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# 3.4: Sum and Difference Identities

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use and identify the sum and difference identities.
• Apply the sum and difference identities to solve trigonometric equations.
• Find the exact value of a trigonometric function for certain angles.

In this section we are going to explore $\cos(a \pm b), \sin (a \pm b)$, and $\tan(a \pm b)$. These identities have very useful expansions and can help to solve identities and equations.

## Sum and Difference Formulas: Cosine

Is $\cos 15^\circ = \cos (45^\circ - 30^\circ)$? Upon appearance, yes, it is. This section explores how to find an expression that would equal $\cos (45^\circ - 30^\circ)$. To simplify this, let the two given angles be $a$ and $b$ where $0 < b < a < 2\pi$.

Begin with the unit circle and place the angles $a$ and $b$ in standard position as shown in Figure A. Point Pt1 lies on the terminal side of $b$, so its coordinates are $(\cos b, \sin b)$ and Point Pt2 lies on the terminal side of $a$ so its coordinates are $(\cos a, \sin a)$. Place the $a - b$ in standard position, as shown in Figure B. The point A has coordinates $(1, 0)$ and the Pt3 is on the terminal side of the angle $a - b$, so its coordinates are $(\cos[a - b], \sin[a - b])$.

Triangles $OP_1P_2$ in figure A and Triangle $OAP_3$ in figure B are congruent. (Two sides and the included angle, $a - b$, are equal). Therefore the unknown side of each triangle must also be equal. That is: $d\ (A, P_3) = d\ (P_1, P_2)$

Applying the distance formula to the triangles in Figures A and B and setting them equal to each other:

$\sqrt{[\cos (a-b)-1]^2 + [\sin (a-b)-0]^2} = \sqrt{(\cos a - \cos b)^2+(\sin a - \sin b)^2}$

Square both sides to eliminate the square root.

$[\cos (a - b) - 1]^2 + [\sin (a-b)-0]^2 = (\cos a - \cos b)^2 + (\sin a - \sin b)^2$

FOIL all four squared expressions and simplify.

$\cos^2 (a-b)-2 \cos(a-b)+ 1 + \sin^2 (a-b) & = \cos^2 a - 2 \cos a \cos b + \cos^2 b + \sin^2 a - 2 \sin a \sin b + \sin^2 b \\\underbrace{\sin^2 (a-b) + \cos^2 (a-b)} - 2 \cos (a-b) + 1 & = \underbrace{\sin^2 a + \cos^2 a} - 2 \cos a \cos b + \underbrace{\sin^2 b + \cos^2 b} - 2 \sin a \sin b \\1-2 \cos (a-b) + 1 & = 1 - 2 \cos a \cos b + 1 - 2 \sin a \sin b \\2-2 \cos (a-b) & = 2-2 \cos a \cos b -2 \sin a \sin b \\-2 \cos (a-b) & = -2 \cos a \cos b - 2 \sin a \sin b \\\cos (a-b) & = \cos a \cos b + \sin a \sin b$

In $\cos(a - b) = \cos a \cos b + \sin a \sin b$, the difference formula for cosine, you can substitute $a - (- b) = a + b$ to obtain: $\cos(a + b) = \cos[a - (- b)]$ or $\cos a \cos (- b) + \sin a \sin(-b)$. since $\cos(-b) = \cos b$ and $\sin (-b) = -\sin b$, then $\cos(a + b) = \cos a \cos b - \sin a \sin b$, which is the sum formula for cosine.

## Using the Sum and Difference Identities of Cosine

The sum/difference formulas for cosine can be used to establish other identities:

Example 1: Find an equivalent form of $\cos \left (\frac{\pi}{2} - \theta \right )$ using the cosine difference formula.

Solution:

$\cos \left (\frac{\pi}{2} - \theta \right ) & = \cos \frac{\pi}{2} \cos \theta + \sin \frac{\pi}{2} \sin \theta \\\cos \left (\frac{\pi}{2} - \theta \right ) & = 0 \times \cos \theta + 1 \times \sin \theta, \ \text{substitute}\ \cos \frac{\pi}{2} = 0 \ \text{and}\ \sin \frac{\pi}{2} = 1 \\\cos \left (\frac{\pi}{2} - \theta \right ) & = \sin \theta$

We know that is a true identity because of our understanding of the sine and cosine curves, which are a phase shift of $\frac{\pi}{2}$ off from each other.

The cosine formulas can also be used to find exact values of cosine that we weren’t able to find before, such as $15^\circ =(45^\circ - 30^\circ), 75^\circ =(45^\circ + 30^\circ)$, among others.

Example 2: Find the exact value of $\cos 15^\circ$

Solution: Use the difference formula where $a = 45^\circ$ and $b = 30^\circ$.

$\cos (45^\circ - 30^\circ) & = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\\cos 15^\circ & = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} \\\cos 15^\circ & = \frac{\sqrt{6} + \sqrt{2}}{4}$

Example 3: Find the exact value of $\cos 105^\circ$.

Solution: There may be more than one pair of key angles that can add up (or subtract to) $105^\circ$. Both pairs, $45^\circ + 60^\circ$ and $150^\circ-45^\circ$, will yield the correct answer.

1.

$\cos 105^\circ & = \cos(45^\circ + 60^\circ) \\& = \cos 45^\circ \cos 60^\circ - \sin 45^\circ \sin 60^\circ,\ \text{substitute in the known values} \\& = \frac{\sqrt{2}}{2} \times \frac{1}{2} - \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} \\& = \frac{\sqrt{2} - \sqrt{6}}{4}$

2.

$\cos 105^\circ & = \cos(150^\circ - 45^\circ) \\& = \cos 150^\circ \cos 45^\circ + \sin 150^\circ \sin 45^\circ \\& = - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \\& = - \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\& = \frac{\sqrt{2} - \sqrt{6}}{4}$

You do not need to do the problem multiple ways, just the one that seems easiest to you.

Example 4: Find the exact value of $\cos \frac{5 \pi}{12}$, in radians.

Solution: $\cos \frac{5 \pi}{12} = \cos \left (\frac{\pi}{4} + \frac{\pi}{6} \right )$, notice that $\frac{\pi}{4} = \frac{3 \pi}{12}$ and $\frac{\pi}{6} = \frac{2 \pi}{12}$

$\cos \left (\frac{\pi}{4} + \frac{\pi}{6} \right ) & = \cos \frac{\pi}{4} \cos \frac{\pi}{6} - \sin \frac{\pi}{4} \sin \frac{\pi}{6} \\\cos \frac{\pi}{4} \cos \frac{\pi}{6} - \sin \frac{\pi}{4} \sin \frac{\pi}{6} & = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2} \\& = \frac{\sqrt{6} - \sqrt{2}}{4}$

## Sum and Difference Identities: Sine

To find $\sin(a + b)$, use Example 1, from above:

$\sin(a + b) & = \cos \left [\frac{\pi}{2} - (a+b) \right ] && \text{Set}\ \theta = a + b\\& = \cos \left [\left (\frac{\pi}{2} - a \right ) - b \right ]&& \text{Distribute the negative} \\ & = \cos \left (\frac{\pi}{2} - a \right ) \cos b + \sin \left (\frac{\pi}{2} - a \right ) \sin b && \text{Difference Formula for cosines} \\ & = \sin a \cos b + \cos a \sin b && \text{Co-function Identities}$

In conclusion, $\sin(a + b) = \sin a \cos b + \cos a \sin b$, which is the sum formula for sine.

To obtain the identity for $\sin(a - b)$:

$\sin(a - b) & = \sin[a + (-b)] \\ & = \sin a \cos(-b) + \cos a \sin (-b) && \text{Use the sine sum formula} \\ \sin(a - b) & = \sin a \cos b - \cos a \sin b && \text{Use}\ \cos(-b) = \cos b,\ \text{and}\ \sin(-b) = -\sin b$

In conclusion, $\sin(a - b) = \sin a \cos b - \cos a \sin b$, so, this is the difference formula for sine.

Example 5: Find the exact value of $\sin \frac{5 \pi}{12}$

Solution: Recall that there are multiple angles that add or subtract to equal any angle. Choose whichever formula that you feel more comfortable with.

$\sin \frac{5 \pi}{12} & = \sin \left (\frac{3 \pi}{12} + \frac{2 \pi}{12} \right ) \\& = \sin \frac{3 \pi}{12} \cos \frac{2 \pi}{12} + \cos \frac{3 \pi}{12} \sin \frac{2 \pi}{12} \\\sin \frac{5 \pi}{12} & = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} \\& = \frac{\sqrt{6} + \sqrt{2}}{4}$

Example 6: Given $\sin \alpha = \frac{12}{13}$, where $\alpha$ is in Quadrant II, and $\sin \beta = \frac{3}{5}$, where $\beta$ is in Quadrant I, find the exact value of $\sin(\alpha + \beta)$.

Solution: To find the exact value of $\sin(\alpha + \beta)$, here we use $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$. The values of $\sin \alpha$ and $\sin \beta$ are known, however the values of $\cos \alpha$ and $\cos \beta$ need to be found.

Use $\sin^2 \alpha + \cos^2 \alpha = 1$, to find the values of each of the missing cosine values.

For $\cos a : \sin^2 \alpha + \cos^2 \alpha = 1$, substituting $\sin \alpha = \frac{12}{13}$ transforms to $\left (\frac{12}{13} \right )^2 + \cos^2 \alpha = \frac{144}{169} + \cos^2 \alpha = 1$ or $\cos^2 \alpha = \frac{25}{169} \cos \alpha = \pm \frac{5}{13}$, however, since $\alpha$ is in Quadrant II, the cosine is negative, $\cos \alpha = - \frac{5}{13}$.

For $\cos \beta$ use $\sin^2\beta + \cos^2\beta = 1$ and substitute $\sin \beta = \frac{3}{5}, \left (\frac{3}{5} \right )^2 + \cos^2 \beta = \frac{9}{25} + \cos^2 \beta = 1$ or $\cos^2 \beta = \frac{16}{25}$ and $\cos \beta = \pm \frac{4}{5}$ and since $\beta$ is in Quadrant I, $\cos \beta = \frac{4}{5}$

Now the sum formula for the sine of two angles can be found:

$\sin (\alpha + \beta)& = \frac{12}{13} \times \frac{4}{5} + \left (- \frac{5}{13} \right ) \times \frac{3}{5}\ \text{or}\ \frac{48}{65} - \frac{15}{65} \\\sin (\alpha + \beta)& = \frac{33}{65}$

## Sum and Difference Identities: Tangent

To find the sum formula for tangent:

$\tan(a + b) & = \frac{\sin(a+b)}{\cos(a+b)} && \text{Using}\ \tan \theta = \frac{\sin \theta}{\cos \theta} \\& = \frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b - \sin a \sin b} && \text{Substituting the sum formulas for sine and cosine} \\& = \frac{\frac{\sin a \cos b + \sin b \cos a}{\cos a \cos b}}{\frac{\cos a \cos b - \sin a \sin b}{\cos a \cos b}} && \text{Divide both the numerator and the denominator by}\ \cos a \cos b \\& = \frac{\frac{\sin a \cos b} {\cos a \cos b} + \frac{\sin b \cos a}{\cos a \cos b}}{\frac{\cos a \cos b} {\cos a \cos b} - \frac{\sin a \sin b}{\cos a \cos b}} && \text{Reduce each of the fractions} \\ & = \frac{\frac{\sin a}{\cos a} + \frac{\sin b}{\cos b}}{1- \frac{\sin a \sin b}{\cos a \cos b}} && \text{Substitute}\ \frac{\sin \theta}{\cos \theta} = \tan \theta \\\tan(a + b) & = \frac{\tan a + \tan b}{1 - \tan a \tan b} && \text{Sum formula for tangent}$

In conclusion, $\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$. Substituting $-b$ for $b$ in the above results in the difference formula for tangent:

$\tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$

Example 7: Find the exact value of $\tan 285^\circ$.

Solution: Use the difference formula for tangent, with $285^\circ = 330^\circ - 45^\circ$

$\tan (330^\circ - 45^\circ) & = \frac{\tan 330^\circ - \tan 45^\circ}{1 + \tan 330^\circ \tan 45^\circ} \\& = \frac{- \frac{\sqrt{3}}{3} - 1}{1 - \frac{\sqrt{3}}{3} \cdot 1} = \frac{-3 - \sqrt{3}}{3-\sqrt{3}} \\& = \frac{-3- \sqrt{3}}{3- \sqrt{3}} \cdot \frac{3+ \sqrt{3}}{3+ \sqrt{3}} \\& = \frac{-9-6 \sqrt{3} -3}{9-3} \\& = \frac{-12 - 6 \sqrt{3}}{6} \\& = - 2 - \sqrt{3}$

To verify this on the calculator, $\tan 285^\circ = -3.732$ and $-2 -\sqrt{3} = -3.732$.

## Using the Sum and Difference Identities to Verify Other Identities

Example 8: Verify the identity $\frac{\cos (x-y)}{\sin x \sin y} = \cot x \cot y + 1$

$\cot x \cot y + 1 & = \frac{\cos(x-y)}{\sin x \sin y} \\& = \frac{\cos x \cos y}{\sin x \sin y} + \frac{\sin x \sin y}{\sin x \sin y} && \text{Expand using the cosine difference formula}. \\& = \frac{\cos x \cos y}{\sin x \sin y} + 1 \\\cot x \cot y +1 & = \cot x \cot y +1 && \text{cotangent equals cosine over sine}$

Example 9: Show $\cos(a + b) \cos(a - b) = \cos^2 a - \sin^2 b$

Solution: First, expand left hand side using the sum and difference formulas:

$&\cos(a + b) \cos(a - b) = (\cos a \cos b - \sin a \sin b)(\cos a \cos b + \sin a \sin b) \\& \qquad \qquad \qquad \qquad \quad = \cos^2 a \cos^2 b - \sin^2 a \sin^2 b \rightarrow \text{FOIL, middle terms cancel out}\\& \text{Substitute} (1 - \sin^2 b) \text{for} \cos^2 b\ \text{and} (1 - \cos^2 a) \text{for} \sin^2 a\ \text{and simplify.}\\& \qquad \qquad \qquad \qquad \cos^2 a(1 - \sin^2 b) - \sin^2 b(1 - \cos^2 a) \\& \qquad \qquad \qquad \qquad \cos^2 a - \cos^2 a \sin^2 b - \sin^2 b + \cos^2 a \sin^2 b \\& \qquad \qquad \qquad \qquad \cos^2 a - \sin^2 b$

## Solving Equations with the Sum and Difference Formulas

Just like the section before, we can incorporate all of the sum and difference formulas into equations and solve for values of $x$. In general, you will apply the formula before solving for the variable. Typically, the goal will be to isolate $\sin x, \cos x$, or $\tan x$ and then apply the inverse. Remember, that you may have to use the identities in addition to the formulas seen in this section to solve an equation.

Example 10: Solve $3 \sin (x-\pi)=3$ in the interval $[0, 2\pi)$.

Solution: First, get $\sin(x - \pi)$ by itself, by dividing both sides by $3$.

$\frac{3 \sin (x- \pi)}{3} & = \frac{3}{3} \\\sin (x - \pi) & = 1$

Now, expand the left side using the sine difference formula.

$\sin x \cos \pi - \cos x \sin \pi & = 1 \\\sin x (-1) - \cos x (0) & = 1 \\- \sin x & = 1 \\\sin x & = - 1$

The $\sin x = -1$ when $x$ is $\frac{3\pi}{2}$.

Example 11: Find all the solutions for $2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1$ in the interval $[0, 2\pi)$.

Solution: Get the $\cos^2 \left (x+ \frac{\pi}{2} \right )$ by itself and then take the square root.

$2 \cos^2 \left (x+ \frac{\pi}{2} \right ) & = 1 \\\cos^2 \left (x+ \frac{\pi}{2} \right ) & = \frac{1}{2} \\\cos \left (x+ \frac{\pi}{2} \right ) & = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Now, use the cosine sum formula to expand and solve.

$\cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \frac{\sqrt{2}}{2} \\\cos x(0) - \sin x (1) & = \frac{\sqrt{2}}{2} \\- \sin x & = \frac{\sqrt{2}}{2} \\\sin x & = - \frac{\sqrt{2}}{2}$

The $\sin x = - \frac{\sqrt{2}}{2}$ is in Quadrants III and IV, so $x = \frac{5 \pi}{4}$ and $\frac{7 \pi}{4}$.

## Points to Consider

• What are the angles that have $15^\circ$ and $75^\circ$ as reference angles?
• Are the only angles that we can find the exact sine, cosine, or tangent values for, multiples of $\frac{\pi}{12}$? (Recall that $\frac{\pi}{2}$ would be $6 \cdot \frac{\pi}{12}$, making it a multiple of $\frac{\pi}{12}$)

## Review Questions

1. Find the exact value for:
1. $\cos \frac{5 \pi}{12}$
2. $\cos \frac{7 \pi}{12}$
3. $\sin 345^\circ$
4. $\tan 75^\circ$
5. $\cos 345^\circ$
6. $\sin \frac{17 \pi}{12}$
2. If $\sin y = \frac{12}{13}$, $y$ is in quad II, and $\sin z = \frac{3}{5}$, $z$ is in quad I find $\cos(y - z)$
3. If $\sin y = - \frac{5}{13}$, $y$ is in quad III, and $\sin z = \frac{4}{5}$, $z$ is in quad II find $\sin(y + z)$
4. Simplify:
1. $\cos 80^\circ \cos 20^\circ + \sin 80^\circ \sin 20^\circ$
2. $\sin 25^\circ \cos 5^\circ + \cos 25^\circ \sin 5^\circ$
5. Prove the identity: $\frac{\cos (m - n)}{\sin m \cos n} = \cot m + \tan n$
6. Simplify $\cos(\pi + \theta) = -\cos \theta$
7. Verify the identity: $\sin(a + b) \sin(a - b) = \cos^2b - \cos^2a$
8. Simplify $\tan(\pi + \theta)$
9. Verify that $\sin \frac{\pi}{2} = 1$, using the sine sum formula.
10. Reduce the following to a single term: $\cos(x + y) \cos y + \sin(x + y) \sin y$.
11. Prove $\frac{\cos (c+d)}{\cos (c-d)} = \frac{1 - \tan c \tan d}{1 + \tan c \tan d}$
12. Find all solutions to $2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1$, when $x$ is between $[0, 2\pi)$.
13. Solve for all values of $x$ between $[0, 2\pi)$ for $2 \tan^2 \left (x+ \frac{\pi}{6} \right ) + 1 = 7$.
14. Find all solutions to $\sin \left (x+ \frac{\pi}{6} \right ) = \sin \left (x- \frac{\pi}{4} \right )$, when $x$ is between $[0, 2\pi)$.

1. $\cos \frac{5 \pi}{12} & = \cos \left (\frac{2 \pi}{12} + \frac{3 \pi}{12} \right ) = \cos \left (\frac{\pi}{6} + \frac{\pi}{4} \right ) = \cos \frac{\pi}{6} \cos \frac{\pi}{4} - \sin \frac{\pi}{6} \sin \frac{\pi}{4} \\& = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$
2. $\cos \frac{7 \pi}{12} & = \cos \left (\frac{4 \pi}{12} + \frac{3 \pi}{12} \right ) = \cos \left (\frac{\pi}{3} + \frac{\pi}{4} \right ) = \cos \frac{\pi}{3} \cos \frac{\pi}{4} - \sin \frac{\pi}{3} \sin \frac{\pi}{4} \\& = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$
3. $\sin 345^\circ & = \sin (300^\circ + 45^\circ) = \sin 300^\circ \cos 45^ \circ + \cos 300^\circ \sin 45^\circ \\& = - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = - \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}$
4. $\tan 75^\circ & = \tan (45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1- \tan 45^\circ \tan 30^\circ} \\& = \frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}} = \frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}} = \frac{3+\sqrt{3}}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{9+6 \sqrt{3} + 3}{9-3} = \frac{12 + 6 \sqrt{3}}{6} = 2 + \sqrt{3}$
5. $\cos 345^\circ & = \cos(315^\circ + 30^\circ) = \cos 315^\circ \cos 30^\circ - \sin 315^\circ \sin 30^\circ \\& = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - (-\frac{\sqrt{2}}{2}) \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$
6. $\sin \frac{17 \pi}{12} & = \sin \left (\frac{9 \pi}{12} + \frac{8 \pi}{12} \right ) = \sin \left (\frac{3\pi}{4} + \frac{2\pi}{3} \right ) = \sin \frac{3\pi}{4} \cos \frac{2\pi}{3} + \cos \frac{3\pi}{4} \sin \frac{2\pi}{3} \\& = \frac{\sqrt{2}}{2} \cdot (-\frac{1}{2}) + - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{-\sqrt{2} - \sqrt{6}}{4}$
1. If $\sin y = \frac{12}{13}$ and in Quadrant II, then by the Pythagorean Theorem $\cos y = - \frac{5}{13}(12^2 + b^2 = 13^2)$. And, if $\sin z = \frac{3}{5}$ and in Quadrant I, then by the Pythagorean Theorem $\cos z = \frac{4}{5} (a^2 + 3^2 = 5^2)$. So, to find $\cos(y - z) = \cos y \cos z + \sin y \sin z$ and $= - \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = - \frac{20}{65} + \frac{36}{65} = \frac{16}{65}$
2. If $\sin y = - \frac{5}{13}$ and in Quadrant III, then cosine is also negative. By the Pythagorean Theorem, the second leg is $12 (5^2 + b^2 = 13^2)$, so $\cos y = - \frac{12}{13}$. If the $\sin z = \frac{4}{5}$ and in Quadrant II, then the cosine is also negative. By the Pythagorean Theorem, the second leg is $3 (4^2 + b^2 = 5^2)$, so $\cos = - \frac{3}{5}$. To find $\sin(y + z)$, plug this information into the sine sum formula. $\sin (y + z) & = \sin y \cos z + \cos y \sin z \\& = - \frac{5}{13} \cdot - \frac{3}{5} + - \frac{12}{13} \cdot \frac{4}{5} = \frac{15}{65} - \frac{48}{65} = - \frac{33}{65}$
1. This is the cosine difference formula, so: $\cos 80^\circ \cos 20^\circ + \sin 80^\circ 20^\circ = \cos (80^\circ - 20^\circ) = \cos 60^\circ = \frac{1}{2}$
2. This is the expanded sine sum formula, so: $\sin 25^\circ \cos 5^\circ + \cos 25^\circ \sin 5^\circ = \sin(25^\circ + 5^\circ) = \sin 30^\circ = \frac{1}{2}$
3. Step 1: Expand using the cosine sum formula and change everything into sine and cosine $\frac{\cos (m - n)}{\sin m \cos n} & = \cot m + \tan n \\\frac{\cos m \cos n + \sin m \sin n}{\sin m \cos n} & = \frac{\cos m}{\sin m} + \frac{\sin n}{\cos n}$ Step 2: Find a common denominator for the right hand side. $= \frac{\cos m \cos n + \sin m \sin n}{\sin m \cos n}$ The two sides are the same, thus they are equal to each other and the identity is true.
4. $\cos (\pi + \theta) = \cos \pi \cos \theta - \sin \pi \sin \theta = - \cos \theta$
5. Step 1: Expand $\sin(a + b)$ and $\sin(a - b)$ using the sine sum and difference formulas. $\sin(a + b) \sin(a - b) = \cos^2 b - \cos^2 a\ (\sin a \cos b + \cos a \sin b) (\sin a \cos b - \cos a \sin b)$ Step 2: FOIL and simplify. $\sin^2 a \cos^2 b - \sin a \cos a \sin b \cos b + \sin a \sin b \cos a \cos b - \cos^2 a \sin^2 b \sin^2 a \cos^2 b - \cos a^2 \sin^2 b$ Step 3: Substitute $(1 - \cos^2 a)$ for $\sin^2 a$ and $(1 - \cos^2 b)$ for $\sin^2 b$, distribute and simplify. $& (1 - \cos^2 a) \cos^2 b - \cos a^2 (1 - \cos^2 b) \\& \cos^2 b - \cos^2 a \cos^2 b - \cos^2 a + \cos^2 a \cos^2 b \\& \cos^2 b - \cos^2 a$
6. $\tan (\pi + \theta) = \frac{\tan \pi + \tan \theta}{1- \tan \pi \tan \theta} = \frac{\tan \theta}{1} = \tan \theta$
7. $\sin \frac{\pi}{2} = \sin \left (\frac{\pi}{4} + \frac{\pi}{4} \right) = \sin \frac{\pi}{4} \cos \frac{\pi}{4} - \cos \frac{\pi}{4} \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{4} - \frac{2}{4} = 0$ This could also be verified by using $60^\circ + 30^\circ$
8. Step 1: Expand using the cosine and sine sum formulas. $\cos(x + y) \cos y + \sin(x + y) \sin y = (\cos x \cos y - \sin x \sin y) \cos y + (\sin x \cos y + \cos x \sin y) \sin y$ Step 2: Distribute $\cos y$ and $\sin y$ and simplify. $& = \cos x \cos^2 y - \sin x \sin y \cos y + \sin x \sin y \cos y + \cos x \sin^2 y \\& = \cos x \cos^2 y + \cos x \sin^2 y \\& = \cos x \underbrace{(\cos^2 y + \sin^2 y)}_{1} \\& = \cos x$
9. Step 1: Expand left hand side using the sum and difference formulas $\frac{\cos (c+d)}{\cos(c-d)} & = \frac{1 - \tan c \tan d}{1 + \tan c \tan d} \\\frac{\cos c \cos d - \sin c \sin d}{\cos c \cos d + \sin c \sin d} & = \frac{1- \tan c \tan d}{1+ \tan c \tan d}$ Step 2: Divide each term on the left side by $\cos c \cos d$ and simplify $\frac{\frac{\cos c \cos d}{\cos c \cos d} - \frac{\sin c \sin d}{\cos c \cos d}}{\frac{\cos c \cos d}{\cos c \cos d} + \frac{\sin c \sin d}{\cos c \cos d}} & = \frac{1 - \tan c \tan d}{1 + \tan c \tan d} \\\frac{1 - \tan c \tan d}{1 + \tan c \tan d} & = \frac{1 - \tan c \tan d}{1 + \tan c \tan d}$
10. To find all the solutions, between $[0, 2\pi)$, we need to expand using the sum formula and isolate the $\cos x$. $2 \cos^2 \left (x + \frac{\pi}{2} \right ) & = 1 \\\cos^2 \left (x + \frac{\pi}{2} \right ) & = \frac{1}{2} \\\cos \left (x + \frac{\pi}{2} \right ) & = \pm\sqrt{\frac{1}{2}} = \pm\frac{\sqrt{2}}{2} \\\cos x \cos \frac{\pi}{2} - \sin x \sin \frac{\pi}{2} & = \pm\frac{\sqrt{2}}{2} \\\cos x \cdot 0 - \sin x \cdot 1 & = \pm\frac{\sqrt{2}}{2} \\- \sin x & = \pm\frac{\sqrt{2}}{2} \\\sin x & = \pm\frac{\sqrt{2}}{2}$ This is true when $x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}$, or $\frac{7 \pi}{4}$
11. First, solve for $\tan (x+ \frac{\pi}{6})$. $2 \tan^2 \left (x+ \frac{\pi}{6} \right ) +1 & = 7 \\2 \tan^2 \left (x+ \frac{\pi}{6} \right ) & = 6 \\\tan^2 \left (x+ \frac{\pi}{6} \right ) & = 3 \\\tan \left (x+ \frac{\pi}{6} \right ) & = \pm\sqrt{3}$ Now, use the tangent sum formula to expand for when $\tan (x+ \frac{\pi}{6}) = \sqrt{3}$. $\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = \sqrt{3} \\\tan x + \tan \frac{\pi}{6} & = \sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\\tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\\tan x + \frac{\sqrt{3}}{3} & = \sqrt{3} - \tan x \\2 \tan x & = \frac{2 \sqrt{3}}{3} \\\tan x & = \frac{\sqrt{3}}{3}$ This is true when $x = \frac{\pi}{6}$ or $\frac{7 \pi}{6}$. If the tangent sum formula to expand for when $\tan (x+ \frac{\pi}{6}) = -\sqrt{3}$, we get no solution as shown. $\frac{\tan x + \tan \frac{\pi}{6}}{1 - \tan x \tan \frac{\pi}{6}} & = -\sqrt{3} \\\tan x + \tan \frac{\pi}{6} & = -\sqrt{3} \left (1 - \tan x \tan \frac{\pi}{6} \right ) \\\tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \sqrt{3} \tan x \cdot \frac{\sqrt{3}}{3} \\\tan x + \frac{\sqrt{3}}{3} & = -\sqrt{3} + \tan x \\\frac{\sqrt{3}}{3} & = -\sqrt{3}\\$ Therefore, the tangent sum formula cannot be used in this case. However, since we know that $\tan(x+\frac{\pi}{6}) = -\sqrt{3}$ when $x+\frac{\pi}{6} = \frac{5\pi}{6}$ or $\frac{11\pi}{6}$, we can solve for $x$ as follows. $x+\frac{\pi}{6}=\frac{5\pi}{6} \\x = \frac{4\pi}{6} \\x = \frac{2\pi}{3} \\ \\x+\frac{\pi}{6}=\frac{11\pi}{6} \\x = \frac{10\pi}{6} \\x = \frac{5\pi}{3}$ Therefore, all of the solutions are $x=\frac{\pi}{6}, \frac{2\pi}{3}, \frac{7 \pi}{6}, \frac{5\pi}{3}$
12. To solve, expand each side: $\sin \left (x + \frac{\pi}{6} \right ) & = \sin x \cos \frac{\pi}{6} + \cos x \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x \\\sin \left (x - \frac{\pi}{4} \right ) & = \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x$ Set the two sides equal to each other: $\frac{\sqrt{3}}{2} \sin x + \frac{1}{2} \cos x & = \frac{\sqrt{2}}{2} \sin x - \frac{\sqrt{2}}{2} \cos x \\\sqrt{3} \sin x + \cos x & = \sqrt{2} \sin x - \sqrt{2} \cos x \\\sqrt{3} \sin x - \sqrt{2} \sin x & = - \cos x - \sqrt{2} \cos x \\\sin x \left (\sqrt{3} - \sqrt{2} \right ) & = \cos x \left (-1 - \sqrt{2} \right ) \\\frac{\sin x}{\cos x} & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\\tan x & = \frac{-1- \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} \\& = \frac{- \sqrt{3} - \sqrt{2} + \sqrt{6} - 2}{3-2} \\& = -2 + \sqrt{6} - \sqrt{3} - \sqrt{2}$ As a decimal, this is $-2.69677$, so $\tan^{-1}(-2.69677) = x, x = 290.35^\circ$ and $110.35^\circ$.

Feb 23, 2012

Dec 15, 2015