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# 3.9: Applications of Sum and Difference Formulas

Difficulty Level: At Grade Created by: CK-12
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Practice Applications of Sum and Difference Formulas
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You are quite likely familiar with the values of trig functions for a variety of angles. Angles such as 30\begin{align*}30^\circ\end{align*}, 60\begin{align*}60^\circ\end{align*}, and 90\begin{align*}90^\circ\end{align*} are common. However, if you were asked to find the value of a trig function for a more rarely used angle, could you do so? Or what if you were asked to find the value of a trig function for a sum of angles? For example, if you were asked to find sin(3π2+π4)\begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*} could you?

Read on, and in this section, you'll get practice with simplifying trig functions of angles using the sum and difference formulas.

### Sum and Difference Formulas

Quite frequently one of the main obstacles to solving a problem in trigonometry is the inability to transform the problem into a form that makes it easier to solve. Sum and difference formulas can be very valuable in helping with this.

Here we'll get some extra practice putting the sum and difference formulas to good use. If you haven't gone through them yet, you might want to review the sections on the Sum and Difference Formulas for sine, cosine, and tangent.

#### Solve using the Sum Formula

Verify the identity cos(xy)sinxsiny=cotxcoty+1\begin{align*}\frac{\cos (x-y)}{\sin x \sin y} = \cot x \cot y + 1\end{align*}

cotxcoty+1cotxcoty+1=cos(xy)sinxsiny=cosxcosysinxsiny+sinxsinysinxsiny=cosxcosysinxsiny+1=cotxcoty+1Expand using the cosine difference formula.cotangent equals cosine over sine\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*}

#### Solve using the Difference Formula

Solve 3sin(xπ)=3\begin{align*}3 \sin (x-\pi)=3\end{align*} in the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

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

3sin(xπ)3sin(xπ)=33=1\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.

sinxcosπcosxsinπsinx(1)cosx(0)sinxsinx=1=1=1=1\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 sinx=1\begin{align*}\sin x = -1\end{align*} when x\begin{align*}x\end{align*} is 3π2\begin{align*}\frac{3\pi}{2}\end{align*}.

#### Solve using the Sum Formula

Find all the solutions for 2cos2(x+π2)=1\begin{align*}2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1\end{align*} in the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

Get the cos2(x+π2)\begin{align*}\cos^2 \left (x+ \frac{\pi}{2} \right )\end{align*} by itself and then take the square root.

2cos2(x+π2)cos2(x+π2)cos(x+π2)=1=12=12=12=22\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.

cosxcosπ2sinxsinπ2cosx(0)sinx(1)sinxsinx=22=22=22=22\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 sinx=22\begin{align*}\sin x = - \frac{\sqrt{2}}{2}\end{align*} is in Quadrants III and IV, so x=5π4\begin{align*}x = \frac{5 \pi}{4}\end{align*} and 7π4\begin{align*}\frac{7 \pi}{4}\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to find sin(3π2+π4)\begin{align*}\sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right)\end{align*}, use the sine sum formula:

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)sin(3π2+π4)=sin(3π2)×cos(π4)+cos(3π2)×sin(π4)=(1)(22)+(0)(22)=22\begin{align*} \sin (a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\\ \sin \left( \frac{3\pi}{2} + \frac{\pi}{4} \right) = \sin \left( \frac{3\pi}{2} \right) \times \cos \left( \frac{\pi}{4} \right) + \cos \left( \frac{3\pi}{2} \right) \times \sin \left( \frac{\pi}{4} \right)\\ = (-1)\left( \frac{\sqrt{2}}{2} \right) + (0)\left( \frac{\sqrt{2}}{2} \right)\\ = -\frac{\sqrt{2}}{2}\\ \end{align*}

#### Example 2

Find all solutions to 2cos2(x+π2)=1\begin{align*}2 \cos^2 \left (x+ \frac{\pi}{2} \right ) = 1\end{align*}, when x\begin{align*}x\end{align*} is between [0,2π)\begin{align*}[0, 2\pi)\end{align*}.

To find all the solutions, between [0,2π)\begin{align*}[0, 2\pi)\end{align*}, we need to expand using the sum formula and isolate the cosx\begin{align*}\cos x\end{align*}.

2cos2(x+π2)cos2(x+π2)cos(x+π2)cosxcosπ2sinxsinπ2cosx0sinx1sinxsinx=1=12=±12=±22=±22=±22=±22=±22\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 x=π4,3π4,5π4\begin{align*}x = \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}\end{align*}, or 7π4\begin{align*}\frac{7 \pi}{4}\end{align*}

#### Example 3

Solve for all values of x\begin{align*}x\end{align*} between [0,2π)\begin{align*}[0, 2\pi)\end{align*} for 2tan2(x+π6)+1=7\begin{align*}2 \tan^2 \left (x+ \frac{\pi}{6} \right ) + 1 = 7\end{align*}.

First, solve for tan(x+π6)\begin{align*}\tan (x+ \frac{\pi}{6})\end{align*}.

2tan2(x+π6)+12tan2(x+π6)tan2(x+π6)tan(x+π6)=7=6=3=±3\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 tan(x+π6)=3\begin{align*}\tan (x+ \frac{\pi}{6}) = \sqrt{3}\end{align*}.

tanx+tanπ61tanxtanπ6tanx+tanπ6tanx+33tanx+332tanxtanx=3=3(1tanxtanπ6)=33tanx33=3tanx=233=33\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*}

#### Example 4

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*}.

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*}.

### Review

Prove each identity.

1. \begin{align*}\cos(3x)+\cos(x)=2\cos(2x)\cos(x)\end{align*}
2. \begin{align*}\cos(3x)=\cos^3(x)-3\sin^2(x)\cos(x)\end{align*}
3. \begin{align*}\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)\end{align*}
4. \begin{align*}\sin(4x)+\sin(2x)=2\sin(3x)\cos(x)\end{align*}
5. \begin{align*}\tan(5x)\tan(3x)=\frac{\tan^2(4x)-\tan^2(x)}{1-\tan^2(4x)\tan^2(x)}\end{align*}
6. \begin{align*}\cos((\frac{\pi}{2}-x)-y)=\sin(x+y)\end{align*}

1. \begin{align*}y=\cos(3)\cos(x)+\sin(3)\sin(x)\end{align*}
2. \begin{align*}y=\cos(x)\cos(\frac{\pi}{2})+\sin(x)\sin(\frac{\pi}{2})\end{align*}
3. \begin{align*}y=\sin(x)\cos(\frac{\pi}{2})+\cos(x)\sin(\frac{\pi}{2})\end{align*}
4. \begin{align*}y=\sin(x)\cos(\frac{3\pi}{2})-\cos(3)\sin(\frac{\pi}{2})\end{align*}
5. \begin{align*}y=\cos(4x)\cos(2x)-\sin(4x)\sin(2x)\end{align*}
6. \begin{align*}y=\cos(x)\cos(x)-\sin(x)\sin(x)\end{align*}

Solve each equation on the interval \begin{align*}[0,2\pi)\end{align*}.

1. \begin{align*}2\sin(x-\frac{\pi}{2})=1\end{align*}
2. \begin{align*}4\cos(x-\pi)=4\end{align*}
3. \begin{align*}2\sin(x-\pi)=\sqrt{2}\end{align*}

To see the Review answers, open this PDF file and look for section 3.9.

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### Vocabulary Language: English

Difference Formula

Trigonometric function difference formulas exist for each of the primary trigonometric functions. For example, the cosine difference formula is $cos(A - B) = cosA cosB + sinA sinB$.

Sum Formula

A sum formula is a formula to help simplify a trigonometric function of the sum of two angles, such as $\sin(a+b)$.

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