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If 0 < x, y < $$\pi$$ and cosx + cosy $$-$$ cos(x + y) = $${3 \over 2}$$, then sinx + cosy is equal to :

JEE · Math · previous-year question

  1. A.$${{1 + \sqrt 3 } \over 2}$$correct
  2. B.$${{1 \over 2}}$$
  3. C.$${{\sqrt 3 } \over 2}$$
  4. D.$${{1 - \sqrt 3 } \over 2}$$

Answer

A. $${{1 + \sqrt 3 } \over 2}$$

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 $${x \over 2} = 30^\circ $$ & $${y \over 2} = 30^\circ $$ $$x = y = 60^\circ $$ $$\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}$$

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