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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Find the value of 32 90^ 17 180^ 17 360^ 17 720^ 17 . If the sum of...

Find the value of 32sin⁡90∘17cos⁡180∘17cos⁡360∘17cos⁡720∘1732 \sin \frac{90^{\circ}}{17} \cos \frac{180^{\circ}}{17} \cos \frac{360^{\circ}}{17} \cos \frac{720^{\circ}}{17}. If the sum of the roots of the equation cos⁡4x+6=7cos⁡2x\cos 4 x+6=7 \cos 2 x in the interval [0,314][0,314] is kπ,k∈Rk \pi, k \in R. Find k.

Correct answer

02

Step-by-step solution

Let θ=(9017)∘⇒17θ=90∘\theta=\left(\frac{90}{17}\right)^{\circ} \Rightarrow 17 \theta=90^{\circ} Now, 32sin⁡90∘17cos⁡180∘17cos⁡360∘17cos⁡720∘17=32sin⁡θcos⁡2θcos⁡4θcos⁡8θ32 \sin \frac{90^{\circ}}{17} \cos \frac{180^{\circ}}{17} \cos \frac{360^{\circ}}{17} \cos \frac{720^{\circ}}{17}=32 \sin \theta \cos 2 \theta \cos 4 \theta \cos 8 \theta =32cos⁡θsin⁡θcos⁡2θcos⁡4θcos⁡8θcos⁡θ=2×sin⁡16θcos⁡θ=2sin⁡(17θ−θ)cos⁡θ=2sin⁡(90∘−θ)cos⁡θ=2\begin{aligned} & =\frac{32 \cos \theta \sin \theta \cos 2 \theta \cos 4 \theta \cos 8 \theta}{\cos \theta}=2 \times \frac{\sin 16 \theta}{\cos \theta} \\ & =\frac{2 \sin (17 \theta-\theta)}{\cos \theta}=\frac{2 \sin \left(90^{\circ}-\theta\right)}{\cos \theta}=2\end{aligned}
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