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3.3.7: Finding Exact Trigonometric Values Using Sum and Difference Formulas

  • Page ID
    14851
  • Convert angles to sum or difference of 30, 45, and 60 degrees to solve.

    You measure an angle with your protractor to be \(165^{\circ} \). How could you find the exact sine of this angle without Using a calculator?

    Sum and Difference Formulas

    You know that \(\sin 30^{\circ} =\dfrac{1}{2}\), \(\cos 135^{\circ} =−\dfrac{\sqrt{2} }{2}\), \(\tan 300^{\circ} =−\sqrt{3}\), etc... from the special right triangles. In this concept, we will learn how to find the exact values of the trig functions for angles other than these multiples of 30^{\circ} ,45^{\circ} , and 60^{\circ} . Using the Sum and Difference Formulas, we can find these exact trig values.

    Sum and Difference Formulas

    \(\begin{aligned} \sin (a\pm b)&=\sin a\cos b\pm \cos a\sin b \\ \cos (a\pm b)&=\cos a\cos b\mp \sin a\sin b \\ \tan (a\pm b)&=\dfrac{\tan a\pm \tan b}{1\mp \tan a\tan b}\end{aligned}\)

    Let's find the following exact values Using the Sum and Difference Formulas.

    1. \(\sin 75^{\circ}\)

    This is an example of where we can use the sine sum formula from above, \(\sin (a+b)=\sin a\cos b+\cos a\sin b\), where \(a=45^{\circ} \) and \(b=30^{\circ} \).

    \(\begin{aligned} \sin 75^{\circ} &=\sin (45^{\circ} +30^{\circ} ) \\&=\sin 45^{\circ} \cos 30^{\circ} +\cos 45^{\circ} \sin 30^{\circ} \\ &=\dfrac{\sqrt{2} }{2}\cdot \dfrac{\sqrt{3} }{2}+\dfrac{\sqrt{2} }{2}\cdot \dfrac{1}{2} \\ &=\dfrac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}\)

    In general, \(\sin (a+b)\neq \sin a+\sin b\) and similar statements can be made for the other sum and difference formulas.

    1. \(\cos \dfrac{11 \pi}{12}\)

    For this problem, we could use either the sum or difference cosine formula, \(\dfrac{11 \pi}{12}=\dfrac{2\pi}{3}+\dfrac{\pi}{4}\) or \(\dfrac{11 \pi}{12}=\dfrac{7\pi}{6}−\dfrac{\pi}{4}\). Let’s use the sum formula.

    \(\begin{aligned} \cos \dfrac{11 \pi}{12}&=\cos (\dfrac{2\pi}{3}+\dfrac{\pi}{4}) \\ &=\cos \dfrac{2\pi}{3}\cos \dfrac{\pi}{4}−\sin \dfrac{2\pi}{3}\sin \dfrac{\pi}{4} \\&=−\dfrac{1}{2}\cdot \dfrac{\sqrt{2} }{2}−\dfrac{\sqrt{3} }{2}\cdot \dfrac{\sqrt{2} }{2} \\&=−\dfrac{\sqrt{2} +\sqrt{6}}{4} \end{aligned}\)

    1. \(\tan \left(−\dfrac{\pi }{12}\right)\)

    This angle is the difference between \(\dfrac{\pi}{4}\) and \(\dfrac{\pi}{3}\).

    \(\begin{aligned} \tan (\dfrac{\pi}{4}−\dfrac{\pi}{3})&=\dfrac{\tan \dfrac{\pi}{4}−\tan \dfrac{\pi}{3}}{1+\tan \dfrac{\pi}{4}\tan \dfrac{\pi}{3}} \\ &=\dfrac{1−\sqrt{3}}{1+\sqrt{3}}\end{aligned}\)

    This angle is also the same as \(\dfrac{23\pi}{12}\). You could have also used this value and done \(\tan \left(\dfrac{\pi}{4}+\dfrac{5 \pi}{3}\right)\) and arrived at the same answer.

    Example \(\PageIndex{1}\)

    Earlier, you were asked to find the exact value of \(\sin 165^{\circ} \) without Using the calculator. 

    Solution

    We can use the sine sum formula, \(\sin (a+b)=\sin a\cos b+\cos a\sin b\), where \(a=120^{\circ} \) and \(b=45^{\circ}\).

    \(\begin{aligned}
    \sin 165^{\circ} &=\sin \left(120^{\circ}+45^{\circ}\right) \\
    &=\sin 120^{\circ} \cos 45^{\circ}+\cos 120^{\circ} \sin 45^{\circ} \\
    &=\dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2}+\dfrac{-1}{2} \cdot \dfrac{\sqrt{2}}{2} \\
    &=\dfrac{\sqrt{6}-\sqrt{2}}{4}
    \end{aligned}\)

     

    Example \(\PageIndex{2}\)

    Find the exact value of \(\cos 15^{\circ} \). 

    Solution

    \(\begin{aligned}
    \cos 15^{\circ} &=\cos \left(45^{\circ}-30^{\circ}\right) \\
    &=\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ} \\
    &=\dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} \\
    &=\dfrac{\sqrt{6}+\sqrt{2}}{4}
    \end{aligned}\)

    Example \(\PageIndex{3}\)

    Find the exact value of \(\tan 255^{\circ} \).

    Solution

    \(\begin{aligned}
    \tan \left(210^{\circ}+45^{\circ}\right) &=\dfrac{\tan 210^{\circ}+\tan 45^{\circ}}{1-\tan 210^{\circ} \tan 45^{\circ}} \\
    &=\dfrac{\dfrac{\sqrt{3}}{3}+1}{1-\dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{\sqrt{3}+3}{3}}{\dfrac{3-\sqrt{3}}{3}}=\dfrac{\sqrt{3}+3}{3-\sqrt{3}}
    \end{aligned}\)

    Review

    Find the exact value of the following trig functions.

    1. \(\sin 15^{\circ}\)
    2. \(\cos \dfrac{5 \pi}{12}\)
    3. \(\tan 345^{\circ}\)
    4. \(\cos (−255^{\circ} )\)
    5. \(\sin \dfrac{13 \pi}{12}\)
    6. \(\sin \dfrac{17\pi}{12}\)
    7. \(\cos 15^{\circ}\)
    8. \(\tan (−15^{\circ} )\)
    9. \(\sin 345^{\circ}\)
    10. Now, use \(\sin 15^{\circ} \) from #1, and find \(\sin 345^{\circ} \). Do you arrive at the same answer? Why or why not?
    11. Using \(\cos 15^{\circ} \) from #7, find \(\cos 165^{\circ} \). What is another way you could find \(\cos 165^{\circ} \)?
    12. Describe any patterns you see between the sine, cosine, and tangent of these “new” angles.
    13. Using your calculator, find the \(\sin 142^{\circ} \). Now, use the sum formula and your calculator to find the \(\sin 142^{\circ}\)  Using \(83^{\circ}\)  and \(59^{\circ}\).
    14. Use the sine difference formula to find \(\sin 142^{\circ} \) with any two angles you choose. Do you arrive at the same answer? Why or why not?
    15. Challenge Using \(\sin (a+b)=\sin a\cos b+\cos a\sin b\) and \(\cos (a+b)=\cos a\cos b−\sin a\sin b\), show that \(\tan (a+b)=\dfrac{\tan a+\tan b}{1−\tan a\tan b}\).

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 14.12. 

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