If sinθ + sin²θ = 1. Find the value of cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ + 2cos⁴θ + 2cos²θ – 2.

Formula: -

(i) sin²θ + cos²θ = 1

(ii) (a + b)³ = a³ + b³ + 3a²b + 3b²a

cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ + 2cos⁴θ + 2cos²θ – 2

sinθ + sin²θ = 1

⇒ sinθ = 1 – sin²θ = cos²θ …… (1)

cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ + 2cos⁴θ + 2cos²θ – 2

taking 2 as common

= cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ + 2(cos⁴θ + cos²θ – 1)

using formula (ii)

= (cos⁴θ + cos²θ)³ + 2(cos⁴θ + cos²θ – 1)

using equation(1)

= (sin²θ + cos²θ)³ + 2(sin²θ + cos²θ – 1)

using formula (i)

= 1 + 2(1 – 1)

= 1

cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ + 2cos⁴θ + 2cos²θ – 2 = 1