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# Solving Trigonometric Equations using Sum and Difference Formulas

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

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Practice 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(π2x)=1\begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*} . What is the value of x\begin{align*}x\end{align*} ?

### 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 0x<2π\begin{align*}0\le x <2\pi\end{align*} .

#### Example A

Solve cos(xπ)=22\begin{align*}\cos (x-\pi)=\frac{\sqrt{2}}{2}\end{align*} .

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

cos(xπ)cosxcosπ+sinxsinπcosxcosx=22=22=22=22

The cosine negative in the 2nd\begin{align*}2^{nd}\end{align*} and 3rd\begin{align*}3^{rd}\end{align*} quadrants. x=3π4\begin{align*}x=\frac{3\pi}{4}\end{align*} and 5π4\begin{align*}\frac{5\pi}{4}\end{align*} .

#### Example B

Solve sin(x+π4)+1=sin(π4x)\begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1=\sin \left(\frac{\pi}{4}-x\right)\end{align*} .

Solution:

sin(x+π4)+1sinxcosπ4+cosxsinπ4+1sinx22+cosx22+12sinxsinx=sin(π4x)=sinπ4cosxcosπ4sinx=22.cosx22sinx=1=12=22

In the interval, x=5π4\begin{align*}x=\frac{5\pi}{4}\end{align*} and 7π4\begin{align*}\frac{7\pi}{4}\end{align*} .

#### Example C

Solve 2sin(x+π3)=tanπ3\begin{align*}2\sin \left(x+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\end{align*} .

Solution:

2sin(x+π3)2(sinxcosπ3+cosxsinπ3)2sinx12+2cosx32sinx+3cosxsinxsin2x1cos2x00=tanπ3=3=3=3=3(1cosx)=3(12cosx+cos2x)square both sides=36cosx+3cos2x  substitute sin2x=1cos2x=4cos2x6cosx+2=2cos2x3cosx+1

At this point, we can factor the equation to be (2cosx1)(cosx1)=0\begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*} . cosx=12\begin{align*}\cos x =\frac{1}{2}\end{align*} , and 1, so x=0,π3,5π3\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 5π3\begin{align*}\frac{5\pi}{3}\end{align*} does not work.

2sin(5π3+π3)2sin2π0=tanπ3=33

Therefore, 5π3\begin{align*}\frac{5\pi}{3}\end{align*} is an extraneous solution.

Concept Problem Revisit

In the previous lesson you solved the expression sin(π2x)\begin{align*}\sin(\frac{\pi}{2} - x)\end{align*} as:

sin(π2x)=sinπ2cosxcosπ2sinx=1cosx0sinx=cosx

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

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

### Guided Practice

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

1. cos(2πx)=12\begin{align*}\cos(2\pi - x)=\frac{1}{2}\end{align*}

2. sin(π6x)+1=sin(x+π6)\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)\end{align*}

3. cos(π2+x)=tanπ4\begin{align*}\cos \left(\frac{\pi}{2}+x\right)=\tan \frac{\pi}{4}\end{align*}

Answers

1.

cos(2πx)cos2πcosx+sin2πsinxcosxx=12=12=12=π3 and 5π3

2.

sin(π6x)+1sinπ6cosxcosπ6sinx+112cosx32sinx+1113x=sin(x+π6)=sinxcosπ6+cosxsinπ6=32sinx+12cosx=3sinx=sinx=sin1(13)=0.6155 and 2.5261 rad

3.

cos(π2+x)cosπ2cosxsinπ2sinxsinxsinxx=tanπ4=1=1=1=3π2

### Explore More

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

1. sin(xπ)=22\begin{align*}\sin (x-\pi)=-\frac{\sqrt{2}}{2}\end{align*}
2. cos(2π+x)=1\begin{align*}\cos(2\pi +x)=-1\end{align*}
3. tan(x+π4)=1\begin{align*}\tan \left(x+\frac{\pi}{4}\right)=1\end{align*}
4. sin(π2x)=12\begin{align*}\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}\end{align*}
5. sin(x+3π4)+sin(x3π4)=1\begin{align*}\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1\end{align*}
6. sin(x+π6)=sin(xπ6)\begin{align*}\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)\end{align*}
7. cos(x+π6)=cos(xπ6)+1\begin{align*}\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1\end{align*}
8. cos(x+π3)+cos(xπ3)=1\begin{align*}\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1\end{align*}
9. tan(x+π)+2sin(x+π)=0\begin{align*}\tan(x+\pi)+2\sin (x+\pi)=0\end{align*}
10. tan(x+π)+cos(x+π2)=0\begin{align*}\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0\end{align*}
11. tan(x+π4)=tan(xπ4)\begin{align*}\tan \left(x+\frac{\pi}{4}\right)=\tan \left(x-\frac{\pi}{4}\right)\end{align*}
12. sin(x5π3)sin(x2π3)=0\begin{align*}\sin \left(x-\frac{5\pi}{3}\right)-\sin \left(x-\frac{2\pi}{3}\right)=0\end{align*}
13. 4sin(x+π)2=2cos(x+π2)\begin{align*}4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)\end{align*}
14. 1+2cos(xπ)+cosx=0\begin{align*}1+2\cos(x-\pi)+\cos x =0\end{align*}
15. Real Life Application The height, h\begin{align*}h\end{align*} (in feet), of two people in different seats on a Ferris wheel can be modeled by h1=50cos3t+46\begin{align*}h_1=50\cos 3t+46\end{align*} and h2=50cos3(t3π4)+46\begin{align*}h_2=50\cos 3\left(t-\frac{3\pi}{4}\right)+46\end{align*} where t\begin{align*}t\end{align*} is the time (in minutes). When are the two people at the same height?

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