# Simplifying Trigonometric Expressions using Sum and Difference Formulas

## Simplify sine, cosine, and tangent of angles that are added or subtracted.

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Simplifying Trig Expressions using Sum and Difference Formulas

As Agent Trigonometry you are given this clue: \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*}. How could you simplify this expression to make solving your case easier?

### Simplifying Trigonometric Expressions

We can also use the sum and difference formulas to simplify trigonometric expressions.

The \begin{align*}\sin a = -\frac{3}{5}\end{align*} and \begin{align*}\cos b =\frac{12}{13}\end{align*}. \begin{align*}a\end{align*} is in the \begin{align*}3^{rd}\end{align*} quadrant and \begin{align*}b\end{align*} is in the \begin{align*}1^{st}\end{align*}. Let's find \begin{align*}\sin(a+b)\end{align*}.

First, we need to find \begin{align*}\cos a\end{align*} and \begin{align*}\sin b\end{align*}. Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, \begin{align*}\cos a=-\frac{4}{5}\end{align*} because it is in the \begin{align*}3^{rd}\end{align*} quadrant and \begin{align*}\sin b = \frac{5}{13}\end{align*}. Now, use the appropriate formulas.

\begin{align*}\sin (a+b) &=\sin a \cos b + \cos a \sin b \\ &= -\frac{3}{5}\cdot \frac{12}{13}+-\frac{4}{5}\cdot \frac{5}{13} \\ &= -\frac{56}{65}\end{align*}

Now, using the information from the previous problem above, let's find \begin{align*}\tan (a+b)\end{align*}.

From the cosine and sine of \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, we know that \begin{align*}\tan a=\frac{3}{4}\end{align*} and \begin{align*}\tan b=\frac{5}{12}\end{align*}.

\begin{align*}\tan (a+b) &=\frac{\tan a +\tan b}{1-\tan a \tan b} \\ &= \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}} \\ &= \frac{\frac{14}{12}}{\frac{11}{16}}=\frac{56}{33}\end{align*}

Finally, let's simplify \begin{align*}\cos (\pi - x)\end{align*}.

Expand this using the difference formula and then simplify.

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

### Examples

#### Example 1

Earlier, you were asked to simplify \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*}.

You can expand the expression using the difference formula and then simplify.

\begin{align*}\sin(\frac{\pi}{2} - x)=\sin \frac{\pi}{2} \cos x - \cos \frac{\pi}{2} \sin x \\ &=1\cdot \cos x - 0\cdot \sin x \\ &=\cos x\end{align*}

#### Example 2

Using the information from the first problem above (where we found \begin{align*}\sin (a+b)\end{align*}), find \begin{align*}\cos(a-b)\end{align*}.

\begin{align*}\cos(a-b) &=\cos a \cos b + \sin a \sin b \\ &=-\frac{4}{5}\cdot \frac{12}{13}+-\frac{3}{5}\cdot\frac{5}{13} \\ &=-\frac{63}{65}\end{align*}

#### Example 3

Simplify \begin{align*}\tan (x+\pi)\end{align*}.

\begin{align*}\tan (x+\pi)&=\frac{\tan x +\tan \pi}{1-\tan x \tan \pi} \\ &=\frac{\tan x +0}{1-\tan 0} \\ &=\tan x\end{align*}

### Review

\begin{align*}\sin a =-\frac{8}{17}, \pi \le a < \frac{3\pi}{2}\end{align*} and \begin{align*}\sin b =-\frac{1}{2}, \frac{3\pi}{2}\le b <2\pi\end{align*}. Find the exact trig values of:

1. \begin{align*}\sin (a+b)\end{align*}
2. \begin{align*}\cos (a+b)\end{align*}
3. \begin{align*}\sin (a-b)\end{align*}
4. \begin{align*}\tan (a+b)\end{align*}
5. \begin{align*}\cos (a-b)\end{align*}
6. \begin{align*}\tan (a-b)\end{align*}

Simplify the following expressions.

1. \begin{align*}\sin (2\pi-x)\end{align*}
2. \begin{align*}\sin \left(\frac{\pi}{2}+x\right)\end{align*}
3. \begin{align*}\cos (x+\pi)\end{align*}
4. \begin{align*}\cos \left(\frac{3\pi}{2}-x\right)\end{align*}
5. \begin{align*}\tan(x+2\pi)\end{align*}
6. \begin{align*}\tan(x-\pi)\end{align*}
7. \begin{align*}\sin \left(\frac{\pi}{6}-x\right)\end{align*}
8. \begin{align*}\tan \left(\frac{\pi}{4}+x\right)\end{align*}
9. \begin{align*}\cos \left(x-\frac{\pi}{3}\right)\end{align*}

Determine if the following trig statements are true or false.

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

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.13.

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