How do you prove #(cos^3beta-sin^3beta)/(cosbeta-sinbeta)=(2+sin2beta)/2#?

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Answer 1 · 2016-09-18T07:39:21

Please see below.

Using the identity #a^3-b^3=(a-b)(a^2+b^2+ab)#

Hence #(cos^3beta-sin^3beta)/(cosbeta-sinbeta)#

= #((cosbeta-sinbeta)(cos^2beta+sin^2beta+sinbetacosbeta))/(cosbeta-sinbeta)#

= #(cos^2beta+sin^2beta+sinbetacosbeta)#

= #(2(1+sinbetacosbeta))/2#

= #(2+2sinbetacosbeta)/2#

= #(2+sin2beta)/2#