# 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 \begin{align*}\cos(a \pm b), \sin (a \pm b)\end{align*}, and \begin{align*}\tan(a \pm b)\end{align*}. These identities have very useful expansions and can help to solve identities and equations.

## Sum and Difference Formulas: Cosine

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

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

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

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

\begin{align*}\sqrt{[\cos (a-b)-1]^2 + [\sin (a-b)-0]^2} = \sqrt{(\cos a - \cos b)^2+(\sin a - \sin b)^2}\end{align*}

Square both sides to eliminate the square root.

\begin{align*}[\cos (a - b) - 1]^2 + [\sin (a-b)-0]^2 = (\cos a - \cos b)^2 + (\sin a - \sin b)^2\end{align*}

FOIL all four squared expressions and simplify.

\begin{align*}\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\end{align*}

In \begin{align*}\cos(a - b) = \cos a \cos b + \sin a \sin b\end{align*}, the difference formula for cosine, you can substitute \begin{align*}a - (- b) = a + b\end{align*} to obtain: \begin{align*}\cos(a + b) = \cos[a - (- b)]\end{align*} or \begin{align*}\cos a \cos (- b) + \sin a \sin(-b)\end{align*}. since \begin{align*}\cos(-b) = \cos b\end{align*} and \begin{align*}\sin (-b) = -\sin b\end{align*}, then \begin{align*}\cos(a + b) = \cos a \cos b - \sin a \sin b\end{align*}, 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 \begin{align*}\cos \left (\frac{\pi}{2} - \theta \right )\end{align*} using the cosine difference formula.

Solution:

\begin{align*}\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\end{align*}

We know that is a true identity because of our understanding of the sine and cosine curves, which are a phase shift of \begin{align*}\frac{\pi}{2}\end{align*} 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 \begin{align*}15^\circ =(45^\circ - 30^\circ), 75^\circ =(45^\circ + 30^\circ)\end{align*}, among others.

Example 2: Find the exact value of \begin{align*}\cos 15^\circ\end{align*}

Solution: Use the difference formula where \begin{align*}a = 45^\circ\end{align*} and \begin{align*}b = 30^\circ\end{align*}.

\begin{align*}\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}\end{align*}

Example 3: Find the exact value of \begin{align*}\cos 105^\circ\end{align*}.

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

1.

\begin{align*}\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}\end{align*}

2.

\begin{align*}\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}\end{align*}

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 \begin{align*}\cos \frac{5 \pi}{12}\end{align*}, in radians.

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

\begin{align*}\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}\end{align*}

## Sum and Difference Identities: Sine

To find \begin{align*}\sin(a + b)\end{align*}, use Example 1, from above:

\begin{align*}\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}\end{align*}

In conclusion, \begin{align*}\sin(a + b) = \sin a \cos b + \cos a \sin b\end{align*}, which is the sum formula for sine.

To obtain the identity for \begin{align*}\sin(a - b)\end{align*}:

\begin{align*}\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\end{align*}

In conclusion, \begin{align*}\sin(a - b) = \sin a \cos b - \cos a \sin b\end{align*}, so, this is the difference formula for sine.

Example 5: Find the exact value of \begin{align*}\sin \frac{5 \pi}{12}\end{align*}

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

\begin{align*}\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}\end{align*}

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

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

Use \begin{align*}\sin^2 \alpha + \cos^2 \alpha = 1\end{align*}, to find the values of each of the missing cosine values.

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

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

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

\begin{align*}\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}\end{align*}

## Sum and Difference Identities: Tangent

To find the sum formula for tangent:

\begin{align*}\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}\end{align*}

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

\begin{align*} \tan(a-b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}\end{align*}

Example 7: Find the exact value of \begin{align*}\tan 285^\circ\end{align*}.

Solution: Use the difference formula for tangent, with \begin{align*}285^\circ = 330^\circ - 45^\circ\end{align*}

\begin{align*}\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}\end{align*}

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

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

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

\begin{align*}\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}\end{align*}

Example 9: Show \begin{align*}\cos(a + b) \cos(a - b) = \cos^2 a - \sin^2 b\end{align*}

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

\begin{align*}&\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\end{align*}

## 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 \begin{align*}x\end{align*}. In general, you will apply the formula before solving for the variable. Typically, the goal will be to isolate \begin{align*}\sin x, \cos x\end{align*}, or \begin{align*}\tan x\end{align*} 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 \begin{align*}3 \sin (x-\pi)=3\end{align*} in the interval \begin{align*}[0, 2\pi)\end{align*}.

Solution: First, get \begin{align*}\sin(x - \pi)\end{align*} by itself, by dividing both sides by \begin{align*}3\end{align*}.

\begin{align*}\frac{3 \sin (x- \pi)}{3} & = \frac{3}{3} \\ \sin (x - \pi) & = 1\end{align*}

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

\begin{align*}\sin x \cos \pi - \cos x \sin \pi & = 1 \\ \sin x (-1) - \cos x (0) & = 1 \\ - \sin x & = 1 \\ \sin x & = - 1 \end{align*}

The \begin{align*}\sin x = -1\end{align*} when \begin{align*}x\end{align*} is \begin{align*}\frac{3\pi}{2}\end{align*}.

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

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

\begin{align*}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}\end{align*}

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

\begin{align*}\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} \end{align*}

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

## Points to Consider

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

## Review Questions

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

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

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: