# 3.3.6: Solving Trigonometric Equations using Sum and Difference Formulas

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

As Agent Trigonometry, you are given a piece of the puzzle: $$\sin \left(\dfrac{\pi}{2}−x\right)=−1$$. What is the value of $$x$$?

### 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 $$0\leq x<2\pi$$.

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

1. $$\cos (x-\pi) =\dfrac{\sqrt{2}}{2}$$

Use the formula to simplify the left-hand side and then solve for x.

\begin{aligned} \cos (x-\pi) &=\dfrac{\sqrt{2}}{2} \\ \cos x \cos \pi+\sin x \sin \pi &=\dfrac{\sqrt{2}}{2} \\ -\cos x &=\dfrac{\sqrt{2}}{2} \\ \cos x &=-\dfrac{\sqrt{2}}{2} \end{aligned}

The cosine negative in the 2nd and 3rd quadrants. $$x=\dfrac{3\pi}{4}$$ and $$\dfrac{5 \pi}{4}$$.

1. $$\sin \left(x+\dfrac{\pi}{4}\right)+1 =\sin \left(\dfrac{\pi}{4}-x\right)$$

\begin{aligned} \sin \left(x+\dfrac{\pi}{4}\right)+1 &=\sin \left(\dfrac{\pi}{4}-x\right) \\ \sin x \cos \dfrac{\pi}{4}+\cos x \sin \dfrac{\pi}{4}+1 &=\sin \dfrac{\pi}{4} \cos x-\cos \dfrac{\pi}{4} \sin x \\ \sin x \cdot \dfrac{\sqrt{2}}{2}+\cos x \cdot \dfrac{\sqrt{2}}{2}+1 &=\dfrac{\sqrt{2}}{2} \cdot \cos x-\dfrac{\sqrt{2}}{2} \cdot \sin x \\ \sqrt{2} \sin x &=-1 \\ \sin x &=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2} \end{aligned}

In the interval, $$x=\dfrac{5 \pi}{4}$$ and $$\dfrac{7\pi}{4}$$.

1. $$2\sin \left(x+\dfrac{\pi}{3}\right)=\tan \dfrac{\pi}{3}$$

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

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

\begin{aligned} 2\sin \left (\dfrac{5 \pi}{3}+\dfrac{\pi}{3}\right)&=\tan \dfrac{\pi}{3} \\ 2\sin 2\pi &=\sqrt{3} \\ 0 &\neq \sqrt{3} \end{aligned}

Therefore, $$\dfrac{5 \pi}{3}$$ is an extraneous solution.

##### Example $$\PageIndex{1}$$

Earlier, you were asked to find the value of x from the equation $$\sin \left(\dfrac{\pi}{2}−x\right)=−1$$.

Solution

First, simplify the expression $$\sin \left(\dfrac{\pi}{2}−x\right)$$ as:

\begin{aligned} \sin \left(\dfrac{\pi}{2}−x\right)&=\sin \dfrac{\pi}{2}\cos x−\cos \dfrac{\pi}{2}\sin x \\ &=1\cdot \cos x−0\cdot \sin x \\&=\cos x \end{aligned}

So what you're now looking for is the value of $$x$$ where $$\cos x=−1$$.

The cosine of $$180^{\circ}$$ is equal to −1.

Solve the following equations in the interval $$0\leq x<2\pi$$.

##### Example $$\PageIndex{2}$$

$$\cos (2\pi −x)=\dfrac{1}{2}$$

Solution

\begin{aligned} \cos (2 \pi-x) &=\dfrac{1}{2} \\ \cos 2 \pi \cos x+\sin 2 \pi \sin x &=\dfrac{1}{2} \\ \cos x &=\dfrac{1}{2} \\ x &=\dfrac{\pi}{3} \text { and } \dfrac{5 \pi}{3} \end{aligned}

##### Example $$\PageIndex{3}$$

$$\sin \left(\dfrac{\pi}{6}−x\right)+1=\sin \left(x+\dfrac{\pi}{6}\right)$$

Solution

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

##### Example $$\PageIndex{4}$$

$$\cos \left(\dfrac{\pi}{2}+x\right)=\tan \dfrac{\pi}{4}$$

Solution

\begin{aligned} \cos \left(\dfrac{\pi}{2}+x \right)&=\tan \dfrac{\pi}{4} \\ \cos \dfrac{\pi}{2}\cos x−\sin \dfrac{\pi}{2}\sin x&=1 \\ −\sin x&=1 \\ \sin x &=−1\\ x&=\dfrac{3 \pi}{2} \end{aligned}

## Review

Solve the following trig equations in the interval $$0\leq x<2\pi$$.

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