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

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

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Answer 1 · 2016-02-29T05:53:54

$cos (\frac{π}{2} + θ) = - sin θ$ $cos(pi/2+theta)=-sintheta$ is proved by the formula $cos (a + b) = cos a cos b - sin a sin b$ $cos (a+b)=cosacosb-sinasinb$ .

Explanation:

$cos (a + b) = cos a cos b - sin a sin b$ $cos (a+b)=cosacosb-sinasinb$

let $a = \frac{π}{2} & b = θ$ $a=pi/2 & b= theta$

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

$\Rightarrow cos (\frac{π}{2} + θ) = (0) cos θ - (1) sin θ$ $=>cos(pi/2+theta)=(0)costheta-(1)sintheta$

$\Rightarrow cos (\frac{π}{2} + θ) = 0 - sin θ$ $=>cos(pi/2+theta)=0-sintheta$

$\Rightarrow cos (\frac{π}{2} + θ) = - sin θ$ $=>cos(pi/2+theta)=-sintheta$

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

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