Using Identities to Find Exact Trig Values Exercise 1

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.
 

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. \[\begin{aligned} \cos{\left(\frac{5\pi}{6}-\frac{\pi}{4}\right)} &= \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{2}}{2}\right) + \left(-\frac{1}{2}\right) \left(-\frac{\sqrt{2}}{2}\right) \\ &= -\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \\ &= \frac{\sqrt{2}-\sqrt{6}}{4} \end{aligned}\]