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Trigonometric Ratio and Identites question

2023 · 8 Apr · Shift 2 · Q33
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Trigonometric Ratio and Identites question

2023 · 8 Apr · Shift 2 · Q33

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
The value of 36(4cos⁡29∘−1)(4cos⁡227∘−1)(4cos⁡281∘−1)(4cos⁡2243∘−1)36\left(4 \cos ^{2} 9^{\circ}-1\right)\left(4 \cos ^{2} 27^{\circ}-1\right)\left(4 \cos ^{2} 81^{\circ}-1\right)\left(4 \cos ^{2} 243^{\circ}-1\right)36(4cos29∘−1)(4cos227∘−1)(4cos281∘−1)(4cos2243∘−1) is :
  1. A
    18
  2. B
    36
  3. C
    54
  4. D
    27
View written solutionFree

Correct answer: B

  1. Rewrite each factor using a standard identity

We use 4cos⁡2θ−1=1+2cos⁡2θ4\cos^2\theta-1=1+2\cos 2\theta4cos2θ−1=1+2cos2θ because 4cos⁡2θ−1=2(1+cos⁡2θ)−1=1+2cos⁡2θ.4\cos^2\theta-1=2(1+\cos 2\theta)-1=1+2\cos 2\theta.4cos2θ−1=2(1+cos2θ)−1=1+2cos2θ.

So the expression becomes 36 (1+2cos⁡18∘)(1+2cos⁡54∘)(1+2cos⁡162∘)(1+2cos⁡486∘).36\,(1+2\cos 18^\circ)(1+2\cos 54^\circ)(1+2\cos 162^\circ)(1+2\cos 486^\circ).36(1+2cos18∘)(1+2cos54∘)(1+2cos162∘)(1+2cos486∘).

Since cosine is periodic with period 360∘360^\circ360∘, cos⁡486∘=cos⁡126∘.\cos 486^\circ=\cos 126^\circ.cos486∘=cos126∘. Hence E=36(1+2cos⁡18∘)(1+2cos⁡54∘)(1+2cos⁡162∘)(1+2cos⁡126∘).E=36(1+2\cos 18^\circ)(1+2\cos 54^\circ)(1+2\cos 162^\circ)(1+2\cos 126^\circ).E=36(1+2cos18∘)(1+2cos54∘)(1+2cos162∘)(1+2cos126∘).

  1. Use the identity

A very useful identity is 1+2cos⁡x=sin⁡3x2sin⁡x2.1+2\cos x=\frac{\sin \frac{3x}{2}}{\sin \frac{x}{2}}.1+2cosx=sin2x​sin23x​​. This follows from

=\sin\frac{x}{2}(\cos x+2\cos x)=\sin\frac{x}{2}(1+2\cos x).$$ Applying it to each factor: - For $x=18^\circ$: $$1+2\cos 18^\circ=\frac{\sin 27^\circ}{\sin 9^\circ}.$$ - For $x=54^\circ$: $$1+2\cos 54^\circ=\frac{\sin 81^\circ}{\sin 27^\circ}.$$ - For $x=162^\circ$: $$1+2\cos 162^\circ=\frac{\sin 243^\circ}{\sin 81^\circ}.$$ - For $x=126^\circ$: $$1+2\cos 126^\circ=\frac{\sin 189^\circ}{\sin 63^\circ}.$$ So $$E=36\cdot \frac{\sin 27^\circ}{\sin 9^\circ}\cdot \frac{\sin 81^\circ}{\sin 27^\circ}\cdot \frac{\sin 243^\circ}{\sin 81^\circ}\cdot \frac{\sin 189^\circ}{\sin 63^\circ}.$$ 3. **Simplify the sines** Use angle relations: $$\sin 243^\circ=\sin(180^\circ+63^\circ)=-\sin 63^\circ,$$ $$\sin 189^\circ=\sin(180^\circ+9^\circ)=-\sin 9^\circ.$$ Substitute: $$E=36\cdot \frac{\sin 27^\circ}{\sin 9^\circ}\cdot \frac{\sin 81^\circ}{\sin 27^\circ}\cdot \frac{-\sin 63^\circ}{\sin 81^\circ}\cdot \frac{-\sin 9^\circ}{\sin 63^\circ}.$$ Everything cancels telescopically: $$E=36.$$ 4. **Check options** - A: $18$ ❌ - B: $36$ ✅ - C: $54$ ❌ - D: $27$ ❌ Therefore, the correct answer is **Option B**.
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