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Trigonometric Ratio and Identites question

2023 · 10 Apr · Shift 1 · Q33

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Trigonometric Ratio and Identites
2023 · 10 Apr · Shift 1 · Q33

Trigonometric Ratio and Identites question

2023 · 10 Apr · Shift 1 · Q33

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1

96\cos {\pi \over {33}}\cos {{2\pi } \over {33}}\cos {{4\pi } \over {33}}\cos {{8\pi } \over {33}}\cos {{16\pi } \over {33}}

is equal to :

A
4
B
2
C
1
D
3

View written solutionFree

Correct answer: D

Let $P=96\cos\frac{\pi}{33}\cos\frac{2\pi}{33}\cos\frac{4\pi}{33}\cos\frac{8\pi}{33}\cos\frac{16\pi}{33}.$

We use the standard identity $\sin 2x=2\sin x\cos x.$ Repeatedly applying it gives $\sin 2^n x=2^n\sin x\prod_{k=0}^{n-1}\cos(2^k x).$

Take $x=\frac{\pi}{33}, \quad n=5.$ Then $\sin\left(32\cdot \frac{\pi}{33}\right)=2^5\sin\frac{\pi}{33}\cos\frac{\pi}{33}\cos\frac{2\pi}{33}\cos\frac{4\pi}{33}\cos\frac{8\pi}{33}\cos\frac{16\pi}{33}.$ So,

=\frac{\sin\frac{32\pi}{33}}{32\sin\frac{\pi}{33}}.$$ 3. Substitute into $P$: $$P=96\cdot \frac{\sin\frac{32\pi}{33}}{32\sin\frac{\pi}{33}} =3\cdot \frac{\sin\frac{32\pi}{33}}{\sin\frac{\pi}{33}}.$$ 4. Now, $$\sin\frac{32\pi}{33}=\sin\left(\pi-\frac{\pi}{33}\right)=\sin\frac{\pi}{33}.$$ Hence, $$P=3\cdot \frac{\sin\frac{\pi}{33}}{\sin\frac{\pi}{33}}=3.$$ 5. Therefore the correct option is: $$\boxed{3}$$ which is option $\boxed{D}$. 6. Verification with stored answer: - Derived answer: $D$ - Stored correct answer: $D$ - They agree.

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