First slide

Trigonometric transformations

Question

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

Moderate

A

x+y=0
B

x=2y
C

x=y
D

2x=y

Solution

$\begin{array}{l} \cos x + \cos y - \cos (x + y) = \frac{3}{2} \\ or 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) - 2 \cos^{2} (\frac{x + y}{2}) + 1 = \frac{3}{2} \\ or 2 \cos^{2} (\frac{x + y}{2}) - 2 \cos (\frac{x + y}{2}) \cos (\frac{x - y}{2}) + \frac{1}{2} = 0 \end{array}$
Now, $\cos (\frac{x + y}{2})$ is always real, then discriminant $\geq$ 0. Thus,
$4 \cos^{2} (\frac{x - y}{2}) - 4 \geq 0$
$\begin{aligned} or \cos^{2} (\frac{x - y}{2}) \geq 1 \\ or \cos^{2} (\frac{x - y}{2}) = 1 \\ or \frac{x - y}{2} = 0 or x = y \end{aligned}$

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