JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 7)
If 0 < x, y < $$\pi$$ and cosx + cosy $$-$$ cos(x + y) = $${3 \over 2}$$, then sinx + cosy is equal to :
$${{1 + \sqrt 3 } \over 2}$$
$${{1 \over 2}}$$
$${{\sqrt 3 } \over 2}$$
$${{1 - \sqrt 3 } \over 2}$$
Explanation
$$2\cos \left( {{{x + y} \over 2}} \right)\cos \left( {{{x - y} \over 2}} \right) - \left[ {2{{\cos }^2}\left( {{{x + y} \over 2}} \right) - 1} \right] = {3 \over 2}$$
$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}$$
$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}$$
$$\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}$$
Possible when $${x \over 2} = 30^\circ $$ & $${y \over 2} = 30^\circ $$
$$x = y = 60^\circ $$
$$\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}$$
$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}$$
$$2\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}$$
$$\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}$$
Possible when $${x \over 2} = 30^\circ $$ & $${y \over 2} = 30^\circ $$
$$x = y = 60^\circ $$
$$\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}$$
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