How to prove that $sin (\frac{π}{2} - θ)$ $sin (pi/2 - theta)$ = $cos θ$ $cos theta$ ?

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How to prove that $sin (\frac{π}{2} - θ)$ $sin (pi/2 - theta)$ = $cos θ$ $cos theta$ ?

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Answer 1 · 2016-03-13T17:07:18

see explanation

Explanation:

using appropriate $Addition formula$ $color(blue)" Addition formula "$

$∙ sin (A \pm B) = sin A cos B \pm cos A sin B$ $• sin(A ± B) = sinAcosB ± cosAsinB$

hence $sin (\frac{π}{2} - θ) = sin (\frac{π}{2}) cos θ - cos (\frac{π}{2}) sin θ$ $sin(pi/2 -theta) = sin(pi/2) costheta - cos(pi/2)sintheta$

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

hence $sin (\frac{π}{2}) cos θ - cos (\frac{π}{2}) sin θ = cos θ - 0$ $sin(pi/2)costheta - cos(pi/2)sintheta = costheta - 0$

$\Rightarrow sin (\frac{π}{2} - θ) = cos θ$ $rArr sin(pi/2 - theta ) = costheta$

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How to prove that $sin (\frac{π}{2} - θ)$ $sin (pi/2 - theta)$ = $cos θ$ $cos theta$ ?

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Explanation:

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