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# Solving Trigonometric Equations using Sum and Difference Formulas

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Solving Trig Equations using Sum and Difference Formulas

As Agent Trigonometry, you are now given another piece of the puzzle: $\sin(\frac{\pi}{2} - x) = -1$ . What is the value of $x$ ?

### Guidance

Lastly, we can use the sum and difference formulas to solve trigonometric equations. For this concept, we will only find solutions in the interval $0\le x <2\pi$ .

#### Example A

Solve $\cos (x-\pi)=\frac{\sqrt{2}}{2}$ .

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

$\cos (x-\pi) &=\frac{\sqrt{2}}{2} \\\cos x \cos \pi +\sin x \sin \pi &=\frac{\sqrt{2}}{2} \\-\cos x &=\frac{\sqrt{2}}{2}\\\cos x &=-\frac{\sqrt{2}}{2}$

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

#### Example B

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

Solution:

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

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

#### Example C

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

Solution:

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

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

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

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

Concept Problem Revisit

In the previous lesson you solved the expression $\sin(\frac{\pi}{2} - x)$ as:

$\sin(\frac{\pi}{2} - x)=\sin \frac{\pi}{2} \cos x - \cos \frac{\pi}{2} \sin x \\&=1\cdot \cos x - 0\cdot \sin x \\&=cos x$

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

The cosine of $180^\circ$ is equal to $-1$ .

### Guided Practice

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

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

2. $\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)$

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

1. $\cos (2\pi-x) &=\frac{1}{2} \\\cos 2\pi \cos x +\sin 2 \pi \sin x &=\frac{1}{2} \\\cos x &=\frac{1}{2} \\x &=\frac{\pi}{3} \ and \ \frac{5\pi}{3}$

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

3. $\cos \left(\frac{\pi}{2}+x\right) &=\tan \frac{\pi}{4} \\\cos \frac{\pi}{2} \cos x - \sin \frac{\pi}{2} \sin x &=1 \\-\sin x &=1 \\\sin x &=-1 \\x &=\frac{3\pi}{2}$

### Explore More

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

1. $\sin (x-\pi)=-\frac{\sqrt{2}}{2}$
2. $\cos(2\pi +x)=-1$
3. $\tan \left(x+\frac{\pi}{4}\right)=1$
4. $\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}$
5. $\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1$
6. $\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)$
7. $\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1$
8. $\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1$
9. $\tan(x+\pi)+2\sin (x+\pi)=0$
10. $\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0$
11. $\tan \left(x+\frac{\pi}{4}\right)=\tan \left(x-\frac{\pi}{4}\right)$
12. $\sin \left(x-\frac{5\pi}{3}\right)-\sin \left(x-\frac{2\pi}{3}\right)=0$
13. $4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)$
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 $h_2=50\cos 3\left(t-\frac{3\pi}{4}\right)+46$ where $t$ is the time (in minutes). When are the two people at the same height?