How do you prove #cosx=pi/2 - sinx #?

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Answer 1 · 2016-03-01T21:31:20

The closest true identities to the given equality would be

#cos(x-pi/2) = sin(x)#
or
#cos(x) = sin(x+pi/2)#

(These are equivalent, as you could, for example, substitute #x = y+pi/2# into the first to obtain the second)

To prove the above, we can use the identity

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

along with sine being an odd function and cosine being an even function. That is
#sin(-x) = -sin(x)#
and
#cos(-x) = cos(x)#

With these, we have

#cos(x-pi/2) = cos(x)cos(-pi/2)-sin(x)sin(-pi/2)#

#=cos(x)cos(pi/2)+sin(x)sin(pi/2)#

#=cos(x)*0+sin(x)*1#

#=sin(x)#