Practicing Success
Practicing Success
CUET Preparation Today
If sin6θ + cos6θ = \(\frac{1}{3}\), 0° < θ < 90°, then find the value of 2sinθ.cosθ |
\(\frac{2\sqrt {2}}{3}\) \(\frac{\sqrt {2}}{3}\) \(\frac{\sqrt {6}}{6}\) \(\frac{\sqrt {7}}{3}\) |
\(\frac{2\sqrt {2}}{3}\) |
⇒ 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 = \(\frac{2\sqrt {2}}{3}\) |
