3. Ifsinθ + sin²θ + sin³θ = 1, prove that cos⁶θ – 4cos´θ + 8cos²θ = 4.
Answers
Answered by
2
sinθ + sin²θ + sin³θ = 1
=>sinθ + sin²θ + sin³θ = 1
=>sinθ + sin²θ + sin³θ = (1-sin²θ)=cos²θ
square both sides you get
=>sin⁶θ + 2sin⁴θ+ sin²θ=cos⁴θ
=> replace sin by cos
=>(1-cos²θ)³ + 2(1-cos²θ)² +(1-cos²θ) - cos⁴θ=0
open the brackets and simplify further to get
=> -cos⁶θ + 4cos⁴θ -8cos²θ +4=0
=>cos⁶θ – 4cos⁴θ + 8cos²θ = 4
hence proved..
=>sinθ + sin²θ + sin³θ = 1
=>sinθ + sin²θ + sin³θ = (1-sin²θ)=cos²θ
square both sides you get
=>sin⁶θ + 2sin⁴θ+ sin²θ=cos⁴θ
=> replace sin by cos
=>(1-cos²θ)³ + 2(1-cos²θ)² +(1-cos²θ) - cos⁴θ=0
open the brackets and simplify further to get
=> -cos⁶θ + 4cos⁴θ -8cos²θ +4=0
=>cos⁶θ – 4cos⁴θ + 8cos²θ = 4
hence proved..
Similar questions