## Precalculus (6th Edition) Blitzer

The exact value of the expression is $-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right)$.
Let us consider the following equation. $x=\sin \frac{11\pi }{4}\cos \frac{5\pi }{6}+\cos \frac{11\pi }{4}\sin \frac{5\pi }{6}$ …… (1) So, in quadrant I, all the trigonometric functions are positive. In quadrant II, sine and cosecant are positive and all other trigonometric functions are negative. In quadrant III, tangent and cotangent are positive and all other trigonometric functions are negative. In quadrant IV, cosine and secant are positive and all other trigonometric functions are negative. The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi$, in quadrant II. Thus, calculate the reference angle as follows \begin{align} & {\theta }'=3\pi -\frac{11\pi }{4} \\ & =\frac{12\pi -11\pi }{4} \\ & =\frac{\pi }{4} \end{align} Hence, $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Here, the sine is positive in quadrant II. So, $\sin \frac{11\pi }{4}=\sin \frac{\pi }{4}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{\pi }{4}$. $\sin \frac{11\pi }{4}=\frac{\sqrt{3}}{2}$ The angle $\frac{5\pi }{6}$ lies between $\frac{\pi }{2}$ and $\pi$ , in quadrant II. And, calculate the reference angle as follows \begin{align} & {\theta }'=\pi -\frac{5\pi }{6} \\ & =\frac{6\pi -5\pi }{6} \\ & =\frac{\pi }{6} \end{align} So, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ Here, the cosine is positive in quadrant II. Therefore, $\cos \frac{5\pi }{6}=\cos \frac{\pi }{6}$ Put $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$. $\cos \frac{5\pi }{6}=\frac{\sqrt{3}}{2}$ The angle $\frac{11\pi }{4}$ lies between $\frac{5\pi }{2}$ and $3\pi$ , in quadrant II. Also, calculate the reference angle as follows \begin{align} & {\theta }'=3\pi -\frac{11\pi }{4} \\ & =\frac{12\pi -11\pi }{4} \\ & =\frac{\pi }{4} \end{align} So, $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Here, the cosine is positive in quadrant II. Therefore, $\cos \frac{11\pi }{4}=\cos \frac{\pi }{4}$ Put $\frac{\sqrt{2}}{2}$ for $\cos \frac{\pi }{4}$. $\cos \frac{11\pi }{4}=\frac{\sqrt{2}}{2}$ The angle $\frac{5\pi }{6}$ lies between $\frac{\pi }{2}$ and $\pi$, in quadrant II. And, calculate the reference angle as follows \begin{align} & {\theta }'=\pi -\frac{5\pi }{6} \\ & =\frac{6\pi -5\pi }{6} \\ & =\frac{\pi }{6} \end{align} Hence, $\sin \frac{\pi }{6}=\frac{1}{2}$ Here, the sine is positive in quadrant II. Therefore, $\sin \frac{5\pi }{6}=\sin \frac{\pi }{6}$ Put $\frac{1}{2}$ for $\sin \frac{\pi }{6}$. $\sin \frac{5\pi }{6}=\frac{1}{2}$ Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{11\pi }{4}$ , $-\frac{\sqrt{3}}{2}$ for $\cos \frac{5\pi }{6}$ , $\frac{1}{2}$ for $\sin \frac{5\pi }{6}$ and $\frac{\sqrt{2}}{2}$ for $\cos \frac{11\pi }{4}$ in equation (1). \begin{align} & x=\frac{\sqrt{2}}{2}\left( -\frac{\sqrt{3}}{2} \right)+\left( -\frac{\sqrt{2}}{2} \right)\left( \frac{1}{2} \right) \\ & =-\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ & =-\left( \frac{\sqrt{6}+\sqrt{2}}{4} \right) \end{align}