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

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129
Jun 1, 2017

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)#

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Write your answer here...
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mason m Share
Jun 1, 2017

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)#

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