Question

Prove it.Step wise.No mistakes.No spanning.​

Answer 1

Question.

Cos^6A+sin^6A=1-3cos^2A+3cos^4A .

Step-by-step explanation:

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L.H.S.

=>cos^6A+sin^6A.

=>(cos^2A)^3+(sin^2A)^3.

=>(cos^2A+sin^2A)(cos^4+sin^4- sin^2A*cos^2A) .

=>(1){(cos^2A)^2+(sin^2A)^2-sin^2A*cos^2A}.

=>{(cos^2A+sin^2A)^2-2cos^2A*sin^2A-sin^2A*cos^2A} .

=>(1)^2-3cos^2A*sin^2A.

=> 1-3cos^2A*sin^2A.

=>1-3cos^2A (1-cos^2A).

=>1-3cos^2A+3cos^4A.

=> R.H.S.

Formula used.

• a^3+b^3=(a+b)(a^2+b^2-ab).
• a^2+b^2=(a+b)^2-2ab.
• sin^2A=(1-cos^2A).
• sin^2A+cos^2A=1.
• (a+b)^2=a^2+b^2+2ab.
• (a-b)^2=a^2+b^2-2ab.