JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 8)
If $$\sin \theta + \cos \theta = {1 \over 2}$$, then 16(sin(2$$\theta$$) + cos(4$$\theta$$) + sin(6$$\theta$$)) is equal to :
23
$$-$$27
$$-$$23
27
Explanation
$$\sin \theta + \cos \theta = {1 \over 2}$$
$${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}$$
$$\sin 2\theta = - {3 \over 4}$$
Now :
$$\cos 4\theta = 1 - 2{\sin ^2}2\theta $$
$$ = 1 - 2{\left( { - {3 \over 4}} \right)^2}$$
$$ = 1 - 2 \times {9 \over {16}} = - {1 \over 8}$$
$$\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta $$
$$ = (3 - 4{\sin ^2}2\theta ).\sin 2\theta $$
$$ = \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)$$
$$ \Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}$$
$$16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]$$
= $$16\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23$$
$${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}$$
$$\sin 2\theta = - {3 \over 4}$$
Now :
$$\cos 4\theta = 1 - 2{\sin ^2}2\theta $$
$$ = 1 - 2{\left( { - {3 \over 4}} \right)^2}$$
$$ = 1 - 2 \times {9 \over {16}} = - {1 \over 8}$$
$$\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta $$
$$ = (3 - 4{\sin ^2}2\theta ).\sin 2\theta $$
$$ = \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)$$
$$ \Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}$$
$$16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]$$
= $$16\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23$$
Comments (0)
