Question

if sin A +sin^3Acos^3A prove that cos^6A-4cos^4A+8cos^2A=4

Answer 1

sinA+sin^3A =cos^2A , squaring both side we get

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

==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2=cos^4A

==> sin^2A +2sin^2sinA*sin^3A +(sin^3A)^2 = cos^4A

we have ,1-sin^2A =cos^2A;

and some math formula like,

(a-b)^2=a^2-2ab+b^2

(a-b)^3=a^3 - 3a^2 *b+3a*b^2 -b^3

(sin^3A)^2 =(sin^2A)^3=(sin^6)

by using this we get,

==>sin^2A +2(sin^2A)^2 +(sin^2A)^3 =cos^4A

==> 1-cos^2A +2(1-cos^2A)^2 +(1-cos^2A)^3 =cos^4A

==> 1-cos^2A +2(1–2cos^2A +cos^4A)+(1–3cos^2A +3cos^4A -cos^6A)=cos^4A

==>1-cos^2A + 2–4cos^2A +2cos^4A +1–3cos^2A +3cos^4A -cos^6A -cos^4A=0

==>4–8cos^2A +4cos^4A -cos^6A

so we rearranging the term we get

==> –8cos^2A +4cos^4A -cos^6A=-4

==> cos^6A -4cos^4A+8cos^2A=4

Do mark brainliest :)