Question

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

• A. $\frac{\left(1+\sqrt{3}\right)}{2}$
• B. $\frac{\left(1-\sqrt{3}\right)}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

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

Explanation for correct option:

Step-1: Simplify the given data.

Given,$\cos \left(x\right)+\cos \left(y\right)-\cos \left(x+y\right)=\frac{3}{2},............\left(i\right)$

We know that $\mathit{c}\mathit{o}\mathit{s}\left(c\right)\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\left(d\right)\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\left(\frac{c+d}{2}\right)\mathit{c}\mathit{o}\mathit{s}\left(\frac{c-d}{2}\right)\&\mathit{c}\mathit{o}\mathit{s}\left(x\right)\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\left(\frac{x}{2}\right)\mathbf{-}\mathbf{1}$

$\Rightarrow$$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},$

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

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

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

Let, $\cos \left(\frac{x+y}{2}\right)=t,$

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

Step-2: Use discriminant $\mathbf{D}\mathbf{=}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}\mathbf{\ge }\mathbf{0}$.

$\Rightarrow$ ${\left[-4\cos \left(\frac{x-y}{2}\right)\right]}^2-4\times 4\times 1\ge 0,$

$\Rightarrow$ ${\cos }^2\left(\frac{x-y}{2}\right)-1\ge 0,$

$\Rightarrow$ ${\cos }^2\left(\frac{x-y}{2}\right)\ge 1,$

$\Rightarrow$ ${\cos }^2\left(\frac{x-y}{2}\right)=1,$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\left(x\right)\mathbf{=}\left[0,1\right]\right]$

$\Rightarrow$ $\frac{x-y}{2}=0$

$\Rightarrow$ $x=y.......\left(ii\right)$

Step-3: Put the value $x=y$ in equation $\left(ii\right)$.

$\Rightarrow$ $\cos \left(x\right)+\cos \left(x\right)-\cos \left(2x\right)=\frac{3}{2}$

$\Rightarrow$ $2\cos \left(x\right)-\left(2{\cos }^2\left(x\right)-1\right)=\frac{3}{2}$

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

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

$\Rightarrow$${\left(2\cos \left(x\right)\right)}^2-2\times 2\times \cos \left(x\right)+{\left(1\right)}^2=0$

$\Rightarrow$ ${\left(2\cos \left(x\right)-1\right)}^2=0$

$\Rightarrow$ $\cos \left(x\right)=\frac{1}{2}$

$\Rightarrow$ $x=60°$

Step-4: Finding value of $\sin \left(x\right)+\cos \left(y\right)$.

$$\begin{gathered} =\sin \left(60°\right)+\cos \left(60°\right) \\ =\frac{\sqrt{3}}{2}+\frac{1}{2} \\ =\frac{\sqrt{\mathbf{3}}\mathbf{+}\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Hence, correct answer is option (A).