Why does $cos(90 - x) = sin(x)$ and $sin(90 - x) = cos(x)$ ?

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Answer 1 · 2015-04-17T00:49:30

Note that the image below is only for $x$ in Q1 (the first quadrant).
If you wish you should be able to draw it with $x$ in any quadrant.

enter image source here

Definition of $sin(x)$

$($ side opposite angle $x)//($ hypotenuse $)$

Definition of $cos(90^@ -x)$

$($ side adjacent to angle $(90^@-x))//($ hypotenuse $)$

but $($ side opposite angle $x) = ($ side adjacent to angle $(90^@-x)$

Therefore

$sin(x) = cos(90^@ -x)$

Similarly

$cos(x) = sin(90^@ - x)$

Answer 2 · 2016-02-07T01:41:27

These can also be proven using the sine and cosine angle subtraction formulas:

$cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)$

$sin(alpha-beta)=sin(alpha)cos(beta)-cos(alpha)sin(beta)$

Applying the former equation to $cos(90^@-x)$ , we see that

$cos(90^@-x)=cos(90^@)cos(x)+sin(90^@)sin(x)$

$cos(90^@-x)=0*cos(x)+1*sin(x)$

$cos(90^@-x)=sin(x)$

Applying the latter to $sin(90^@-x)$ , we can also prove that

$sin(90^@-x)=sin(90^@)cos(x)-cos(90^@)sin(x)$

$sin(90^@-x)=1*cos(x)-0*sin(x)$

$sin(90^@-x)=cos(x)$

Why does cos(90 - x) = sin(x)cos(90 - x) = sin(x) and sin(90 - x) = cos(x)sin(90 - x) = cos(x)?