3.Trigonometrical Ratios, Functions and Identities
hard

If $0 <  x , y < \pi$ and $\cos x +\cos y-\cos ( x + y )=\frac{3}{2},$ then $\sin x+\cos y$ is equal to ...... .

A

$\frac{1}{2}$

B

$\frac{1+\sqrt{3}}{2}$

C

$\frac{\sqrt{3}}{2}$

D

$\frac{1-\sqrt{3}}{2}$

(JEE MAIN-2021)

Solution

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$

$\cos ^{2}\left(\frac{x+y}{2}\right)-\cos \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)$

$+\frac{1}{4} \cdot \cos ^{2}\left(\frac{x-y}{2}\right)+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow\left(\cos \left(\frac{x+y}{2}\right)-\frac{1}{2} \cos \left(\frac{x-y}{2}\right)\right)^{2}+\frac{1}{4} \sin ^{2}\left(\frac{x-y}{2}\right)=0$

$\Rightarrow \sin \left(\frac{x-y}{2}\right)=0$ and $\cos \left(\frac{x+y}{2}\right)=\frac{1}{2} \cos \left(\frac{x-y}{2}\right)$

$\Rightarrow x=y$ and $\cos x=\frac{1}{2}=\cos y$

$\therefore \sin x=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin x+\cos y=\frac{1+\sqrt{3}}{2}$

Standard 11
Mathematics

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