6 A-4cos- coSIf sinA +sin3A-cos2A, prove that cos6A-4cos4A +8cos2A 4

1.	6 A-4cos- coSIf sinA +sin3A-cos2A, prove that cos6A-4cos4A +8cos2A 4
Answer» Given, sinA+sin^3A =cos^2A , squaring both side we get (sinA+sin^3A)^2 = (cos^2A)^2 sin^2A +2sin^2sinAsin^3A +(sin^3A)^2=cos^4A sin^2A +2sin^2sinAsin^3A +(sin^3A)^2 = cos^4A we have,1-sin^2A =cos^2A;[using(a-b)^2=a^2-2ab+b^2(a-b)^3=a^3 - 3a^2 b+3ab^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

Answer»

Given, 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;[using(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

Discussion