If 0<x,y<π and cos⁡x+cos⁡y−cos⁡(x+y)=32, then sin⁡x+cos⁡y=

cos⁡x+cos⁡y−cos⁡(x+y)=32⇒2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2+1=32⇒cos⁡x+y2cos⁡x−y2−cos2⁡x+y2=14=14cos2⁡x−y2+sin2⁡x−y2⇒cos2⁡x+y2+14cos2⁡x−y2−cos⁡x+y2cos⁡x−y2+14sin2⁡x−y2=0⇒cos⁡x+y2−12cos⁡x−y22+14sin2⁡x−y2=0⇒cos⁡x+y2=12cos⁡x−y2,sin⁡x−y2=0⇒x=y,cos⁡x=12=cos⁡y⇒sin⁡x=32⋅ ∴sin⁡x+cos⁡y=32+12=3+12

If 0<x,y<π and cos⁡x+cos⁡y−cos⁡(x+y)=32, then sin⁡x+cos⁡y=

cos⁡x+cos⁡y−cos⁡(x+y)=32⇒2cos⁡x+y2cos⁡x−y2−2cos2⁡x+y2+1=32⇒cos⁡x+y2cos⁡x−y2−cos2⁡x+y2=14=14cos2⁡x−y2+sin2⁡x−y2⇒cos2⁡x+y2+14cos2⁡x−y2−cos⁡x+y2cos⁡x−y2+14sin2⁡x−y2=0⇒cos⁡x+y2−12cos⁡x−y22+14sin2⁡x−y2=0⇒cos⁡x+y2=12cos⁡x−y2,sin⁡x−y2=0⇒x=y,cos⁡x=12=cos⁡y⇒sin⁡x=32⋅ ∴sin⁡x+cos⁡y=32+12=3+12

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$If 0 < x, y < π and \cos x + \cos y - \cos (x + y) = \frac{3}{2}, then \sin x + \cos y =$

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

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

$\frac{1}{2}$

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

answer is D.

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Detailed Solution

$\begin{array}{l} \cos x + \cos y - \cos (x + y) = \frac{3}{2} \Rightarrow 2 \cos \frac{x + y}{2} \cos \frac{x - y}{2} - 2 \cos^{2} \frac{x + y}{2} + 1 = \frac{3}{2} \\ \Rightarrow \cos \frac{x + y}{2} \cos \frac{x - y}{2} - \cos^{2} \frac{x + y}{2} = \frac{1}{4} = \frac{1}{4} (\cos^{2} \frac{x - y}{2} + \sin^{2} \frac{x - y}{2}) \\ \Rightarrow \cos^{2} \frac{x + y}{2} + \frac{1}{4} \cos^{2} \frac{x - y}{2} - \cos \frac{x + y}{2} \cos \frac{x - y}{2} + \frac{1}{4} \sin^{2} \frac{x - y}{2} = 0 \\ \Rightarrow {(\cos \frac{x + y}{2} - \frac{1}{2} \cos \frac{x - y}{2})}^{2} + \frac{1}{4} \sin^{2} \frac{x - y}{2} = 0 \Rightarrow \cos \frac{x + y}{2} = \frac{1}{2} \cos \frac{x - y}{2}, \sin (\frac{x - y}{2}) = 0 \\ \Rightarrow x = y, \cos x = \frac{1}{2} = \cos y \Rightarrow \sin x = \frac{\sqrt{3}}{2} \cdot ∴ \sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \end{array}$