Question #eabbb

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Question #eabbb

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Answer 1 · 2017-03-02T15:50:26

see below

Explanation:

Use the formulas $cos x + cos y = 2 cos (\frac{1}{2} (x + y)) cos (\frac{1}{2} (x - y))$ $cos x+cos y =2 cos (1/2 (x+y))cos(1/2(x-y))$ and $cos (- x) = cos x$ $cos(-x)=cosx$

Right Hand Side:
$\frac{1}{2} [cos x + cos 3 x] = \frac{1}{2} [2 cos (\frac{1}{2} (x + 3 x)) cos (\frac{1}{2} (x - 3 x))]$ $1/2 [cosx+cos 3x]=1/2 [2cos (1/2 (x+3x))cos(1/2(x-3x))]$

$=1/cancel2 cancel2cos (1/2 (x+3x))cos(1/2(x-3x))$

$=cos (1/2 (4x))cos(1/2(-2x))$

$=cos 2x*cos(-x)$

$=cos2x cos x$

$=cosx cos 2x$

$:.=$ Left Hand Side

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Question #eabbb

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