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

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

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Answer 1 · 2017-02-03T02:57:08

see explanation below

Explanation:

expand $sin (A + B) = sin A cos A + sin B cos A$ $sin(A+B)=sinAcosA+sin BcosA$ and,
$sin (A - B) = sin A cos A - sin B cos A$ $sin(A-B)=sinAcosA-sin BcosA$

Therefore
$sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x)$ $sin(pi/6+x) + sin(pi/6-x)$ $= sin(pi/6)cosx + cancel (sinx cos(pi/6))+ sin(pi/6)cosx - cancel (sinx cos(pi/6))$

$= 2 sin(pi/6)cosx$ , where $sin(pi/6) =1/2$

therefore,

$= 2 sin(pi/6)cosx=2(1/2)cos x = cos x$ ,

Answer 2 · 2017-02-03T03:06:02

Explained below.

Explanation:

Expand the left hand side,

$(sin (pi/6) cos x + cos (pi/6) sinx) +(sin (pi/6) cosx - cos (pi/6) sin x)$

= $2 sin (pi/6) cos x$

= $2 (1/2)cos x$

= cos x = right hand side

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

2 Answers

Explanation:

Explanation:

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How do you verify the identity sin(pi/6+x)+sin(pi/6-x)=cosxsin(π6+x)+sin(π6−x)=cosxsin(pi/6+x)+sin(pi/6-x)=cosx?

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Explanation:

Explanation:

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