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JEE Main 2020
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

For any θ(π4,π2)\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right), the expression 3{(\cos \theta - \sin \theta )^4}$$$$ + 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta equals :

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Solution

Given, 3(sinθ\theta - cosθ\theta ) 4 + 6(sinθ\theta + cosθ\theta ) 2 + 4sin 6 θ\theta = 3[(sinθ\theta - cosθ\theta ) 2 ] 2 + 6 (sin 2 θ\theta + cos 2 θ\theta + 2sinθ\theta cosθ\theta ) + 4sin 6 θ\theta = 3[sin 2 θ\theta + cos 2 θ\theta -2sinθ\theta cosθ\theta ] 2 + 6(1 + sin2θ\theta ) + 4sin 6 θ\theta = 3(1 - sin2θ\theta ) 2 + 6(1 + sin2θ\theta ) + 4sin 6 θ\theta = 3 (1 - 2 sin2θ\theta + sin 2 2θ\theta ) + 6 + 6sin2θ\theta + 4sin 6 θ\theta = 3 - 6sin2θ\theta + 3sin 2 2θ\theta + 6 + 6sin2θ\theta + 4sin 6 θ\theta = 9 + 3sin 2 2θ\theta + 4 sin 6 θ\theta = 9 + 3(2sinθ\theta cosθ\theta ) 2 + 4(1 - cos 2 θ\theta ) 3 = 9 + 12sin 2 θ\theta cos 2 θ\theta + 4 (1 - cos 6 θ\theta - 3cos 2 θ\theta + 3cos 4 θ\theta ) = 13 + 12 (1 - cos 2 θ\theta - 4cos 6 θ\theta - 12cosθ\theta + 12 cos 4 θ\theta = 13 + 12 cos 2 θ\theta - 12 cos 4 θ\theta - 4cos 6 θ\theta - 12 cos 2 θ\theta + 12 cos 4 θ\theta = 13 - 4 cos 6 θ\theta

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