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

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Aicharya · Accepted Answer

$\cos x+\cos y-\cos (x+y)=\frac{3}{2}$ $\begin{array}{r} 	herefore \quad 2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)-\left(2 \cos ^2\left(\frac{x+y}{2}ight)-1ight)=\frac{3}{2} \ \quad \cdots\left[\begin{array}{l} \because \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}ight) \cos \left(\frac{\alpha-\beta}{2}ight) 	ext { and } \ \cos 	heta=2 \cos ^2\left(\frac{	heta}{2}ight)-1 \end{array}ight] \end{array}$ $\begin{aligned} & 	herefore \quad 2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)-2 \cos ^2\left(\frac{x+y}{2}ight)=\frac{3}{2}-1 \ & 	herefore \quad 2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)-2 \cos ^2\left(\frac{x+y}{2}ight)=\frac{1}{2} \ & 	herefore \quad 4 \cos ^2\left(\frac{x+y}{2}ight)-4 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)+1=0\end{aligned}$ Substituting $\cos \left(\frac{x+y}{2}ight)=\mathrm{t}$, we get $4 \mathrm{t}^2-4 \mathrm{t} \cos \left(\frac{x-y}{2}ight)+1=0$ As $t$ is real, we get $b^2-4 a c \geq 0$ $\begin{aligned} & \Rightarrow\left[-4 \cos \left(\frac{x-y}{2}ight)ight]^2-4 	imes 4 	imes 1 \geq 0 \ & \Rightarrow 16 \cos ^2\left(\frac{x-y}{2}ight)-16 \geq 0 \ & \Rightarrow \cos ^2\left(\frac{x-y}{2}ight) \geq 1 \ & \Rightarrow \cos ^2\left(\frac{x-y}{2}ight)=1 \end{aligned}$ $\ldots[\because-1 \leq \cos 	heta \leq 1$, for all values of $	heta]$ $\begin{aligned} & \Rightarrow \frac{x-y}{2}=0 \ & \Rightarrow x=y \end{aligned}$

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

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