Question

If cos θ + cos² θ = 1, prove that sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1

Answer 1

Given `cos theta + cos^2 theta = 1`

We have to prove sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1

From the given equation, we have

`cos theta + cos^2 theta = 1`

`=> cos theta = 1 - cos^2 theta`

`=> ccos theta = sin^2 theta`

`=> sin^2 theta = cos theta`

Therefore, we have

sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2

`= (sin^12 theta + 3 sin^10 theta + 3 sin^8 theta + sin^6 theta) + (2 sin^4 theta + 2 sin^2 theta) - 2` 

`= {(sin^4 theta)^3 + 3(sin^4 theta)^2 sin^2 theta + 3 sin^4 theta(sin^2 theta)^2 + (sin^2 theta)^3} + 2(sin^4 theta + sin^2 theta) - 2`

`= (sin^4 theta  + sin^2 theta)^3 + 2 (sin^4 theta + sin^2 theta) - 2`

`= (cos^2 theta + cos theta)^3 + 2 (cos^2 theta + cos theta) - 2`

= (1)^3 + 2(1) - 2

= 1

hence proved