How do you evaluate #(sin^2 63 + sin^2 27)/(cos^2 17 + cos^2 73)#?

1 Answer
ProfLayton
Aug 4, 2017

1

Explanation:

Note that #sinx=cos(90-x)#
#(sin^2(63)+sin^2(27))/(cos^2(17)+cos^2(73))=(sin^2(63)+cos^2(90-27))/(cos^2(17)+sin^2(90-73))#
#=(sin^2(63)+cos^2(63))/(cos^2(17)+sin^2(17))#

And using the fact that #sin^2x+cos^2x=1#:

#=1/1#
#=1#

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