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Question

Prove thatSin3 A+cos3 Asin A+cos A+Sin3 Acos3 Asin Acos A=2 [4 MARKS]

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Solution

Formula: 2 Marks
Proof: 2 Marks

Using a3+b3(a+b)(a2ab+b2) and a3b3=(ab)(a2+ab+b2), we have
LHS=Sin3 A+cos3 Asin A+cos A+Sin3 Acos3 Asin Acos A=(sin A+cos A)(sin2sin A cos A+cos2A)(sin A+cos A)+(sin Acos A)(sin2 A+sin A cos a+cos2A)(sin Acos A)=1sin A cos A+1+sin A cos A=2=RHS

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