Question bank/JEE Main/Mathematics

JEE MainMathematicsEasyTrigonometry

For any θ∈(π4,π2)\theta \in \left( \frac{\pi}{4},\frac{\pi}{2} \right),
the expression
3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin⁡6θ3(sin\theta - cos\theta)^{4} + 6(sin\theta + cos\theta)^{2} + 4\sin^{6}\theta
equals

Show answer and solution
3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin⁡6θ3(sin\theta - cos\theta)^{4} + 6(sin\theta + cos\theta)^{2} + 4\sin^{6}\theta
=3(1−sin2θ)2+6(1+sin2θ)+4(1−cos⁡2θ)3= 3(1 - sin2\theta)^{2} + 6(1 + sin2\theta) + 4\left( 1 - \cos^{2}\theta \right)^{3}
=3(1−2sin2θ+sin⁡22θ)+6(1+sin2θ)+4(1−3cos⁡2θ+3cos⁡4θ−cos⁡6θ)= 3\left( 1 - 2sin2\theta + \sin^{2}2\theta \right) + 6(1 + sin2\theta) + 4\left( 1 - 3\cos^{2}\theta + 3\cos^{4}\theta - \cos^{6}\theta \right)
=13+3sin⁡22θ+12cos⁡2θ(cos⁡2θ−1)−4cos⁡6θ= 13 + 3\sin^{2}2\theta + 12\cos^{2}\theta\left( \cos^{2}\theta - 1 \right) - 4\cos^{6}\theta
=13+12sin⁡2θcos⁡2θ−12sin⁡2θcos⁡2θ−4cos⁡6θ= 13 + 12\sin^{2}\theta\cos^{2}\theta - 12\sin^{2}\theta\cos^{2}\theta - 4\cos^{6}\theta
=13−4cos⁡6θ= 13 - 4\cos^{6}\theta

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans