Question #df561

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Answer 1 · 2017-12-04T00:16:54

Please see below.

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We will use the identities:

#cos(2theta)=2cos^2theta-1#

#sin^2theta+cos^2theta=1#, so #sin^2theta=1-cos^2theta#

#(sin^2(y/2)-cos^2(y/2))/cosy=(1-cos^2(y/2)-cos^2(y/2))/cosy=#

#(1-2cos^2(y/2))/cosy=(-(2cos^2(y/2)-1))/cosy=-cos(2(y/2))/cosy#

#-cos(cancelcolor(red)2(y/cancelcolor(red)2))/cosy=(-cosy)/cosy=-1#