Given that sin(2θ)=2/3, the value of

Question

sin^6θ+cos^6θ

Answer 1

Given that

To prove

Apply sum of the cubes formula

               (We know that)

It has given that 

Cancel 2 on both sides

In the formula of

                 We know that 

Already we have from equation 1

Substitute the values from equation 2 and 3

Therefore the value of

 

Answer 2

sin (2θ) = 2/3.

⇒ cos (2θ) = √(1 - sin²(2θ))  = √(1 - (2/3)² = √(1 - 4/9) = √(9 - 4)/3 = √5/3.

sin⁶(θ) + cos⁶(θ) = [ sin²(θ) ]³ + [ cos²(θ) ]³ .

Double angle formula : cos (2θ) = 1 - 2sin²(θ).

cos (2θ) = 1 - 2sin²(θ) = √5/3

1 - 2sin²(θ) = √5/3

2sin²(θ) = 1 - √5/3

sin²(θ) = (3 - √5)/6.

Pythagorean identity : sin² θ + cos² θ = 1.

cos² θ = 1 - (3 - √5)/6 = (6 - 3 + √5)/6 = (3 + √5)/6.

⇒ sin⁶(θ) + cos⁶(θ) = [ sin²(θ) ]³ + [ cos²(θ) ]³

= [ (3 - √5)/6 ]³ + [(3 + √5)/6 ]³

= [ 2(3³ + 3*3*5)] / 6 ³               [since, (a - b)³ + (a + b)³ = 2[a³ + 3ab²] ]

= 144/216

= 2/3.

Therefore, sin⁶(θ) + cos⁶(θ) = 2/3.