# Solving Trigonometric Equations using Sum and Difference Formulas

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

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Solving Trig Equations using Sum and Difference Formulas

As Agent Trigonometry, you are given a piece of the puzzle: \begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}. What is the value of \begin{align*}x\end{align*}?

### Solving Trigonometric Functions

We can use the sum and difference formulas to solve trigonometric equations. For this concept, we will only find solutions in the interval \begin{align*}0\le x <2\pi\end{align*}.

Let's solve the following functions using the sum and difference formulas.

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

Use the formula to simplify the left-hand side and then solve for \begin{align*}x\end{align*}.

\begin{align*}\cos (x-\pi) &=\frac{\sqrt{2}}{2} \\ \cos x \cos \pi +\sin x \sin \pi &=\frac{\sqrt{2}}{2} \\ -\cos x &=\frac{\sqrt{2}}{2}\\ \cos x &=-\frac{\sqrt{2}}{2}\end{align*}

The cosine negative in the \begin{align*}2^{nd}\end{align*} and \begin{align*}3^{rd}\end{align*} quadrants. \begin{align*}x=\frac{3\pi}{4}\end{align*} and \begin{align*}\frac{5\pi}{4}\end{align*}.

1. \begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1=\sin \left(\frac{\pi}{4}-x\right)\end{align*}

\begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1 &=\sin \left(\frac{\pi}{4}-x\right) \\ \sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}+1 &=\sin \frac{\pi}{4}\cos x -\cos \frac{\pi}{4}\sin x \\ \sin x \cdot \frac{\sqrt{2}}{2}+\cos x \cdot \frac{\sqrt{2}}{2}+1 &=\frac{\sqrt{2}}{2}.\cos x -\frac{\sqrt{2}}{2} \cdot \sin x \\ \sqrt{2} \sin x &=-1 \\ \sin x &=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{align*}

In the interval, \begin{align*}x=\frac{5\pi}{4}\end{align*} and \begin{align*}\frac{7\pi}{4}\end{align*}.

1. \begin{align*}2\sin \left(x+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\end{align*}

\begin{align*}2\sin\left(x+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right) &=\sqrt{3} \\ 2\sin x \cdot \frac{1}{2}+2\cos x \cdot \frac{\sqrt{3}}{2} &=\sqrt{3} \\ \sin x +\sqrt{3}\cos x &=\sqrt {3} \\ \sin x &=\sqrt{3}(1-\cos x) \\ \sin ^2x &=3(1-2\cos x+\cos ^2x) \qquad \text{square both sides} \\ 1-\cos ^2x &=3-6\cos x + 3\cos ^2 x \qquad \ \ \text{substitute} \ \sin ^2{x} = 1-\cos ^2x \\ 0 &=4\cos^2x-6\cos x +2 \\ 0 &=2\cos^2x-3\cos x +1\end{align*}

At this point, we can factor the equation to be \begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*}. \begin{align*}\cos x =\frac{1}{2}\end{align*}, and 1, so \begin{align*}x=0, \frac{\pi}{3}, \frac{5 \pi}{3}\end{align*}. Be careful with these answers. When we check these solutions it turns out that \begin{align*}\frac{5\pi}{3}\end{align*} does not work.

\begin{align*}2\sin\left(\frac{5\pi}{3}+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\sin 2\pi &=\sqrt{3} \\ 0 &\ne \sqrt{3}\end{align*}

Therefore, \begin{align*}\frac{5\pi}{3}\end{align*} is an extraneous solution.

### Examples

#### Example 1

Earlier, you were asked to find the value of x from the equation \begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}

First, simplify the expression \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*} as:

\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*}

So what you're now looking for is the value of \begin{align*}x\end{align*} where \begin{align*}\cos x = -1\end{align*}.

The cosine of \begin{align*}180^\circ\end{align*} is equal to \begin{align*}-1\end{align*}.

Solve the following equations in the interval \begin{align*}0\le x<2\pi\end{align*}.

#### Example 2

\begin{align*}\cos(2\pi - x)=\frac{1}{2}\end{align*}
\begin{align*}\cos (2\pi-x) &=\frac{1}{2} \\ \cos 2\pi \cos x +\sin 2 \pi \sin x &=\frac{1}{2} \\ \cos x &=\frac{1}{2} \\ x &=\frac{\pi}{3} \ and \ \frac{5\pi}{3}\end{align*}

#### Example 3

\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)\end{align*}

\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 &=\sin \left(x+\frac{\pi}{6}\right)\\ \sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x +1 &=\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6} \\ \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x+1 &=\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x \\ 1 &=\sqrt{3}\sin x \\ \frac{1}{\sqrt{3}} &=\sin x \\ x &=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=0.6155 \ and \ 2.5261 \ \text{rad}\end{align*}

#### Example 4

\begin{align*}\cos \left(\frac{\pi}{2}+x\right)=\tan \frac{\pi}{4}\end{align*}
\begin{align*}\cos \left(\frac{\pi}{2}+x\right) &=\tan \frac{\pi}{4} \\ \cos \frac{\pi}{2} \cos x - \sin \frac{\pi}{2} \sin x &=1 \\ -\sin x &=1 \\ \sin x &=-1 \\ x &=\frac{3\pi}{2}\end{align*}

### Review

Solve the following trig equations in the interval \begin{align*}0\le x < 2\pi\end{align*}.

1. \begin{align*}\sin (x-\pi)=-\frac{\sqrt{2}}{2}\end{align*}
2. \begin{align*}\cos(2\pi +x)=-1\end{align*}
3. \begin{align*}\tan \left(x+\frac{\pi}{4}\right)=1\end{align*}
4. \begin{align*}\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}\end{align*}
5. \begin{align*}\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1\end{align*}
6. \begin{align*}\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)\end{align*}
7. \begin{align*}\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1\end{align*}
8. \begin{align*}\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1\end{align*}
9. \begin{align*}\tan(x+\pi)+2\sin (x+\pi)=0\end{align*}
10. \begin{align*}\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0\end{align*}
11. \begin{align*}\tan \left(x+\frac{\pi}{4}\right)=\tan \left(x-\frac{\pi}{4}\right)\end{align*}
12. \begin{align*}\sin \left(x-\frac{5\pi}{3}\right)-\sin \left(x-\frac{2\pi}{3}\right)=0\end{align*}
13. \begin{align*}4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)\end{align*}
14. \begin{align*}1+2\cos(x-\pi)+\cos x =0\end{align*}
15. Real Life Application The height, \begin{align*}h\end{align*} (in feet), of two people in different seats on a Ferris wheel can be modeled by \begin{align*}h_1=50\cos 3t+46\end{align*} and \begin{align*}h_2=50\cos 3\left(t-\frac{3\pi}{4}\right)+46\end{align*} where \begin{align*}t\end{align*} is the time (in minutes). When are the two people at the same height?