How do you verify $sinx=cos(pi/2 - x)$ ?

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How do you verify $sinx=cos(pi/2 - x)$ ?

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Answer 1 · 2016-03-11T11:12:54

see below

Explanation:

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In the above figure $Delta$ ABC is a rt angled triangle where $/_ABC=pi/2$ , $/_BAC=x$ and $/_ACB=pi/2-x$

Now $cos/_ACB=(BC)/(AC)$ ( side opposite to the angle considered for trigonometric ratio is taken as perpendicular and the side lying vertical to it is taken as base)
$=>cos(pi/2-x)=(BC)/(AC)$

Again $sin/_CAB=(BC)/(AC)$
$=>sinx=(BC)/(AC)$
Comparing these two relation we can say
$sinx=cos(pi/2-x)$

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How do you verify $sinx=cos(pi/2 - x)$ ?

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