Question

Prove that cos 72°. cos18°- sin 72°.sin 18°=0​

Answer 1

Answer:

$\mathrm{\cos \left(72^{\circ \:}\right)\cos \left(18^{\circ \:}\right)-\sin \left(72^{\circ \:}\right)\sin \left(18^{\circ \:}\right)=0 \ True}$

Step-by-step explanation:

$\cos \left(72^{\circ \:}\right)\cos \left(18^{\circ \:}\right)-\sin \left(72^{\circ \:}\right)\sin \left(18^{\circ \:}\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)=\cos \left(s+t\right)$

$\cos \left(72^{\circ \:}\right)\cos \left(18^{\circ \:}\right)-\sin \left(72^{\circ \:}\right)\sin \left(18^{\circ \:}\right)=\cos \left(72^{\circ \:}+18^{\circ \:}\right)$

$=\cos \left(72^{\circ \:}+18^{\circ \:}\right)$

$=\cos \left(90^{\circ \:}\right)$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(90^{\circ \:}\right)=0$

$=0$

$\mathrm{Hence\ Proved}$

$\mathrm{ZARA}$