If sin6θ + cos6θ = \(\frac{1}{3}\), 0° < θ < 90°, then find the value of 2sinθ.cosθ
Answer & explanation
Correct answer: option 1
⇒ sin6θ + cos6θ = 1 - 3 sin2θ.cos2θ
⇒ \(\frac{1}{3}\) = 1 - 3 sin2θ.cos2θ
⇒ sin2θ.cos2θ = \(\frac{2}{9}\)
⇒ sinθ.cosθ = \(\frac{\sqrt {2}}{3}\)
Now
⇒ 2 sinθ.cosθ = \(\frac{2\sqrt {2}}{3}\)