<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

3.9: Applications of Sum and Difference Formulas

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated11 minsto complete
%
Progress
Practice Applications of Sum and Difference Formulas
 
 
 
MEMORY METER
This indicates how strong in your memory this concept is
Practice
Progress
Estimated11 minsto complete
%
Estimated11 minsto complete
%
Practice Now
MEMORY METER
This indicates how strong in your memory this concept is
Turn In

You are quite likely familiar with the values of trig functions for a variety of angles. Angles such as \begin{align*}30^\circ\end{align*}, \begin{align*}60^\circ\end{align*}, and \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 \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 \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*}

Solve using the Difference Formula

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

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*}.

Solve using the Sum Formula

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*}.

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*}.

Examples

Example 1

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

\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 \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*}.

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*}

Example 3

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*}.

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*}

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*}

Use sum and difference formulas to help you graph each function.

  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*}

Review (Answers)

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

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Vocabulary

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).

Image Attributions

Show Hide Details
Description
Difficulty Level:
At Grade
Subjects:
Grades:
Date Created:
Sep 26, 2012
Last Modified:
Mar 23, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
MAT.TRG.338.L.1
Here