $\sin (2 x) = 2 \sin (x) \cos (x) = 2 (\frac{3}{4}) (\frac{\sqrt{7}}{4}) = \frac{3 \sqrt{7}}{8}$ .

[12.] Show that $\frac{1 - \sin (2 θ)}{\cos (2 θ)} = \frac{\cot (θ) - 1}{\cot (θ) + 1}$ .

I’m going to work the right side of this…and I’ll start by multiplying the top and bottom by $\sin (θ)$ .

$\frac{\cot (θ) - 1}{\cot (θ) + 1} \cdot \frac{\sin (θ)}{\sin (θ)} = \frac{\cos (θ) - \sin (θ)}{\cos (θ) + \sin (θ)} \cdot \frac{\cos (θ) - \sin (θ)}{\cos (θ) - \sin (θ)} = \frac{\cos^{2} (θ) - 2 \sin (θ) \cos (θ) + \sin^{2} (θ)}{\cos^{2} (θ) - \sin^{2} (θ)} = \frac{1 - 2 \sin (θ) \cos (θ)}{\cos^{2} (θ) - \sin^{2} (θ)} = \frac{1 - \sin (2 θ)}{\cos (2 θ)}$ . QED.

[13.] Evaluate $\tan (\frac{1}{2} \cos^{- 1} (\frac{2}{3}))$ .

$\tan (\frac{1}{2} \cos^{- 1} (\frac{2}{3})) = \frac{1 - \cos (\cos^{- 1} (\frac{2}{3}))}{1 + \cos (\cos^{- 1} (\frac{2}{3}))} = \frac{1 - \frac{2}{3}}{1 + \frac{2}{3}} = \frac{\frac{1}{3}}{\frac{5}{3}} = \frac{1}{5}$

§6.6: Product/Sum Formulas

Turning Products into Sums

You know that $\cos (α - β) = \cos (α) \cos (β) + \sin (α) \sin (β)$ and that $\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β)$ . Subtract these two and you get $\cos (α - β) - \cos (α + β) = 2 \sin (α) \sin (β)$ …which can be rearranged to $\sin (α) \sin (β) = \frac{1}{2} [\cos (α - β) - \cos (α + β)]$ . A little algebra can be employed to make two other version of this!

$\cos (α) \cos (β) = \frac{1}{2} [\cos (α - β) + \cos (α + β)]$

$\sin (α) \cos (β) = \frac{1}{2} [\sin (α + β) + \sin (α - β)]$

Turning Sums into Products

Rearranging those formulas, and replacing the angles with $\frac{α + β}{2}$ and $\frac{α - β}{2}$ results in formulas that turn addition into multiplication.

$\sin (α) \pm \sin (β) = 2 \sin (\frac{α \pm β}{2}) \cos (\frac{α \mp β}{2})$

$\cos (α) + \cos (β) = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2})$

$\cos (α) - \cos (β) = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})$

Examples

[14.] Express as a sum of sines: $\sin (x) \cos (3 x)$ .

$\sin (x) \cos (3 x) = \frac{1}{2} [\sin (x + 3 x) + \sin (x - 3 x)] = \frac{1}{2} [\sin (4 x) + \sin (- 2 x)] = \frac{1}{2} [\sin (4 x) - \sin (2 x)]$

[15.]Express as a sum of cosines: $\sin (x) \sin (3 x)$ .

$\sin (x) \sin (3 x) = \frac{1}{2} [\cos (x + 3 x) - \cos (x - 3 x)] = \frac{1}{2} [\cos (4 x) - \cos (- 2 x)] = \frac{1}{2} [\cos (4 x) - \cos (2 x)]$

[16.] Express as a product of cosines: $\cos (- x) + \cos (5 x)$ .

$\cos (- x) + \cos (5 x) = 2 \cos (\frac{- x + 5 x}{2}) \cos (\frac{- x - 5 x}{2}) = 2 \cos (\frac{- 4 x}{2}) \cos (\frac{- 6 x}{2}) 2 \cos (- 2 x) \cos (- 3 x) = 2 \cos (2 x) \cos (3 x)$

§6.7: Solving Trigonometric Equations

Finding the General Solution

For many equation involving trigonometric functions, there are an infinite number of solutions…in these cases, we usually find all solutions over on period of the function, and then add a little notation at the end that says “add or subtract any number of periods you want.”

If the problem says to give the general solution, or if the problem says to find all solutions, then you will use this method (illustrated below).

Solving on an Interval

One of the most common instructions is to solve over some given interval. This is typically much easier! The exception is if the angle used in the trig function is a multiple ( $2 x$ or $3 x$ , for example).

Examples

[17.] Give the general solution to $\tan (x) = \sqrt{3}$ .

First, the period of tangent is $π$ …so I’ll find all solutions in the interval $[0., π]$ : $\tan (x) = \sqrt{3} \Rightarrow x = \frac{π}{3}$ . Now, to that I need to add any (integer) number of periods…here’s how that looks: $x = \frac{π}{3} + k π$ , $k \in ℤ$ .

[18.] Solve $\sin (x) = \cos (x)$ for $x \in [0, 2 π)$ .

Since an interval is given, I just need to find all of the solutions on the given interval. $\sin (x) = \cos (x) \Rightarrow \tan (x) = 1 \Rightarrow x \in {\frac{π}{4}, \frac{5 π}{4}}$ .

[19.] Find all solutions to $\sin (2 x) = \sin (x)$ .

The period of $\sin (2 x)$ is $π$ , and the period of $\sin (x)$ is $2 π$ …I’ll use the larger one for my initial solutions. $\sin (2 x) = \sin (x) \Rightarrow 2 \sin (x) \cos (x) = \sin (x) \Rightarrow 2 \sin (x) \cos (x) - \sin (x) = 0 \Rightarrow \sin (x) (2 \cos (x) - 1) = 0$ . $\sin (x) = 0 \Rightarrow x \in {0, π}$ ; $2 \cos (x) - 1 = 0 \Rightarrow \cos (x) = \frac{1}{2} \Rightarrow x \in {\frac{π}{3}, \frac{5 π}{3}}$ . Thus, my final solution will be $x \in {0, \frac{π}{3}, π, \frac{5 π}{3}} + 2 k π$ , $k \in ℤ$ .

[20.] Solve $\tan^{2} (2 x) = 3$ for $x \in [0, π)$ .

This is where you must be careful…since $x \in [0, π)$ , $2 x \in [0, 2 π)$ ! $\tan^{2} (2 x) = 3 \Rightarrow \tan (2 x) = \pm \sqrt{3} \Rightarrow 2 x \in {\frac{π}{6}, \frac{5 π}{6}, \frac{7 π}{6}, \frac{11 π}{6}}$ . That makes $x \in {\frac{π}{12}, \frac{5 π}{12}, \frac{7 π}{12}, \frac{11 π}{12}}$ .

§6.8: Solving More Trigonometric Equations

Quadratic Trigonometric Equations

Do you remember when we saw that $e^{2 x} - e^{x} + 1$ was really a quadratic expression? Well, that’s about to happen again…this time with trig functions.

Other Methods

…in other words, this section will require solving methods other that just quadratic…

Examples

[21.] Solve $2 \cos^{2} (x) + \cos (x) = 1$ for $x \in [0, 2 π)$ .

$2 \cos^{2} (x) + \cos (x) = 1 \Rightarrow 2 \cos^{2} (x) + \cos (x) - 1 = 0$ . Let $u = \cos (x)$ , making the equation $2 u^{2} + u - 1 = 0 \Rightarrow (2 u - 1) (u + 1) = 0$ , so either $2 u - 1 = 0 \Rightarrow u = \frac{1}{2}$ or $u + 1 = 0 \Rightarrow u = - 1$ . Since $u = \cos (x)$ , that gives us $\cos (x) = \frac{1}{2} \Rightarrow x \in {\frac{π}{3}, \frac{5 π}{3}}$ or $\cos (x) = - 1 \Rightarrow x = π$ .

The solutions are $x \in {\frac{π}{3}, π, \frac{5 π}{3}}$ .

[22.] Solve $\cos^{2} (x) = \sin^{2} (x)$ for $x \in [0, 2 π)$ .

$\cos^{2} (x) = \sin^{2} (x) \Rightarrow \cos^{2} (x) - \sin^{2} (x) = 0 \Rightarrow \cos (2 x) = 0$ . Since $x \in [0, 2 π)$ , $2 x \in [0, 4 π)$ ! $\cos (2 x) = 0 \Rightarrow 2 x \in {\frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}}$ …which means that $x \in {\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}}$ .

(I can think of one other way to do this…)

Page last validated 2011-02-07