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# 3.5: Double Angle Identities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the double angle identities to solve other identities.
• Use the double angle identities to solve equations.

## Deriving the Double Angle Identities

One of the formulas for calculating the sum of two angles is:

$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

If $\alpha$ and $\beta$ are both the same angle in the above formula, then

$\sin (\alpha + \alpha) & = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\\sin 2 \alpha & = 2 \sin \alpha \cos \alpha$

This is the double angle formula for the sine function. The same procedure can be used in the sum formula for cosine, start with the sum angle formula:

$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

If $\alpha$ and $\beta$ are both the same angle in the above formula, then

$\cos (\alpha + \alpha )& = \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\\cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha$

This is one of the double angle formulas for the cosine function. Two more formulas can be derived by using the Pythagorean Identity, $\sin^2 \alpha + \cos^2 \alpha = 1$.

$\sin^2 \alpha = 1 - \cos^2 \alpha$ and likewise $\cos^2 \alpha = 1 - \sin^2 \alpha$

$\text{Using}\ \sin^2 \alpha & = 1 - \cos^2 \alpha: && \text{Using}\ \cos^2 \alpha = 1 - \sin^2 \alpha: \\\cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha && \qquad \ \ \cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha \\& = \cos^2 \alpha - (1 - \cos^2 \alpha) && \qquad \qquad \quad \ = (1 - \sin^2 \alpha) - \sin^2 \alpha \\& = \cos^2 \alpha - 1 + \cos^2 \alpha && \qquad \qquad \quad \ = 1 - \sin^2 \alpha - \sin^2 \alpha \\& = 2 \cos^2 \alpha - 1 && \qquad \qquad \quad \ = 1 - 2 \sin^2 \alpha$

Therefore, the double angle formulas for $\cos 2 a$ are:

$\cos 2 \alpha & = \cos^2 \alpha - \sin^2 \alpha \\\cos 2 \alpha & = 2 \cos^2 \alpha -1 \\\cos 2 \alpha & = 1 - 2 \sin^2 \alpha$

Finally, we can calculate the double angle formula for tangent, using the tangent sum formula:

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

If $\alpha$ and $\beta$ are both the same angle in the above formula, then

$\tan (\alpha + \alpha) & = \frac{\tan \alpha + \tan \alpha}{1 - \tan \alpha \tan \alpha} \\\tan 2 \alpha & = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}$

## Applying the Double Angle Identities

Example 1: If $\sin a = \frac{5}{13}$ and $a$ is in Quadrant II, find $\sin 2a$, $\cos 2a$, and $\tan 2a$.

Solution: To use $\sin 2a = 2 \sin a \cos a$, the value of $\cos a$ must be found first.

$& =\cos^2 a + \sin^2 a = 1\\& = \cos^2 a + \left (\frac{5}{13} \right )^2 = 1\\& = \cos^2 a + \frac{25}{169} = 1\\& = \cos^2 a = \frac{144}{169}, \cos a = \pm \frac{12}{13}$.

However since $a$ is in Quadrant II, $\cos a$ is negative or $\cos a = - \frac{12}{13}$.

$\sin 2a = 2 \sin a \cos a = 2 \left (\frac{5}{13} \right ) \times \left (- \frac{12}{13} \right ) = \sin 2a = - \frac{120}{169}$

For $\cos 2a$, use $\cos(2a) = \cos^2 a - \sin^2 a$

$\cos(2a) & = \left (- \frac{12}{13} \right )^2 - \left (\frac{5}{13} \right )^2 \ \text{or}\ \frac{144-25}{169} \\\cos (2a) & = \frac{119}{169}$

For $\tan 2a$, use $\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}$. From above, $\tan a = \frac{\frac{5}{13}}{-\frac{12}{13}} = - \frac{5}{12}$.

$\tan(2a) = \frac{2 \cdot \frac{-5}{12}}{1 - \left (\frac{-5}{12} \right )^2} = \frac{\frac{-5}{6}}{1 - \frac{25}{144}} = \frac{\frac{-5}{6}}{\frac{119}{144}} = - \frac{5}{6} \cdot \frac{144}{119} = - \frac{120}{119}$

Example 2: Find $\cos 4 \theta$.

Solution: Think of $\cos 4 \theta$ as $\cos (2 \theta + 2 \theta)$.

$\cos 4 \theta = \cos (2 \theta + 2 \theta) = \cos 2 \theta \cos 2 \theta - \sin 2 \theta \sin 2 \theta = \cos^2 2 \theta - \sin^2 2 \theta$

Now, use the double angle formulas for both sine and cosine. For cosine, you can pick which formula you would like to use. In general, because we are proving a cosine identity, stay with cosine.

$& = (2 \cos^2 \theta - 1)^2 - (2 \sin \theta \cos \theta)^2 \\& = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 \sin^2 \theta \cos^2 \theta \\& = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 (1 - \cos^2 \theta) \cos^2 \theta \\& = 4 \cos^4 \theta - 4 \cos^2 \theta + 1 - 4 \cos^2 \theta + 4 \cos^4 \theta \\& = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$

Example 3: If $\cot x = \frac{4}{3}$ and $x$ is an acute angle, find the exact value of $\tan 2x$ .

Solution: Cotangent and tangent are reciprocal functions, $\tan x = \frac{1}{\cot x}$ and $\tan x = \frac{3}{4}$.

$\tan 2x & = \frac{2 \tan x}{1 - \tan^2 x} \\& = \frac{2\cdot \frac{3}{4}}{1 - \left (\frac{3}{4} \right )^2} \\& = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} \\& = \frac{3}{2} \cdot \frac{16}{7} = \frac{24}{7}$

Example 4: Given $\sin (2x) = \frac{2}{3}$ and $x$ is in Quadrant I, find the value of $\sin x$.

Solution: Using the double angle formula, $\sin 2x = 2 \sin x \cos x$. Because we do not know $\cos x$, we need to solve for $\cos x$ in the Pythagorean Identity, $\cos x = \sqrt{1 - \sin^2 x}$. Substitute this into our formula and solve for $\sin x$.

$\sin 2x & = 2 \sin x \cos x \\\frac{2}{3} & = 2 \sin x \sqrt{1 - \sin^2 x} \\\left (\frac{2}{3} \right)^2 & = \left (2 \sin x \sqrt{1 - \sin^2 x} \right )^2 \\\frac{4}{9} & = 4 \sin ^2 x (1 - \sin^2 x) \\\frac{4}{9} & = 4 \sin^2 x - 4 \sin^4 x$

At this point we need to get rid of the fraction, so multiply both sides by the reciprocal.

$& \frac{9}{4} \left (\frac{4}{9} = 4 \sin^2 x - 4 \sin^4 x \right ) \\& \qquad 1 = 9 \sin^2 x - 9 \sin^4 x \\& \qquad 0 = 9 \sin^4 x - 9 \sin^2 x + 1$

Now, this is in the form of a quadratic equation, even though it is a quartic. Set $a = \sin^2 x$, making the equation $9 a^2 - 9a + 1 = 0$. Once we have solved for $a$, then we can substitute $\sin^2 x$ back in and solve for $x$. In the Quadratic Formula, $a = 9, b = -9, c = 1$.

$\frac{9 \pm \sqrt{(-9)^2 - 4(9)(1)}}{2(9)} = \frac{9 \pm \sqrt{81 - 36}}{18} = \frac{9 \pm \sqrt{45}}{18} = \frac{9 \pm 3 \sqrt{5}}{18} = \frac{3 \pm \sqrt{5}}{6}$

So, $a = \frac{3 + \sqrt{5}}{6} \approx 0.873$ or $\frac{3 - \sqrt{5}}{6} \approx .1273$. This means that $\sin^2 x \approx 0.873$ or $.1273$ so $\sin x \approx 0.934$ or $\sin x \approx .357$.

Example 5: Prove $\tan \theta = \frac{1 - \cos 2 \theta}{\sin 2 \theta}$

Solution: Substitute in the double angle formulas. Use $\cos 2 \theta = 1 - 2 \sin^2 \theta$, since it will produce only one term in the numerator.

$\tan \theta & = \frac{1 - (1 - 2 \sin^2 \theta)}{2 \sin \theta \cos \theta} \\& = \frac{2 \sin^2 \theta}{2 \sin \theta \cos \theta} \\& = \frac{\sin \theta}{\cos \theta} \\& = \tan \theta$

## Solving Equations with Double Angle Identities

Much like the previous sections, these problems all involve similar steps to solve for the variable. Isolate the trigonometric function, using any of the identities and formulas you have accumulated thus far.

Example 6: Find all solutions to the equation $\sin 2x = \cos x$ in the interval $[0, 2 \pi]$

Solution: Apply the double angle formula $\sin 2x = 2 \sin x \cos x$

$2 \sin x \cos x & = \cos x \\2 \sin x \cos x - \cos x & = \cos x - \cos x \\2 \sin x \cos x - \cos x & = 0 \\\cos x (2 \sin x - 1) & = 0 \ \text{Factor out}\ \cos x \\\text{Then}\ \cos x & = 0\ \text{or}\ 2 \sin x - 1 = 0 \\\cos x & = 0\ \text{or}\ 2 \sin x - 1 + 1 = 0 + 1 \\\frac{2}{2} \sin x & = \frac{1}{2} \\\sin x & = \frac{1}{2}$

The values for $\cos x = 0$ in the interval $[0, 2\pi]$ are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ and the values for $\sin x = \frac{1}{2}$ in the interval $[0, 2\pi]$ are $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$. Thus, there are four solutions.

Example 7: Solve the trigonometric equation $\sin 2x = \sin x$ such that $(- \pi \le x < \pi)$

Solution: Using the sine double angle formula:

$& \qquad \qquad \quad \ \ \sin 2x = \sin x \\& \qquad \quad 2 \sin x \cos x \ = \sin x \\& \ 2 \sin x \cos x - \sin x = 0 \\& \ \ \ \sin x (2 \cos x -1) = 0 \\& \quad \Bigg\downarrow \qquad \ \ \searrow \\& \qquad \qquad \qquad 2 \cos x - 1 = 0 \\& \qquad \qquad \qquad \quad \ \ 2 \cos x = 1 \\& \sin x = 0 \\& \quad \ \ x = 0, - \pi \qquad \ \cos x = \frac{1}{2} \\& \qquad \qquad \qquad \qquad \qquad x = \frac{\pi}{3}, - \frac{\pi}{3}$

Example 8: Find the exact value of $\cos 2x$ given $\cos x = - \frac{13}{14}$ if $x$ is in the second quadrant.

Solution: Use the double-angle formula with cosine only.

$\cos 2x & = 2 \cos^2 x - 1 \\\cos 2x &= 2 \left (- \frac{13}{14} \right )^2 - 1 \\\cos 2x &= 2 \left (\frac{169}{196} \right )-1 \\\cos 2x & = \left (\frac{338}{196} \right )-1 \\\cos 2x &= \frac{338}{196} - \frac{196}{196} \\\cos 2x &= \frac{142}{196} = \frac{71}{98}$

Example 9: Solve the trigonometric equation $4 \sin \theta \cos \theta = \sqrt{3}$ over the interval $[0, 2\pi)$.

Solution: Pull out a 2 from the left-hand side and this is the formula for $\sin 2x$.

$4 \sin \theta \cos \theta & = \sqrt{3} \\2 (2 \sin \theta \cos \theta) & = \sqrt{3} \\2 (2 \sin \theta \cos \theta) & = 2 \sin 2 \theta \\2 \sin 2 \theta & = \sqrt{3} \\\sin 2 \theta & = \frac{\sqrt{3}}{2}$

The solutions for $2 \theta$ are $\frac{\pi}{3},\frac{2\pi}{3},\frac{7\pi}{3},\frac{8\pi}{3}$, dividing each of these by 2, we get the solutions for $\theta$, which are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{8\pi}{6}$.

## Points to Consider

• Are there similar formulas that can be derived for other angles?
• Can technology be used to either solve these trigonometric equations or to confirm the solutions?

## Review Questions

1. If $\sin x = \frac{4}{5}$ and $x$ is in Quad II, find the exact values of $\cos 2x, \sin 2x$ and $\tan 2x$
2. Find the exact value of $\cos^2 15^\circ - \sin^2 15^\circ$
3. Verify the identity: $\cos 3 \theta = 4 \cos^3\theta - 3 \cos \theta$
4. Verify the identity: $\sin 2t - \tan t = \tan t \cos 2t$
5. If $\sin x = - \frac{9}{41}$ and $x$ is in Quad III, find the exact values of $\cos 2x, \sin 2x$ and $\tan 2x$
6. Find all solutions to $\sin 2x + \sin x = 0$ if $0 \le x < 2 \pi$
7. Find all solutions to $\cos^2 x - \cos 2x = 0$ if $0 \le x < 2 \pi$
8. If $\tan x = \frac{3}{4}$ and $0^\circ < x < 90^\circ$, use the double angle formulas to determine each of the following:
1. $\tan 2x$
2. $\sin 2x$
3. $\cos 2x$
9. Use the double angle formulas to prove that the following equations are identities.
1. $2 \csc 2x = \csc^2 x \tan x$
2. $\cos^4 \theta - \sin^4 \theta = \cos 2 \theta$
3. $\frac{\sin 2x}{1 + \cos 2x} = \tan x$
10. Solve the trigonometric equation $\cos 2x - 1 = \sin^2 x$ such that $[0, 2\pi)$
11. Solve the trigonometric equation $\cos 2x = \cos x$ such that $0 \le x < \pi$
12. Prove $2 \csc 2x \tan x = \sec^2 x$.
13. Solve $\sin 2x - \cos 2x = 1$ for $x$ in the interval $[0, 2\pi)$.
14. Solve the trigonometric equation $\sin^2 x - 2 = \cos 2x$ such that $0 \le x < 2 \pi$

1. If $\sin x = \frac{4}{5}$ and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem, the third side is $3(b = \sqrt{5^2 - 4^2})$. So, $\cos x = - \frac{3}{5}$ and $\tan x = - \frac{4}{3}$. Using this, we can find $\sin 2x, \cos 2x$, and $\tan 2x$. $& && \ \cos 2x = 1 - \sin^2 x && \ \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \\& && \qquad \quad = 1 -2 \cdot \left (\frac{4}{5} \right)^2 && \qquad \quad \ = \frac{2 \cdot - \frac{4}{3}}{1 - \left (- \frac{4}{3} \right )^2} \\& \sin 2x = 2 \sin x \cos x && \qquad \quad = 1 - 2 \cdot \frac{16}{25} && \qquad \quad \ = \frac{-\frac{8}{3}}{1 - \frac{16}{9}} = - \frac{8}{3} \div - \frac{7}{9} \\& \qquad \ \ = 2 \cdot \frac{4}{5} \cdot - \frac{3}{5} && \qquad \quad = 1 - \frac{32}{25} && \qquad \quad \ = - \frac{8}{3} \cdot - \frac{9}{7} \\& \qquad \ \ = - \frac{24}{25} && \qquad \quad =- \frac{7}{25} && \qquad \quad \ = \frac{24}{7}$
2. This is one of the forms for $\cos 2x$. $\cos^2 15^\circ - \sin^2 15^\circ & = \cos (15^\circ \cdot 2) \\& = \cos 30^ \circ \\& = \frac{\sqrt{3}}{2}$
3. Step 1: Use the cosine sum formula $\cos 3 \theta & = 4 \cos^3 \theta - 3 \cos \theta \\\cos (2 \theta + \theta) & = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta$ Step 2: Use double angle formulas for $\cos 2\theta$ and $\sin 2\theta$ $= (2 \cos^2 \theta - 1) \cos \theta - (2 \sin \theta \cos \theta) \sin \theta$ Step 3: Distribute and simplify. $& = 2 \cos^3 \theta - \cos \theta - 2 \sin^2 \theta \cos \theta \\& = - \cos \theta (-2 \cos^2 \theta + 2 \sin^2 \theta + 1) \\& = - \cos \theta [- 2 \cos^2 \theta + 2 (1 - \cos^2 \theta) + 1] && \rightarrow \text{Substitute}\ 1 - \cos^2 \theta \ \text{for}\ \sin^2 \theta \\& = - \cos \theta [- 2 \cos^2 \theta + 2 - 2 \cos^2 \theta + 1] \\& = - \cos \theta (-4 \cos^2 \theta + 3) \\& = 4 \cos^3 \theta - 3 \cos \theta$
4. Step 1: Expand $\sin 2t$ using the double angle formula. $\sin 2t - \tan t & = \tan t \cos 2t \\2 \sin t \cos t - \tan t & = \tan t \cos 2t$ Step 2: change $\tan t$ and find a common denominator. $& 2 \sin t \cos t - \frac{\sin t}{\cos t} \\& \ \frac{2 \sin t \cos^2 t - \sin t}{\cos t} \\& \quad \frac{\sin t (2 \cos^2 t - 1)}{\cos t} \\& \frac{\sin t}{\cos t} \cdot (2 \cos^2 t - 1) \\& \quad \ \tan t \cos 2t$
5. If $\sin x = - \frac{9}{41}$ and in Quadrant III, then $\cos x = - \frac{40}{41}$ and $\tan x = \frac{9}{40}$ (Pythagorean Theorem, $b = \sqrt{41^2 - (-9)^2}$). So, $& && \cos 2x \ = 2 \cos^2 x - 1 && \\& \ \sin 2x = 2 \sin x \cos x &&\qquad \quad = 2 \left (- \frac{40}{41} \right )^2 -1 && \tan 2x = \frac{\sin 2x}{\cos 2x} \\& \qquad \quad = 2 \cdot - \frac{9}{41} \cdot - \frac{40}{41} && \qquad \quad = \frac{3200}{1681} - \frac{1681}{1681} && \qquad \quad = \frac{\frac{720}{1681}}{\frac{1519}{1681}} \\& \qquad \quad = \frac{720}{1681} && \qquad \quad = \frac{1519}{1681} && \qquad \quad = \frac{720}{1519}$
6. Step 1: Expand $\sin 2x$ $\sin 2x + \sin x & = 0 \\2 \sin x \cos x + \sin x & = 0 \\\sin x (2 \cos x + 1) & = 0$ Step 2: Separate and solve each for $x$. $& && 2 \cos x + 1 = 0 \\& \sin x = 0 && \qquad \ \cos x = - \frac{1}{2} \\& \quad \ \ x = 0, \pi \qquad \qquad \text{or} && \qquad \qquad x = \frac{2 \pi}{3}, \frac{4 \pi}{3}$
7. Expand $\cos 2x$ and simplify $\cos^2 x - \cos 2x & = 0 \\\cos^2 x - (2 \cos^2 x - 1) & = 0 \\- \cos^2 x + 1 & = 0 \\\cos^2 x & = 1 \\\cos x & = \pm1$ $\cos x = 1$ when $x = 0$, and $\cos x = -1$ when $x = \pi$. Therefore, the solutions are $x=0, \pi$.
8. a. $3.429$ b. $0.960$ c. $0.280$
9. a.$2 \csc x\ 2x & = \frac{2}{\sin 2x} \\2 \csc x\ 2x & = \frac{2}{2 \sin x \cos x} \\2 \csc x\ 2x & = \frac{1}{\sin x \cos x} \\2 \csc x\ 2x & = \left (\frac{\sin x}{\sin x} \right ) \left (\frac{1}{\sin x \cos x} \right ) \\2 \csc x\ 2x & = \frac{\sin x}{\sin^2 x \cos x} \\2 \csc x\ 2x & = \frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} \\2 \csc x\ 2x & = \csc^2 x \tan x$ b.$\cos^4 \theta - \sin^4 \theta & = (\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta) \\\cos^4 - \sin^4 \theta & = 1 (\cos^2 \theta - \sin^2 \theta) \\\cos 2 \theta & = \cos^2 \theta - \sin^2 \theta \\\therefore \cos^4 \theta - \sin^4 \theta & = \cos 2 \theta$ c.$\frac{\sin 2x}{1 + \cos 2x} & = \frac{2 \sin x \cos x}{1 + (1 - 2 \sin^2 x)} \\\frac{\sin 2x}{1 + \cos 2x} & = \frac{2 \sin x \cos x}{2 - 2 \sin^2 x} \\\frac{\sin 2x}{1 + \cos 2x} & = \frac{2 \sin x \cos x}{2 (1 - \sin^2 x)} \\\frac{\sin 2x}{1 + \cos 2x} & = \frac{2 \sin x \cos x}{2 \cos^2 x} \\\frac{\sin 2x}{1 + \cos 2x} & = \frac{\sin x}{\cos x} \\\frac{\sin 2x}{1 + \cos 2x} & = \tan x$
10. $\cos 2x - 1 = \sin^2 x$ $(1-2 \sin^2 x)-1 &= \sin^2 x \\-2 \sin^2 x & = \sin^2 x \\0 & = 3 \sin^2 x \\0 & = \sin^2 x \\0 & = \sin x \\x & = 0, \pi$
11. $& \qquad \qquad \qquad \quad \ \cos 2x = \cos x \\& \qquad \qquad \quad 2 \cos^2 x - 1 = \cos x \\& \quad \ \ 2 \cos^2 x - \cos x - 1 = 0 \\& (2 \cos x + 1)(\cos x - 1) = 0 \\& \qquad \searrow \qquad \qquad \searrow \\& 2 \cos x + 1 = 0 \ \ \text{or}\ \ \cos x - 1 = 0 \\& \quad \ \ 2 \cos x = -1 \qquad \quad \ \cos x = 1 \\& \qquad \cos x = - \frac{1}{2}$ $\cos x = 1$ when $x = 0$ and $\cos x = - \frac{1}{2}$ when $x = \frac{2\pi}{3}$.
12. $2 \csc 2x \tan x & = \sec^2 x \\\frac{2}{\sin 2x} \cdot \frac{\sin x}{\cos x} & = \frac{1}{\cos^2 x} \\\frac{2}{2 \sin x \cos x} \cdot \frac{\sin x}{\cos x} & = \frac{1}{\cos^2 x} \\\frac{1}{\cos^2 x} & = \frac{1}{\cos^2 x}$
13. $\sin 2x - \cos 2x = 1$ $& 2 \sin x \cos x - (1 - 2 \sin^2 x) = 1 \\& \quad 2 \sin x \cos x - 1 + 2 \sin^2 x = 1 \\& \qquad \quad 2 \sin x \cos x + 2 \sin^2 x = 2 \\& \qquad \qquad \ \sin x \cos x + \sin^2 x = 1 \\& \qquad \qquad \qquad \qquad \sin x \cos x = 1 - \sin^2 x \\& \qquad \qquad \qquad \qquad \sin x \cos x = \cos^2 x \\& \qquad \qquad \left (\pm\sqrt{1 - \cos^2 x} \right ) \cos x = \cos^2 x \\& \qquad \qquad \ \ \left (1 - \cos^2 x \right ) \cos^2 x = \cos^4 x \\& \qquad \qquad \qquad \ \cos^2 x - \cos^4 x = \cos^4 x \\& \qquad \qquad \quad \ \ \cos^2 x - 2 \cos^4 x = 0 \\& \qquad \qquad \ \cos^2 x (1 - 2 \cos^2 x) = 0 \\& \qquad \qquad \quad \swarrow \qquad \qquad \searrow \\& \qquad \qquad \qquad \qquad 1 - 2 \cos^2 x = 0 \\& \cos^2 x = 0 \qquad \qquad -2 \cos^2 x = -1 \\& \ \cos x = 0 \qquad \text{or}\quad \qquad \cos^2 x = \frac{1}{2} \\& \qquad x = \frac{\pi}{2}, \frac{3 \pi}{2} \qquad \qquad \ \cos x = \pm\frac{\sqrt{2}}{2} \\& \qquad \qquad \qquad \qquad \qquad \qquad \ \ x = \frac{\pi}{4}, \frac{5\pi}{4}$ Note: If we go back to the equation $\sin x \cos x = \cos^2 x$, we can see that $\sin x \cos x$ must be positive or zero, since $\cos^2 x$ is always positive or zero. For this reason, $\sin x$ and $\cos x$ must have the same sign (or one of them must be zero), which means that $x$ cannot be in the second or fourth quadrants. This is why $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$ are not valid solutions.
14. Use the double angle identity for $\cos 2x$. $\sin^2 x - 2 & = \cos 2x \\\sin^2 x - 2 & = \cos 2x \\\sin^2 x - 2 & = 1 - 2 \sin^2 x \\3 \sin^2 x & = 3 \\\sin^2 x & = 1 \\\sin x & = \pm 1 \\x & = \frac{\pi}{2}, \frac{3 \pi}{2}$

Feb 23, 2012

Dec 15, 2015