How do you prove $sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos x$ $sin(pi/6 + x) + sin (pi/6 - x)= cosx$ ?

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How do you prove $sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos x$ $sin(pi/6 + x) + sin (pi/6 - x)= cosx$ ?

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Answer 1 · 2016-03-19T22:08:32

$sin (\frac{π}{6} + x) = sin (\frac{π}{6}) . cos x + sin x . cos (\frac{π}{6})$ $sin (pi/6 + x) = sin (pi/6).cos x + sin x.cos (pi/6)$ (1)
$sin (\frac{π}{6} - x) = sin (\frac{π}{6}) . cos x - sin x . cos (\frac{π}{6})$ $sin (pi/6 - x) = sin (pi/6).cos x - sin x.cos (pi/6)$ (2).

Add up (1) and (2) -->
$S u m = 2 sin (\frac{π}{6}) . cos x$ $Sum = 2sin (pi/6).cos x$ .
Trig table gives --> $sin (\frac{π}{6}) = \frac{1}{2}$ $sin (pi/6) = 1/2$
Therefor,
$S = 2 (\frac{1}{2}) cos x = cos x .$ $S = 2(1/2)cos x = cos x.$

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How do you prove $sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos x$ $sin(pi/6 + x) + sin (pi/6 - x)= cosx$ ?

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How do you prove $sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos x$ $sin(pi/6 + x) + sin (pi/6 - x)= cosx$ ?