How do you simplify sin^3θ-cos^3θ/sinθ-cosθ ?

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Answer 1 · 2018-03-26T04:58:46

#(sin^3theta-cos^3theta)/(sintheta-costheta)==1+sinthetacostheta=1+(sin2theta)/2#

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

Hence, #(sin^3theta-cos^3theta)/(sintheta-costheta)#

= #((sintheta-costheta)(sin^2theta+sinthetacostheta+cos^2theta))/(sintheta-costheta)#

= #(cancel((sintheta-costheta))(sin^2theta+sinthetacostheta+cos^2theta))/(cancel((sintheta-costheta))#

= # sin^2theta+sinthetacostheta+cos^2theta#

= #1+sinthetacostheta#

= #1+(sin2theta)/2#