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tags:
Cosine Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Using Identities to Find Exact Trig Values Exercise 1

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Use Sum and Difference Identities to find the exact value of
\[\cos\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)\]

Solution: \(\dfrac{\sqrt{2}-\sqrt{6}}{4}\) 

Solution Method: We use sum and difference identities to find the value of a non-special angle defined by the sum or difference of two special angles. These two angles are special angles, but if I try to subtract them, the lowest common denominator would be a 12, so the difference won’t be a special angle so I can’t use my knowledge of trig values on the unit circle to calculate it. So that’s why we use a difference identity.
 

Sum and Difference Identities
\(\sin{(A+B)}= \sin{A}\cos{B}+\cos{A}\sin{B}\)
 \(\sin{(A-B)} = \sin{A}\cos{B} - \cos{A}\sin{B} \)
 \(\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}\)
 \(\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\)
 \(\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}\)
 \(\tan{(A-B)}=\frac{\tan{A}-\tan{B}}{1+\tan{A}\tan{B}}\)

We clearly have a difference of angles and a cosine function so we’ll use the identity \[\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}\] So we substitute \(A=\frac{5\pi}{6}\) and \(B=\frac{\pi}{4}\). \[\begin{aligned} \cos{\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)}=\cos{\frac{5\pi}{6}}\cos{\frac{\pi}{4}}+\sin{\frac{5\pi}{6}}\sin{\frac{\pi}{4}} \end{aligned}\] And now we only have sine and cosine functions of special angles, which I know from the unit circle. \[\begin{aligned} \cos{\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)} &= \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right) + \left(-\frac{1}{2}\right) \left(-\frac{\sqrt{2}}{2}\right) \\ &= -\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{2}-\sqrt{6}}{4} \end{aligned}\]

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