Prove the following

Question:

If $\$ 0$

  1. (1) $\frac{1+\sqrt{3}}{2}$

  2. (2) $\frac{1-\sqrt{3}}{2}$

  3. (3) $\frac{\sqrt{3}}{2}$

  4. (4) $\frac{1}{2}$


Correct Option: 1

Solution:

$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)-\left[2 \cos ^{2}\left(\frac{x+y}{2}\right)-1\right]=\frac{3}{2}$

$2 \cos \left(\frac{x+y}{2}\right)\left[\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)\right]=\frac{1}{2}$

$2 \cos \left(\frac{x+y}{2}\right)\left[2 \sin \left(\frac{x}{2}\right) \cdot \sin \left(\frac{y}{2}\right)\right]=\frac{1}{2}$

$\cos \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x}{2}\right) \cdot \sin \left(\frac{y}{2}\right)=\frac{1}{8}$

Possible when $\frac{x}{2}=30^{\circ} \& \frac{y}{2}=30^{\circ}$

$x=y=60^{\circ}$

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

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