Solve the trigonometric problem: If A + B + C = 3π/2, find cos 2A + cos 2B + cos 2C. Detailed solution

MHT-CET Trigonometric Problem Solution

Question: If \( A + B + C = \frac{3\pi}{2} \), then \( \cos 2A + \cos 2B + \cos 2C \) is equal to:

Options:

a) \( 1 - 4 \cos A \cos B \cos C \)

b) \( 4 \sin A \sin B \sin C \)

c) \( 1 + 2 \cos A \cos B \cos C \)

d) \( 1 - 4 \sin A \sin B \sin C \)

Detailed Solution

Let’s solve this step-by-step. We’re given \( A + B + C = \frac{3\pi}{2} \) (which is 270°), and we need to find \( \cos 2A + \cos 2B + \cos 2C \). The options suggest a trigonometric identity, so let’s test it with examples.

Example 1: \( A = B = C = \frac{\pi}{2} \)

Check: \( \frac{\pi}{2} + \frac{\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \), which is true.

Compute:

- \( \cos 2A = \cos \pi = -1 \) - \( \cos 2B = \cos \pi = -1 \) - \( \cos 2C = \cos \pi = -1 \)

Sum: \( -1 + (-1) + (-1) = -3 \)

Options:

- a) \( 1 - 4 \cos \frac{\pi}{2} \cos \frac{\pi}{2} \cos \frac{\pi}{2} = 1 - 4 \cdot 0 \cdot 0 \cdot 0 = 1 \)

- b) \( 4 \sin \frac{\pi}{2} \sin \frac{\pi}{2} \sin \frac{\pi}{2} = 4 \cdot 1 \cdot 1 \cdot 1 = 4 \)

- c) \( 1 + 2 \cos \frac{\pi}{2} \cos \frac{\pi}{2} \cos \frac{\pi}{2} = 1 + 2 \cdot 0 \cdot 0 \cdot 0 = 1 \)

- d) \( 1 - 4 \sin \frac{\pi}{2} \sin \frac{\pi}{2} \sin \frac{\pi}{2} = 1 - 4 \cdot 1 \cdot 1 \cdot 1 = -3 \)

Option d) matches: \(-3\).

Example 2: \( A = \pi, B = \frac{\pi}{4}, C = \frac{\pi}{4} \)

Check: \( \pi + \frac{\pi}{4} + \frac{\pi}{4} = \frac{3\pi}{2} \), which is true.

Compute:

- \( \cos 2A = \cos 2\pi = 1 \) - \( \cos 2B = \cos \frac{\pi}{2} = 0 \) - \( \cos 2C = \cos \frac{\pi}{2} = 0 \)

Sum: \( 1 + 0 + 0 = 1 \)

Option d): \( 1 - 4 \sin \pi \sin \frac{\pi}{4} \sin \frac{\pi}{4} = 1 - 4 \cdot 0 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1 - 0 = 1 \)

Matches again.

Conclusion

Testing multiple cases shows \( \cos 2A + \cos 2B + \cos 2C = 1 - 4 \sin A \sin B \sin C \). The correct answer is option d).

Final Answer: \( 1 - 4 \sin A \sin B \sin C \)