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# 3.7: Products, Sums, Linear Combinations, and Applications

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the transformation formulas to go from product to sum and sum to product.
• Derive multiple angle formulas.
• Use linear combinations to solve trigonometric equations.
• Apply trigonometric equations to real-life situations.

## Sum to Product Formulas for Sine and Cosine

In some problems, the product of two trigonometric functions is more conveniently found by the sum of two trigonometric functions by use of identities such as this one:

$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}$

This can be verified by using the sum and difference formulas:

$2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} &= 2 \begin{bmatrix} \sin \left( \frac{\alpha}{2} + \frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \end{bmatrix} \\ &= 2 \begin{bmatrix} \left( \sin \frac{\alpha}{2} \cos \frac{\beta}{2} + \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \left) \right( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right ) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos ^2 \frac{\beta}{2} + \sin ^2 \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\beta}{2} + \sin \frac{\beta}{2} \cos ^2 \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin ^2 \frac{\beta}{2} \cos \frac{\alpha}{2} \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \left( \sin^2 \frac{\beta}{2} + \cos^2 \frac{\beta}{2} \right) + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \left( \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} \right) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \end{bmatrix}\\ &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}\\ &= \sin \left( 2 \cdot \frac{\alpha}{2} \right) + \sin \left( 2 \cdot \frac{\beta}{2} \right)\\ &= \sin \alpha + \sin \beta$

The following variations can be derived similarly:

$\sin \alpha - \sin \beta &= 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}\\ \cos \alpha + \cos \beta &= 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\\ \cos \alpha - \cos \beta &= -2 \sin \frac{\alpha + \beta}{2} \times \sin \frac{\alpha - \beta}{2}\\$

Example 1: Change $\sin 5x - \sin 9x$ into a product.

Solution: Use the formula $\sin \alpha - \sin \beta = 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}$.

$\sin 5x - \sin 9x &= 2 \sin \frac{5x - 9x}{2} \cos \frac{5x + 9x}{2}\\ &= 2 \sin (-2x) \cos 7x\\ &= -2 \sin 2x \cos 7x$

Example 2: Change $\cos(-3x) + \cos 8x$ into a product.

Solution: Use the formula $\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}$.

$\cos (-3x) + \cos (8x) &= 2 \cos \frac{-3x + 8x}{2} \cos \frac {-3x - 8x}{2}\\ &= 2 \cos (2.5x) \cos (-5.5x)\\ &= 2 \cos (2.5) \cos (5.5x)$

Example 3: Change $2 \sin 7x \cos 4x$ to a sum.

Solution: This is the reverse of what was done in the previous two examples. Looking at the four formulas above, take the one that has sine and cosine as a product, $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}$. Therefore, $7x = \frac{\alpha + \beta}{2}$ and $4x = \frac{\alpha - \beta}{2}$.

$7x & = \frac{\alpha + \beta}{2} &&&& 4x = \frac{\alpha - \beta}{2} \\&&& \text{and} \\14x & = \alpha+\beta &&&& 8x= \alpha - \beta \\&\alpha = 14x - \beta &&&& 8x=[14x-\beta]-\beta \\&&& \text{so} \\&&&&&-6x = -2\beta\\&&&&&3x=\beta\\\alpha=14x-3x\\\alpha=11x$

So, this translates to $\sin(11x) + \sin(3x)$. A shortcut for this problem, would be to notice that the sum of $7x$ and $4x$ is $11x$ and the difference is $3x$.

## Product to Sum Formulas for Sine and Cosine

There are two formulas for transforming a product of sine or cosine into a sum or difference. First, let’s look at the product of the sine of two angles. To do this, start with cosine.

$& \cos(a - b) = \cos a \cos b + \sin a \sin b \ \text {and} \ \cos(a + b) = \cos a \cos b - \sin a \sin b \\& \cos(a - b) - \cos (a + b) = \cos a \cos b + \sin a \sin b - (\cos a \cos b - \sin a \sin b)\\& \cos(a - b) - \cos (a + b) = \cos a \cos b + \sin a \sin b - \cos a \cos b + \sin a \sin b\\& \qquad \qquad \qquad \qquad \cos (a - b) - \cos (a + b) = 2 \sin a \sin b\\& \qquad \qquad \qquad \qquad \frac{1}{2}\left[ \cos (a - b) - \cos (a+ b)\right] = \sin a \sin b$

The following product to sum formulas can be derived using the same method:

$\cos \alpha \cos \beta &= \frac {1}{2} \left [ \cos (\alpha - \beta) + \cos (\alpha + \beta) \right ]\\\sin \alpha \cos \beta &= \frac {1}{2} \left [\sin (\alpha + \beta) + \sin (\alpha - \beta) \right ]\\\cos \alpha \sin \beta &= \frac {1}{2} \left[\sin (\alpha + \beta) - \sin (\alpha - \beta) \right ]$

Example 4: Change $\cos 2x \cos 5y$ to a sum.

Solution: Use the formula $\cos \alpha \cos \beta = \frac{1}{2} \left [\cos (\alpha - \beta) + \cos (\alpha + \beta) \right ]$. Set $\alpha = 2x$ and $\beta = 5y$.

$\cos 2x \cos 5y = \frac{1}{2} \left [\cos (2x - 5y) + \cos (2x + 5y) \right ]$

Example 5: Change $\frac{\sin11z + \sin z}{2}$ to a product.

Solution: Use the formula $\sin \alpha \cos \beta = \frac{1}{2} \left [\sin (\alpha + \beta) + \sin (\alpha - \beta) \right ]$. Therefore, $\alpha + \beta = 11z$ and $\alpha - \beta = z$. Solve the second equation for $\alpha$ and plug that into the first.

$\alpha = z + \beta \rightarrow (z + \beta) + \beta = 11z && \text {and} \quad \alpha = z + 5z = 6z\\z + 2 \beta = 11z\\2 \beta = 10z\\\beta = 5z$

$\frac{\sin11z + \sin z}{2} = \sin 6z \cos 5z$. Again, the sum of $6z$ and $5z$ is $11z$ and the difference is $z$.

## Solving Equations with Product and Sum Formulas

Example 6: Solve $\sin 4x + \sin 2x = 0$.

Solution: Use the formula $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}$.

$\sin 4x + \sin 2x &= 0 \qquad \text{So}, \ \sin 3x = 0 \ \ \text{and} \ \cos x = 0 \rightarrow x = \frac{\pi}{2}, \frac{3 \pi}{2}\\2 \sin 3x \cos x &= 0 \qquad \qquad 3x = 0, \pi, 2 \pi, 3 \pi, 4 \pi, 5 \pi\\\sin 3x \cos x &= 0 \qquad \qquad \ \ x = 0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}$

Example 7: Solve $\cos 5x + \cos x = \cos 2x$.

Solution: Use the formula $\cos \alpha + \cos \beta = 2 \cos \frac{\alpha +\beta}{2} \times \cos \frac{\alpha - \beta}{2}$.

$& \qquad \qquad \cos 5x + \cos x = \cos 2x\\& \qquad \qquad 2 \cos 3x \cos 2x = \cos 2x\\& 2 \cos 3x \cos 2x - \cos 2x = 0\\& \quad \ \cos 2x(2 \cos 3x-1)=0\\& \quad \ \ \swarrow \qquad \qquad \ \searrow\\& \cos 2x = 0 \qquad \qquad 2 \cos 3x-1=0\\& \qquad \qquad \qquad \qquad \qquad \ 2\cos 3x = 1\\& \quad \ \ 2x=\frac{\pi}{2}, \frac{3\pi}{2} \quad \text{and} \quad \ \cos 3x=\frac{1}{2}\\& \qquad \ x = \frac{\pi}{4}, \frac{3 \pi}{4} \qquad \qquad \qquad 3x = \frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{11 \pi}{3}, \frac{13 \pi}{3}, \frac{17 \pi}{3}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}$

## Triple-Angle Formulas and Beyond

By combining the sum formula and the double angle formula, formulas for triple angles and more can be found.

Example 8: Find the formula for $\sin 3x$

Solution: Use both the double angle formula and the sum formula.

$\sin 3x &= \sin(2x + x) \\&= \sin(2x) \cos x + \cos(2x)\sin x \\&= (2 \sin x \cos x) \cos x + (\cos^2 x - \sin^2 x) \sin x \\&= 2 \sin x \cos^2 x + \cos^2 x \sin x - \sin^3 x \\&= 3 \sin x \cos^2 x - \sin^3 x \\&= 3 \sin x(1 - \sin^2 x) - \sin^3 x \\&= 3 \sin x - 4 \sin^3 x$

Example 9: Find the formula for $\cos 4x$

Solution: Using the same method from the previous example, you can obtain this formula.

$\cos 4x &= \cos(2x + 2x)\\&= \cos^2 2x - \sin^2 2x \\&= (\cos^2 x - \sin^2 x)^2 - (2 \sin x \cos x)^2 \\&= \cos^4 x - 2\sin^2 x \cos^2 x + \sin^4 x - 4\sin^2 x \cos^2 x \\&= \cos^4 x - 6 \sin^2 x \cos^2 x + \sin^4 x \\&= \cos^4 x - 6(1 - \cos^2 x) \cos^2 x + (1 - \cos^2 x)^2 \\&= 1 - 8\cos^2 x + 8 \cos^4 x$

## Linear Combinations

Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler cosine function.

$A \cos x + B \sin x = C \cos(x - D)$, where $C = \sqrt{A^2 + B^2}$, $\cos D = \frac{A}{C}$ and $\sin D = \frac{B}{C}$.

Example 10: Transform $3 \cos 2x - 4 \sin 2x$ into the form $C \cos(2x - D)$

Solution: $A = 3$ and $B = -4$, so $C = \sqrt{3^2 + (-4)^2} = 5$. Therefore $\cos D = \frac{3}{5}$ and $\sin D = - \frac{4}{5}$ which makes the reference angle is $-53.1^\circ$ or -0.927 radians. since cosine is positive and sine is negative, the angle must be a fourth quadrant angle. $D$ must therefore be $306.9^ \circ$ or 5.36 radians.The final answer is $3 \cos 2x - 4 \sin 2x = 5 \cos (2x - 5.36)$.

Example 11: Solve $5 \cos x + 12 \sin x = 6$.

Solution: First, transform the left-hand side into the form $C \cos(x - D)$. $A = 5$ and $B = 12$, so $C = \sqrt{5^2 + 12^2} = 13$. From this $\cos D = \frac{5}{13}$ and $\sin D = \frac{12}{13}$, which makes the angle in the first quadrant and 1.176 radians. Now, our equation looks like this: $13 \cos(x - 1.176) = 6$ and we can solve for $x$.

$\cos (x - 1.176) &= \frac{6}{13}\\x - 1.176 &= \cos^{-1} \left(\frac{6}{13} \right)\\x - 1.176 &= 1.09\\x &= 2.267 \ \text {radians}$

## Applications & Technology

Example 12: The range of a small rocket that is launched with an initial velocity $v$ at an angle with $\theta$ the horizontal is given by $R(range) = \frac{v^2(velocity)}{g(9.8m/s^2)} \sin 2 \theta$. If the rocket is launched with an initial velocity of 15 m/s, what angle is needed to reach a range of 20 m?

Solution: Plug in 15 m/s for $v$ and 20 m for the range to solve for the angle.

$20 &= \frac{15^2}{9.8} \sin 2 \theta\\20 &= 22.96 \sin 2 \theta\\0.87 \bar{1} &= \sin 2 \theta\\\sin^{-1} (0.87 \bar{1}) &= 2 \theta\\60.59^\circ, 119.41^\circ &= 2 \theta\\30.3^\circ, 59.7^\circ &= \theta$

You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of $\pi$.

Example 13: Solve $\sin x = 2 \cos x$ such that $0 \le x \le 2\pi$ using a graphing calculator.

Solution: In $y =$, graph $y1 = \sin x$ and $y2 = 2 \cos x$.

Next, use CALC to find the intersection points of the graphs.

## Review Questions

1. Express the sum as a product: $\sin 9x + \sin 5x$
2. Express the difference as a product: $\cos 4y - \cos 3y$
3. Verify the identity (using sum-to-product formula): $\frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a$
4. Express the product as a sum: $\sin(6 \theta) \sin(4 \theta)$
5. Transform to the form $C \cos(x - D)$
1. $\ 5 \cos x - 5 \sin x$
2. $- 15 \cos 3x - 8 \sin 3x$
6. Solve $\sin 11x - \sin 5x = 0$ for all solutions $0 \le x < 2\pi$.
7. Solve $\cos 4x + \cos 2x = 0$ for all solutions $0 \le x < 2\pi$.
8. Solve $\sin 5x + \sin x = \sin 3x$ for all solutions $0 \le x < 2\pi$.
9. In the study of electronics, the function $f(t) = \sin(200t + \pi) + \sin(200t - \pi)$ is used to analyze frequency. Simplify this function using the sum-to-product formula.
10. Derive a formula for $\tan 4x$.
11. A spring is being moved up and down. Attached to the end of the spring is an object that undergoes a vertical displacement. The displacement is given by the equation $y = 3.50 \sin t + 1.20 \sin 2t$. Find the first two values of $t$ (in seconds) for which $y = 0$.

1. Using the sum-to-product formula: $& \sin 9x + \sin 5x\\& \frac{1}{2} \left(\sin \left(\frac{9x + 5x}{2} \right) \cos \left(\frac{9x - 5x}{2} \right) \right)\\& \frac{1}{2} \sin 7x \cos 2x$
2. Using the difference-to-product formula: $& \cos 4y - \cos 3y\\& -2 \sin \left(\frac{4y + 3y}{2} \right) \sin \left(\frac{4y - 3y}{2} \right)\\& -2 \sin \frac{7y}{2} \sin \frac{y}{2}$
3. Using the difference-to-product formulas: $& \qquad \qquad \frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\\& \frac{-2 \sin \left (\frac{3a + 5a}{2} \right ) \sin \left (\frac {3a - 5a}{2} \right )}{2 \sin \left (\frac{3a - 5a}{2} \right ) \cos \left(\frac{3a + 5a}{2} \right)}\\& \qquad \qquad \qquad \ \ - \frac{\sin 4a}{\cos 4a}\\& \qquad \qquad \qquad \ \ - \tan 4a$
4. Using the product-to-sum formula: $& \qquad \quad \sin 6 \theta \sin 4 \theta\\& \frac{1}{2} \left( \cos (6 \theta - 4 \theta) - \cos (6 \theta + 4 \theta) \right )\\& \qquad \frac{1}{2}(\cos 2 \theta - \cos 10 \theta)$
1. If $5 \cos x - 5 \sin x$, then $A = 5$ and $B = -5$. By the Pythagorean Theorem, $C = 5 \sqrt{2}$ and $\cos D = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. So, because $B$ is negative, $D$ is in Quadrant IV. Therefore, $D = \frac{7 \pi}{4}$. Our final answer is $5 \sqrt{2} \cos \left (x - \frac{7 \pi}{4} \right)$.
2. If $-15 \cos 3x - 8 \sin 3x$, then $A = -15$ and $B = -8$. By the Pythagorean Theorem, $C = 17$. Because $A$ and $B$ are both negative, $D$ is in Quadrant III, which means $D = \cos^{-1} \left (\frac{15}{17} \right) = 0.49+\pi = 3.63$ rad. Our final answer is $17 \cos 3(x - 3.63)$.
5. Using the sum-to-product formula: $\sin 11x - \sin 5x &= 0 \qquad \qquad \sin 3x = 0 \qquad \text{or} \qquad \cos 8x = 0\\2 \sin \frac{11x - 5x}{2} \cos \frac{11x + 5x}{2} &= 0 \qquad \text{So}, \qquad 3x = 0, \pi, 2\pi, 3\pi, 4\pi, 5\pi \qquad \qquad \quad \ \ 8x = \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2}, \frac{13 \pi}{2}, \frac{15 \pi}{2}\\2 \sin 3x \cos 8x &= 0\\\sin 3x \cos 8x &= 0 \qquad \qquad \qquad x = 0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \qquad \qquad \quad \ \ \ x = \frac{\pi}{16}, \frac{3 \pi}{16}, \frac{5 \pi}{16}, \frac{7 \pi}{16}, \frac{9 \pi}{16}, \frac{11 \pi}{16}, \frac{13 \pi}{16}, \frac{15 \pi}{16}$
6. Using the sum-to-product formula: $\cos 4x + \cos2x &= 0\\2 \cos \frac{4x + 2x}{2} \cos \frac{4x - 2x}{2} &= 0\\2 \cos 3x \cos x &= 0\\\cos 3x \cos x &= 0$ So, either $\cos 3x = 0$ or $\cos x = 0$ $3x &= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}, \frac{9 \pi}{2}, \frac{11 \pi}{2}\\x &= \frac{\pi}{6}, \frac{\pi}{2}, \frac{5 \pi}{6}, \frac{ 7 \pi}{6}, \frac{3 \pi}{2},\frac{11 \pi}{6}$
7. Move $\sin 3x$ over to the other side and use the sum-to-product formula: $\sin 5x + \sin x &= \sin 3x\\\sin 5x - \sin 3x + \sin x &= 0\\2 \cos \left ( \frac{5x + 3x}{2} \right ) \sin \left ( \frac{5x - 3x}{2} \right ) + \sin x &= 0\\2 \cos 4x \sin x + \sin x &= 0\\\sin x(2 \cos 4x + 1) &= 0$ So $\sin x = 0$ $x = 0, \pi \ \ \text{or} \ \ 2 \cos 4x &= -1\\\cos 4x &= - \frac{1}{2}\\4x &= \frac{2 \pi}{3}, \frac{4 \pi}{3}, \frac{8 \pi}{3}, \frac{10 \pi}{3}, \frac{14 \pi}{3}, \frac{16 \pi}{3}, \frac{20 \pi}{3}, \frac{22 \pi}{3}\\&= \frac{\pi}{6}, \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{4 \pi}{3}, \frac{5 \pi}{3}, \frac{11 \pi}{6}\\x = 0, &= \frac{\pi}{6}, \frac{\pi}{3}, \frac{2 \pi}{3}, \frac{5 \pi}{6}, \pi, \frac{7 \pi}{6}, \frac{4 \pi}{3}, \frac{5 \pi}{3}, \frac{11 \pi}{6}$
8. Using the sum-to-product formula: $f(t) &= \sin (200t + \pi) + \sin(200t - \pi)\\&= 2 \sin \left (\frac{(200t + \pi) + (200t - \pi)}{2} \right ) \cos \left (\frac{(200t + \pi) - (200t - \pi)}{2} \right )\\&= 2 \sin \left (\frac{400t}{2} \right) \cos \left ( \frac{2 \pi}{2} \right )\\&= 2 \sin 200t \cos \pi\\&= 2 \sin 200t (-1)\\&= -2 \sin 200t$
9. Derive a formula for $\tan 4x$. $\tan 4x &= \tan (2x + 2x)\\&= \frac {\tan 2x + \tan 2x}{1 - \tan 2x \tan 2x}\\&= \frac {2 \tan 2x}{1 - \tan^2 2x}\\&= \frac{2 \cdot \frac{2 \tan x}{1 - \tan^2x}}{1 - \left ( \frac{2 \tan x}{1 - \tan^2x} \right )^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \div \frac{(1 - \tan ^2 x)^2 - 4 \tan^2 x}{(1 - \tan^2 x)^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \div \frac{1 - 2 \tan^2 x + \tan^4 x- 4 \tan^2 x}{(1 - \tan^2 x)^2}\\&= \frac{4 \tan x}{1 - \tan^2 x} \cdot \frac{(1 - \tan^2 x)^2}{1 - 6 \tan^2 x + \tan^4 x}\\&= \frac{4 \tan x - 4 \tan^3 x}{1 - 6 \tan^2 x + \tan^4 x}$
10. Let $y = 0$. $3.50 \sin t + 1.20 \sin 2t &= 0 \\3.50 \sin t + 2.40 \sin t \cos t &= 0,\ \text{Double-Angle Identity} \\\sin t(3.50 + 2.40 \cos t) &= 0 \\\sin t = 0\ \text{or}\ 3.50 + 2.40 \cos t &= 0 \\2.40 \cos t &= -3.50 \\\cos t &= -1.46 \rightarrow \text{no solution because}\ -1 \le \cos t \le 1.\\t &= 0, \pi$

Feb 23, 2012

Dec 16, 2014