Checkpoint 8.2.
Show that [latex]\sin(60° + 30°)[/latex] is not equal to [latex]\sin 60° + \sin 30°{.}[/latex]
Solution
[latex]\sin(60° + 30°)=1{,}[/latex] but [latex]\sin 60° + \sin 30° = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}[/latex]
Chapter 8: More Functions and Identities
Review the following skills you will need for this section.
Learning Objectives
In Chapter 5 we studied identities that relate the three trigonometric functions sine, cosine, and tangent.
Pythagorean identity: [latex]{\sin^2 \theta + \cos^2 \theta = 1}[/latex]
Tangent identity: [latex] {\tan \theta = \dfrac{\sin \theta}{\cos \theta}}[/latex]
All of the identities that relate the trig ratios of different angles are derived from the sum and difference formulas. Let’s see why we need these formulas.
Is it true that [latex]\cos (\alpha + \beta)~~~~{and}~~~~\cos \alpha + \cos \beta[/latex] are equal for any values of [latex]\alpha[/latex] and [latex]\beta[/latex] ? We can test this hypothesis by evaluating both expressions for some specific values of [latex]\alpha[/latex] and [latex]\beta{,}[/latex] say [latex]\alpha=45°[/latex] and [latex]\beta=30°{,}[/latex] as shown below.
From the figure, you should be able to see that [latex]\cos 75°[/latex] is in fact smaller than either [latex]\cos 45°[/latex] or [latex]\cos 30°{,}[/latex] so it cannot be true that [latex]\cos 75°[/latex] is equal to [latex]\cos 45° + \cos 30°{.}[/latex]
Verify that [latex]\cos (45° + 30°)[/latex] is not equal to [latex]\cos 45° + \cos 30°{.}[/latex]
Use your calculator to evaluate each expression.
[latex]\cos (45° + 30°) = \cos 75° = 0.2588\\ {but} \cos 45° + \cos 30° = 0.7071 + 0.8660 = 1.5731[/latex]
The two expressions are not equal.
Show that [latex]\sin(60° + 30°)[/latex] is not equal to [latex]\sin 60° + \sin 30°{.}[/latex]
[latex]\sin(60° + 30°)=1{,}[/latex] but [latex]\sin 60° + \sin 30° = \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}[/latex]
Because the values of the expressions in the previous example and exercise are different, it is not true that [latex]\cos (\alpha + \beta)[/latex] is equal to [latex]\cos \alpha + \cos \beta[/latex] for all angles [latex]\alpha[/latex] and [latex]\beta{,}[/latex] or that [latex]\sin (\alpha + \beta)[/latex] is equal to [latex]\sin \alpha + \sin \beta{.}[/latex]
It turns out that there is a relationship between the trig ratios for [latex]\alpha + \beta[/latex] and the trig ratios of [latex]\alpha[/latex] and [latex]\beta{,}[/latex] but it is a little more complicated.
[latex]{\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta}[/latex]
[latex]{\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta}[/latex]
Notice that to find the sine or cosine of [latex]\alpha + \beta[/latex], we must know (or be able to find) both trig ratios for both [latex]\alpha[/latex] and [latex]\beta{.}[/latex]
The sum and difference formulas can be used to find exact values for trig ratios of various angles.
Find an exact value for [latex]\cos 105°{.}[/latex]
We can write [latex]105°[/latex] as the sum of two special angles: [latex]105° = 60° + 45°{.}[/latex] Now apply the sum of angles identity for cosine.
[latex]\cos (60° + 45°) = \cos 60° \cos 45° - \sin 60° \sin 45°\\ = \dfrac{1}{2} \dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2} \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}-\sqrt{6}}{4}[/latex]
Thus, [latex]\cos 105° = \dfrac{\sqrt{2}-\sqrt{6}}{4}{.}[/latex] You can check that your calculator gives the same decimal approximation of about [latex]-0.2588[/latex] for both [latex]\cos 105°[/latex] and [latex]\dfrac{\sqrt{2}-\sqrt{6}}{4}{.}[/latex]
Find an exact value for [latex]\sin 75°{.}[/latex]
[latex]\dfrac{\sqrt{6}+\sqrt{2}}{4}[/latex]
Suppose that [latex]\sin \theta = 0.6[/latex] and [latex]\cos \theta = -0.8{.}[/latex] Find an exact value for [latex]\sin\left(\theta + \dfrac{2\pi}{3}\right){.}[/latex]
Recall that [latex]\sin \dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}[/latex] and [latex]\cos \dfrac{2\pi}{3} = \dfrac{-1}{2}{.}[/latex] Substituting these values into the sum formula for sine, we find
[latex]\sin\left(\theta + \dfrac{2\pi}{3}\right) = \sin \theta \cos \dfrac{2\pi}{3} +\cos \theta \sin \dfrac{2\pi}{3}\\ = 0.6 \left(\dfrac{-1}{2}\right) +(-0.8)\left(\dfrac{\sqrt{3}}{2}\right)\\ = \dfrac{3}{5} \cdot \dfrac{-1}{2} + \dfrac{-4}{5} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{-3-4\sqrt{3}}{10}[/latex]
Suppose that [latex]\sin \theta = \dfrac{-5}{13}[/latex] and [latex]\cos \theta = \dfrac{12}{13}{.}[/latex] Find an exact value for [latex]\cos\left(\dfrac{\pi}{4} + \theta\right){.}[/latex]
[latex]\dfrac{17}{13\sqrt{2}} = \dfrac{17\sqrt{2}}{26}[/latex]
The difference formulas for sine and cosine can be derived easily from the sum formulas using the identities for negative angles. Note that the difference formulas are identical to the corresponding sum formulas, except for the signs.
[latex]{\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta}[/latex]
[latex]{\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta}[/latex]
Use the fact that [latex]\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}[/latex] to evaluate [latex]\cos \dfrac{\pi}{12}[/latex] exactly.
Remember that [latex]\cos \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2},~~\cos\dfrac{\pi}{6} =\dfrac{\sqrt{3}}{2},~~\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}{,}[/latex] and [latex]\sin \dfrac{\pi}{6} = \dfrac{1}{2}{.}[/latex] Substituting all these values into the difference formula for cosine, we obtain the following.
[latex]\cos\dfrac{\pi}{12} = \cos \left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)\\ = \cos \dfrac{\pi}{4} \cos \dfrac{\pi}{6} + \sin \dfrac{\pi}{4} \sin \dfrac{\pi}{6}\\ = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} +\dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}= \dfrac{\sqrt{6}+\sqrt{2}}{4}[/latex]
You can check that your calculator gives the same decimal approximation of about 0.9659 for both [latex]\cos \dfrac{\pi}{12}[/latex] and [latex]\dfrac{\sqrt{6}+\sqrt{2}}{4}{.}[/latex]
Evaluate [latex]\sin \dfrac{\pi}{12}[/latex] exactly.
[latex]\dfrac{\sqrt{3}-1}{2\sqrt{2}} = \dfrac{\sqrt{6}-\sqrt{2}}{4}[/latex]
There are also sum and difference formulas for the tangent.
[latex]{\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}}[/latex]
[latex]{\tan (\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}}[/latex]
Find an exact value for [latex]\tan 75°{.}[/latex]
We observe that [latex]75° = 45° + 30°{,}[/latex] so [latex]\tan 75° = \tan (45° + 30°){.}[/latex] We can apply the sum formula for tangent.
[latex]\tan(45° + 30°) = \dfrac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°}\\ = \dfrac{1+\dfrac{1}{\sqrt{3}}}{1-1\left(\dfrac{1}{\sqrt{3}}\right)} \cdot {\dfrac{\sqrt{3}}{\sqrt{3}}} =\dfrac{\sqrt{3}+1}{\sqrt{3}-1}[/latex]
Evaluate [latex]\tan \dfrac{\pi}{12}[/latex] exactly.
[latex]\dfrac{\sqrt{3}-1}{\sqrt{3}+1}[/latex]
There are a number of other very useful identities that can be derived from the sum and difference formulas. In particular, if we set [latex]\alpha = \beta = \theta[/latex] in the sum of angles identities (also called addition formulas), we obtain the double angle formulas. These identities are used frequently and should be memorized!
[latex]{\cos 2\theta = \cos^2 \theta - \sin^2 \theta}[/latex]
[latex]{\sin 2\theta = 2\sin \theta \cos \theta}[/latex]
[latex]{\tan 2\theta = \dfrac {2\tan \theta}{1-\tan^2 \theta}}[/latex]
You can also justify the identities to yourself by graphing both sides of the formula to see that the graphs are identical.
The first thing we can learn from the double angle formulas is that [latex]\sin 2\theta[/latex] is not equal to [latex]2\sin \theta{!}[/latex] You can check this very easily by choosing a value for [latex]\theta{,}[/latex] say [latex]45°{.}[/latex] Then
[latex]\sin 2\theta = \sin (90°) =1\\ {but}~~~ 2\sin \theta = 2\sin (45°) = 2\left(\dfrac{\sqrt{2}}{2}\right) = \sqrt{2}[/latex]
and [latex]\sqrt{2} \not= 1{.}[/latex] Similarly, [latex]\cos 2\theta \not= 2\cos \theta[/latex] and [latex]\tan 2\theta \not= 2\tan \theta[/latex]
Find [latex]\sin 2\theta[/latex] for the angle [latex]\theta[/latex] shown.
We start by using the Pythagorean theorem to find the hypotenuse of the triangle.
[latex]c^2 = 2^2 + 3^2 = 13[/latex]
so [latex]c = \sqrt{13}{.}[/latex] Thus, [latex]\cos \theta = \dfrac{3}{\sqrt{13}}[/latex] and [latex]\sin \theta = \dfrac{2}{\sqrt{13}}{.}[/latex] Now we can use these values in the double angle identity to find [latex]\sin 2\theta{.}[/latex]
[latex]\sin 2\theta = 2\sin \theta \cos \theta\\ = 2\left(\dfrac{2}{\sqrt{13}}\right)\left(\dfrac{3}{\sqrt{13}}\right) = \dfrac{12}{13}[/latex]
Find [latex]\cos 2\theta[/latex] and [latex]\tan 2\theta[/latex] for the angle [latex]\theta[/latex] shown in the previous example.
[latex]\cos 2\theta = \dfrac{5}{13},~~\tan 2\theta = \dfrac{12}{5}[/latex]
We can work with algebraic expressions instead of numerical values for the trig ratios.
Use the figure to express [latex]\cos 2\phi[/latex] in terms of [latex]a{.}[/latex]
We use the Pythagorean theorem to find an expression for the third side of the triangle.
[latex]a^2 + b^2 = 3^2 {{Solve for}~~b.}\\ b = \sqrt{9 - a^2}[/latex]
Now we can write expressions for the sine and cosine of [latex]\phi{.}[/latex]
[latex]\cos \phi = \dfrac{a}{3} ~~~~{and}~~~~\sin \phi = \dfrac{\sqrt{9 - a^2}}{3}[/latex]
Finally, we substitute these expressions into the double angle identity.
[latex]\cos 2\phi = \cos^2 \phi - \sin^2 \phi\\ = \left(\dfrac{a}{3}\right)^2 - \left(\dfrac{\sqrt{9 - a^2}}{3}\right)^2\\ = \dfrac{a^2}{9} - \dfrac{9 - a^2}{9} = \dfrac{a^2-9}{9}[/latex]
For the triangle in the previous example, find expressions for [latex]\sin 2\phi[/latex] and [latex]\tan 2\phi{.}[/latex]
[latex]\sin 2\phi = \dfrac{2a\sqrt{9-a^2}}{9},~~\tan 2\phi = \dfrac{2a\sqrt{9-a^2}}{2a^2 - 9}[/latex]
[latex]{\cos 2\theta} = {\cos^2 \theta - \sin^2 \theta}\\ {= 2\cos^2 \theta -1}\\ {= 1-2\sin^2 \theta}[/latex]
Find an expression for [latex]\cos 2\beta[/latex] if you know that [latex]\cos \beta = \dfrac{x}{4}{.}[/latex]
We use the identity [latex]\cos 2\theta = 2\cos^2 \theta -1[/latex]
[latex]\cos 2\beta = 2\cos^2 \beta -1\\ = 2\left(\dfrac{x}{4}\right)^2 - 1 = \dfrac{2x^2}{16} - 1\\ = \dfrac{2x^2 - 16}{16} = \dfrac{x^2 - 8}{8}[/latex]
Find an expression for [latex]\cos 2\alpha[/latex] if you know that [latex]\sin \alpha = \dfrac{6}{w}{.}[/latex]
[latex]\dfrac{w^2 - 76}{w^2}[/latex]
If a trigonometric equation involves more than one angle, we use identities to rewrite the equation in terms of a single angle.
Solve [latex]~~\sin 2x - \cos x~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
We first use the double angle formula to write [latex]\sin 2x[/latex] in terms of trig functions of [latex]x[/latex] alone.
[latex]\sin 2x -\cos x = 0\\ 2\sin x \cos x -\cos x = 0[/latex]
Once we have all the trig functions in terms of a single angle, we try to write the equation in terms of a single trig function. In this case, we can factor the left side to separate the trig functions.
[latex]2\sin x \cos x -\cos x = 0[/latex]
[latex]\cos x (2\sin x - 1) = 0[/latex] Set each factor equal to zero
[latex]\cos x = 0 \qquad 2\sin x - 1 = 0[/latex] Solve each equation
[latex]\sin x = \dfrac{1}{2}[/latex]
[latex]x = \dfrac{\pi}{2},~\dfrac{3\pi}{2} \qquad x = \dfrac{\pi}{6},~\dfrac{5\pi}{6}[/latex]
There are four solutions, [latex]x = \dfrac{\pi}{2},~\dfrac{3\pi}{2},~\dfrac{\pi}{6},[/latex] and [latex]\dfrac{5\pi}{6}{.}[/latex]
Solve [latex]~~\cos 2t = \cos t~~[/latex] for [latex]0 \le x \le 2\pi{.}[/latex]
[latex]t = 0,~ \dfrac{2\pi}{3},~\dfrac{4\pi}{3}[/latex]
[latex]\cos (-\theta) = \cos \theta[/latex]
[latex]\sin (-\theta) = -\sin \theta[/latex]
[latex]\tan (-\theta) = -\tan \theta[/latex]
[latex]\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta[/latex]
[latex]\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta[/latex]
[latex]\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}[/latex]
[latex]\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta[/latex]
[latex]\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta[/latex]
[latex]\tan (\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}[/latex]
[latex]\cos 2\theta = \cos^2 \theta - \sin^2 \theta \sin 2\theta = 2\sin \theta \cos \theta\\ = 2\cos^2 \theta -1 \tan 2\theta = \dfrac {2\tan \theta}{1-\tan^2 \theta}\\ = 1-2\sin^2 \theta[/latex]