Question

If sin x+sin²x =1 Then
Cos^12 x + 3 cos^10 x + 3 cos^8 x + Cos^6 x + 2 cos^4 x + cos²x-2=

=​

Answer 1

Step-by-step explanation:

The question is very simple

sin x+sin²x =1

sinx = 1 - sin²x

sinx = cos²x ----- equation 1

Now,

Cos^12 x + 3 cos^10 x + 3 cos^8 x + Cos^6 x + 2 cos^4 x + cos²x-2

Taking cos^6x common from first four terms

cos^6x ( Cos^6 x + 3 cos^4 x + 3 cos^2 x + 1 ) + 2 cos^4 x + cos²x-2

Now the part inside the bracket is in the form (a+b)^3

cos^6x ( cos^2x+1 )^3 + 2 cos^4 x + cos²x-2

Now again taking cos^2x

((cos^2x)^2+cos^2x)^3 -1 + 2 cos^4 x + cos²x-2

Now we know that sin x = cos^2x

(sin^2 x + cos^2x)^3 -1 + 2 cos^4 x + cos²x-2

1-1 + 2 cos^4 x + cos²x-2

2 cos^4 x + cos²x - 2

2 (cos^2x)^2 + cos²x -2

Again putting cos²x= sin x

2 sin ^2 x + cos^2 x -2

1 + sin^2x -2

sin^2 x -1