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

Given:

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

To do:

We have to evaluate \( \left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right) \).

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

Therefore,

$\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)=(\sin (90^{\circ}-18^{\circ})+\cos 18^{\circ})(\sin (90^{\circ}-18^{\circ})-\cos 18^{\circ})$

$=(cos 18^{\circ}+cos 18^{\circ})(\cos 18^{\circ}-\cos 18^{\circ})$

$=(2 cos 18^{\circ})\times 0 $

$=0$

Therefore, $\left(\sin 72^{\circ}+\cos 18^{\circ}\right)\left(\sin 72^{\circ}-\cos 18^{\circ}\right)=0$.   

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Updated on: 10-Oct-2022

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