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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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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*}sin(π2x)=1. What is the value of \begin{align*}x\end{align*}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 \begin{align*}0\le x <2\pi\end{align*}0x<2π.

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

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

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

\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*}cos(xπ)cosxcosπ+sinxsinπcosxcosx=22=22=22=22

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

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

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

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

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

\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*}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 \begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*}(2cosx1)(cosx1)=0. \begin{align*}\cos x =\frac{1}{2}\end{align*}cosx=12, and 1, so \begin{align*}x=0, \frac{\pi}{3}, \frac{5 \pi}{3}\end{align*}x=0,π3,5π3. Be careful with these answers. When we check these solutions it turns out that \begin{align*}\frac{5\pi}{3}\end{align*}5π3 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?

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.14. 

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