Skip to main content
K12 LibreTexts

3.3.6: Solving Trigonometric Equations using Sum and Difference Formulas

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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

    As Agent Trigonometry, you are given a piece of the puzzle: \(\sin \left(\dfrac{\pi}{2}−x\right)=−1\). What is the value of \(x\)?

    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 \(0\leq x<2\pi\).

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

    1. \(\cos (x-\pi) =\dfrac{\sqrt{2}}{2}\)

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

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

    The cosine negative in the 2nd and 3rd quadrants. \(x=\dfrac{3\pi}{4}\) and \(\dfrac{5 \pi}{4}\).

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

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

    In the interval, \(x=\dfrac{5 \pi}{4}\) and \(\dfrac{7\pi}{4}\).

    1. \(2\sin \left(x+\dfrac{\pi}{3}\right)=\tan \dfrac{\pi}{3}\)

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

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

    \(\begin{aligned} 2\sin \left (\dfrac{5 \pi}{3}+\dfrac{\pi}{3}\right)&=\tan \dfrac{\pi}{3} \\ 2\sin 2\pi &=\sqrt{3} \\ 0 &\neq \sqrt{3} \end{aligned}\)

    Therefore, \(\dfrac{5 \pi}{3}\) is an extraneous solution.

    Example \(\PageIndex{1}\)

    Earlier, you were asked to find the value of x from the equation \(\sin \left(\dfrac{\pi}{2}−x\right)=−1\).


    First, simplify the expression \(\sin \left(\dfrac{\pi}{2}−x\right)\) as:

    \(\begin{aligned} \sin \left(\dfrac{\pi}{2}−x\right)&=\sin \dfrac{\pi}{2}\cos x−\cos \dfrac{\pi}{2}\sin x \\ &=1\cdot \cos x−0\cdot \sin x \\&=\cos x \end{aligned}\)

    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.

    Solve the following equations in the interval \(0\leq x<2\pi\).

    Example \(\PageIndex{2}\)

    \(\cos (2\pi −x)=\dfrac{1}{2}\)


    \cos (2 \pi-x) &=\dfrac{1}{2} \\
    \cos 2 \pi \cos x+\sin 2 \pi \sin x &=\dfrac{1}{2} \\
    \cos x &=\dfrac{1}{2} \\
    x &=\dfrac{\pi}{3} \text { and } \dfrac{5 \pi}{3}

    Example \(\PageIndex{3}\)

    \(\sin \left(\dfrac{\pi}{6}−x\right)+1=\sin \left(x+\dfrac{\pi}{6}\right)\)


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

    Example \(\PageIndex{4}\)

    \(\cos \left(\dfrac{\pi}{2}+x\right)=\tan \dfrac{\pi}{4}\)


    \(\begin{aligned} \cos \left(\dfrac{\pi}{2}+x \right)&=\tan \dfrac{\pi}{4} \\ \cos \dfrac{\pi}{2}\cos x−\sin \dfrac{\pi}{2}\sin x&=1 \\ −\sin x&=1 \\ \sin x &=−1\\ x&=\dfrac{3 \pi}{2} \end{aligned}\)


    Solve the following trig equations in the interval \(0\leq x<2\pi\).

    1. \(\sin \left (x−\pi \right)=−\dfrac{\sqrt{2}}{2}\)
    2. \(\cos \left(2\pi +x\right)=−1\)
    3. \(\tan \left (x+\dfrac{\pi}{4}\right)=1\)
    4. \(\sin \left (\dfrac{\pi}{2}−x\right)=\dfrac{1}{2}\)
    5. \(\sin \left (x+3\dfrac{\pi}{4}\right)+\sin \left (x−\dfrac{3 \pi}{4}\right)=1\)
    6. \(\sin \left (x+\dfrac{\pi}{6}\right)=−\sin \left (x−\dfrac{\pi}{6}\right)\)
    7. \(\cos \left(x+\dfrac{\pi}{6}\right)=\cos \left(x−\dfrac{\pi}{6}\right)+1\)
    8. \(\cos \left(x+\dfrac{\pi}{3}\right)+\cos \left(x−\dfrac{\pi}{3}\right)=1\)
    9. \(\tan \left (x+\pi \right)+2\sin \left (x+\pi \right)=0\)
    10. \(\tan \left (x+\pi \right)+\cos \left(x+\dfrac{\pi}{2}\right)=0\)
    11. \(\tan \left (x+\dfrac{\pi}{4}\right)=\tan \left (x−\dfrac{\pi}{4}\right)\)
    12. \(\sin \left (x−\dfrac{5 \pi}{3}\right)−\sin \left (x−\dfrac{2 \pi}{3}\right)=0\)
    13. \(4\sin \left (x+\pi \right)−2=2\cos \left(x+\dfrac{\pi}{2}\right)\)
    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 \(h_2=50\cos 3\left(t−\dfrac{3 \pi}{4}\right)+46\) where \(t\) 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.

    This page titled 3.3.6: Solving Trigonometric Equations using Sum and Difference Formulas is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    CK-12 Foundation
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License