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 $(cos^3beta-sin^3beta)/(cosbeta-sinbeta)$

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

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

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

= $(2+2sinbetacosbeta)/2$

= $(2+sin2beta)/2$

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