JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 11)
The value of 2sin (12$$^\circ$$) $$-$$ sin (72$$^\circ$$) is :
$${{\sqrt 5 (1 - \sqrt 3 )} \over 4}$$
$${{1 - \sqrt 5 } \over 8}$$
$${{\sqrt 3 (1 - \sqrt 5 )} \over 2}$$
$${{\sqrt 3 (1 - \sqrt 5 )} \over 4}$$
Explanation
$$2\sin 12^\circ - \sin 72^\circ $$
$$ = \sin 12^\circ + ( - 2\cos 42^\circ \,.\,\sin 30^\circ )$$
$$ = \sin 12^\circ - \cos 42^\circ $$
$$ = \sin 12^\circ - \sin 48^\circ $$
$$ = 2\sin 18^\circ \,.\,\cos 30^\circ $$
$$ = - 2\left( {{{\sqrt 5 - 1} \over 4}} \right)\,.\,{{\sqrt 3 } \over 2}$$
$$ = {{\sqrt 3 \left( {1 - \sqrt 5 } \right)} \over 4}$$
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