MCQEasyJEE 2023Trigonometric Equations

JEE Mathematics 2023 Question with Solution

96cos(π33)cos(2π33)cos(4π33)cos(8π33)cos(16π33)96 \cos\left(\frac{\pi}{33}\right) \cos\left(\frac{2\pi}{33}\right) \cos\left(\frac{4\pi}{33}\right) \cos\left(\frac{8\pi}{33}\right) \cos\left(\frac{16\pi}{33}\right) is equal to:

  • A

    44

  • B

    22

  • C

    33

  • D

    11

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

96cos(π33)cos(2π33)cos(4π33)cos(8π33)cos(16π33)96 \cos\left(\frac{\pi}{33}\right) \cos\left(\frac{2\pi}{33}\right) \cos\left(\frac{4\pi}{33}\right) \cos\left(\frac{8\pi}{33}\right) \cos\left(\frac{16\pi}{33}\right)

Find: The value of the given expression.

Use the identity

cosAcos2Acos22Acos2n1A=sin(2nA)2nsinA\cos A \cos 2A \cos 2^2 A \dots \cos 2^{n-1} A = \frac{\sin(2^n A)}{2^n \sin A}

Here,

A=π33,n=5A = \frac{\pi}{33}, \qquad n = 5

So the expression becomes

96sin(25π33)25sin(π33)96 \cdot \frac{\sin\left(2^5 \cdot \frac{\pi}{33}\right)}{2^5 \sin\left(\frac{\pi}{33}\right)}

Since

25=322^5 = 32

we get

96sin(32π33)32sin(π33)96 \cdot \frac{\sin\left(\frac{32\pi}{33}\right)}{32 \sin\left(\frac{\pi}{33}\right)}

Now use

sin(πx)=sinx\sin(\pi - x) = \sin x

Therefore,

sin(32π33)=sin(π33)\sin\left(\frac{32\pi}{33}\right) = \sin\left(\frac{\pi}{33}\right)

Substituting,

96sin(π33)32sin(π33)=9632=396 \cdot \frac{\sin\left(\frac{\pi}{33}\right)}{32 \sin\left(\frac{\pi}{33}\right)} = \frac{96}{32} = 3

Therefore, the correct option is C.

Common mistakes

  • Using the product identity with the wrong value of nn. There are 5 cosine factors, so n=5n = 5. Counting incorrectly changes the power of 22 in the denominator and gives a wrong result.

  • Not recognizing that sin(32π33)=sin(π33)\sin\left(\frac{32\pi}{33}\right) = \sin\left(\frac{\pi}{33}\right) because 32π33=ππ33\frac{32\pi}{33} = \pi - \frac{\pi}{33}. Use the identity sin(πx)=sinx\sin(\pi - x) = \sin x before simplifying.

  • Cancelling terms before rewriting the numerator correctly. First transform the product into a sine ratio, then apply the identity, and only after that cancel the common sin(π33)\sin\left(\frac{\pi}{33}\right) factor.

Practice more Trigonometric Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Related questions