First slide

Trigonometric transformations

Question

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

Difficult

A

$\frac{1}{2}$
B

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

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

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

Solution

$\begin{aligned} 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - [2 \cos^{2} (\frac{x + y}{2}) - 1] = \frac{3}{2} \\ \Rightarrow 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - 2 \cos^{2} (\frac{x + y}{2}) = \frac{1}{2} \end{aligned}$
$\begin{array}{l} \Rightarrow 4 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - 4 \cos^{2} (\frac{x + y}{2}) = 1 \\ = \cos^{2} (\frac{x - y}{2}) + \sin^{2} (\frac{x - y}{2}) \\ \Rightarrow 4 \cos^{2} (\frac{x + y}{2}) + \cos^{2} (\frac{x - y}{2}) - 4 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) + \sin^{2} (\frac{x - y}{2}) = 0 \end{array}$
$\begin{array}{l} \Rightarrow {(\cos (\frac{x - y}{2}) - 2 \cos (\frac{x + y}{2}))}^{2} + \sin^{2} (\frac{x - y}{2}) = 0 \\ \Rightarrow \sin (\frac{x - y}{2}) = 0 \Rightarrow x = y \\ \Rightarrow \cos (\frac{x - y}{2}) = 2 \cos (\frac{x + y}{2}) Now. put x=y \\ \Rightarrow \cos x = \frac{1}{2} = \cos y \\ ∴ \sin x + \cos y = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \end{array}$

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