Chapter 14.5, Problem 71SR

To determine

To find :Find exact value of $\cos \frac{7\pi }{12}$

Answer to Problem 71SR

  $0.2588$

Explanation of Solution

Given information :

Expression $\text{cos}\frac{7\pi }{12}$

Calculation :

The given equation is

  $\cos \frac{7\pi }{12}$

We can write, $\cos \left(\frac{3\pi }{12}+\frac{4\pi }{12}\right)$

  $\begin{array}{l}=\cos \left(\frac{\pi }{4}+\frac{\pi }{3}\right)\text{(Using }\cos (A+B)=\cos A\cos B-\sin A\sin B\text{)}\\ =\cos \frac{\pi }{4}\cdot \cos \frac{\pi }{3}-\sin \frac{\pi }{4}\cdot \sin \frac{\pi }{3}\\ =\frac{1}{\sqrt{2}}\cdot \frac{1}{12}-\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}=\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}\\ =\frac{1-\sqrt{3}}{2\sqrt{2}}\left[\begin{array}{r}\because \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}},\cos \frac{\pi }{3}=\frac{1}{2}\\ \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}},\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\end{array}\right]\\ =\frac{-0.732}{2.828}=0.2588\end{array}$

  $\begin{array}{l}\text{The given equation is }\\ \cos \frac{7\pi }{12}\\ \text{We can write,}\cos \left(\frac{3\pi }{12}+\frac{4\pi }{12}\right)\\ =\cos \left(\frac{\pi }{4}+\frac{\pi }{3}\right)\text{(Using }\cos (A+B)=\cos A\cos B-\sin A\sin B\text{)}\\ =\cos \frac{\pi }{4}\cdot \cos \frac{\pi }{3}-\sin \frac{\pi }{4}\cdot \sin \frac{\pi }{3}\\ =\frac{1}{\sqrt{2}}\cdot \frac{1}{12}-\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}=\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}\\ =\frac{1-\sqrt{3}}{2\sqrt{2}}\left[\begin{array}{r}\because \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}},\cos \frac{\pi }{3}=\frac{1}{2}\\ \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}},\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\end{array}\right]\\ =\frac{-0.732}{2.828}=0.2588\end{array}$