How do you simplify #\sin ^{2}(\Theta )+\cos ^{2}(\Theta )#?

1 Answer
Cesareo R.
Sep 30, 2016

See below.

Explanation:

As we know, in a circle with radius #r# the point #x,y# in the circumference obeys #sin theta = y/r# and #cos theta = x/r# with #r = sqrt(x^2+y^2)# so

#sin^2theta+cos^2theta = (y/r)^2+(x/r)^2 = (x^2+y^2)/(x^2+y^2) = 1#

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