Question

Find the value of:
cos 2π/3 × cosπ/4 - sin 2π/3 × sinπ/4

Answer 1

Answer:

$\cos \left(\dfrac{2\pi }{3}\right)\cos \left(\dfrac{\pi }{4}\right)-\sin \left(\dfrac{2\pi }{3}\right)\sin \left(\dfrac{\pi }{4}\right)=\dfrac{-\sqrt{2}-\sqrt{6}}{4}\quad \begin{pmatrix}\mathrm{Decimal:}&-0.96593\dots \end{pmatrix}$

Step-by-step explanation:

$\cos \left(\dfrac{2\pi }{3}\right)\cos \left(\dfrac{\pi }{4}\right)-\sin \left(\dfrac{2\pi }{3}\right)\sin \left(\dfrac{\pi }{4}\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \sin \left(x\right)=\cos \left(\dfrac{\pi }{2}-x\right)$

$\sin \left(\dfrac{2\pi }{3}\right)=\cos \left(\dfrac{\pi }{2}-\dfrac{2\pi }{3}\right)$

$=\cos \left(\dfrac{2\pi }{3}\right)\cos \left(\dfrac{\pi }{4}\right)-\cos \left(\dfrac{\pi }{2}-\dfrac{2\pi }{3}\right)\sin \left(\dfrac{\pi }{4}\right)$

$\mathrm{Join}\:\dfrac{\pi }{2}-\dfrac{2\pi }{3}:\quad -\dfrac{\pi }{6}$

$=\cos \left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{2\pi }{3}\right)-\sin \left(\dfrac{\pi }{4}\right)\cos \left(-\dfrac{\pi }{6}\right)$

$\mathrm{Use\:the\:following\:property:}\:\cos \left(-x\right)=\cos \left(x\right)$

$\cos \left(-\dfrac{\pi }{6}\right)=\cos \left(\dfrac{\pi }{6}\right)$

$=\cos \left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{2\pi }{3}\right)-\sin \left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{\pi }{6}\right)$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(\dfrac{\pi }{4}\right)=\dfrac{\sqrt{2}}{2}$

$\cos \left(\dfrac{2\pi }{3}\right)$

$\mathrm{Write}\:\cos \left(\dfrac{2\pi }{3}\right)\:\mathrm{as}\:\cos \left(\dfrac{\pi }{6}+\dfrac{\pi }{2}\right)$

$=\cos \left(\dfrac{\pi }{6}+\dfrac{\pi }{2}\right)$

$\mathrm{Using\:the\:summation\:identity}:\quad \cos \left(x+y\right)=\cos \left(x\right)\cos \left(y\right)-\sin \left(x\right)\sin \left(y\right)$

$=\cos \left(\dfrac{\pi }{6}\right)\cos \left(\dfrac{\pi }{2}\right)-\sin \left(\dfrac{\pi }{6}\right)\sin \left(\dfrac{\pi }{2}\right)$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(\dfrac{\pi }{6}\right)=\dfrac{\sqrt{3}}{2}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(\dfrac{\pi }{6}\right)=\dfrac{1}{2}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(\dfrac{\pi }{2}\right)=0$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(\dfrac{\pi }{2}\right)=1$

$=\dfrac{\sqrt{3}}{2}\cdot \:0-\dfrac{1}{2}\cdot \:1$

$\mathrm{Simplify}$

$=-\dfrac{1}{2}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(\dfrac{\pi }{4}\right)=\dfrac{\sqrt{2}}{2}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(\dfrac{\pi }{6}\right)=\dfrac{\sqrt{3}}{2}$

$=\dfrac{\sqrt{2}}{2}\left(-\dfrac{1}{2}\right)-\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}$

$\dfrac{\sqrt{2}}{2}\left(-\dfrac{1}{2}\right)-\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=-\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}-\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}$

$\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt{2}}{4}$

$\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{6}}{4}$

$=-\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{6}}{4}$

$\mathrm{Apply\:rule}\:\dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{-\sqrt{2}-\sqrt{6}}{4}$