How do you show $cos (x + \frac{π}{2}) + cos (x - \frac{π}{2}) = 0$ $cos(x+pi/2)+cos(x-pi/2)=0$ ?

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How do you show $cos (x + \frac{π}{2}) + cos (x - \frac{π}{2}) = 0$ $cos(x+pi/2)+cos(x-pi/2)=0$ ?

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Answer 1 · 2018-03-07T12:49:56

We need to use the trig identity:
$cos (A \pm B) = cos A cos B \mp sin A sin B$ $cos(A+-B)=cosAcosB∓sinAsinB$

Using this, we get:
$cos (x + \frac{π}{2}) + cos (x - \frac{π}{2}) = (cos x cos (\frac{π}{2}) + sin x sin (\frac{π}{2})) + (cos x cos (\frac{π}{2}) - sin x sin (\frac{π}{2}))$ $cos(x+pi/2)+cos(x-pi/2)=(cosxcos(pi/2)+sinxsin(pi/2))+(cosxcos(pi/2)-sinxsin(pi/2))$

$cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$
$sin (\frac{π}{2}) = 1$ $sin(pi/2)=1$

$cos (x + \frac{π}{2}) + cos (x - \frac{π}{2}) = (0 cos x + 1 sin x) + (0 cos x - 1 sin x) = sin x - sin x = 0$ $cos(x+pi/2)+cos(x-pi/2)=(0cosx+1sinx)+(0cosx-1sinx)=sinx-sinx=0$

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How do you show $cos (x + \frac{π}{2}) + cos (x - \frac{π}{2}) = 0$ $cos(x+pi/2)+cos(x-pi/2)=0$ ?

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