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Solving Trigonometric Equations using Sum and Difference Formulas

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Solving Trig Equations using Sum and Difference Formulas
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As Agent Trigonometry, you are now given another piece of the puzzle: \sin(\frac{\pi}{2} - x) = -1 . What is the value of x ?


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) .


\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} .


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?

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