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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 now given another piece of the puzzle: \begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}. What is the value of \begin{align*}x\end{align*}?

Solving Trigonometric Functions

Lastly, 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*}.

Solve the following functions using the sum and difference formulas

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

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

\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*}

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

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

\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*}

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

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

\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*}

At this point, we can factor the equation to be \begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*}. \begin{align*}\cos x =\frac{1}{2}\end{align*}, and 1, so \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 \begin{align*}\frac{5\pi}{3}\end{align*} 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*}

In the previous lesson you solved 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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