Using identity: $\cos^2 A - \cos^2 B = \sin(A+B)\sin(B-A)$
$$\cos^2 15° - \cos^2 75° = \sin 90° \cdot \sin 60° = 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
Back to JEE Main 2021 Trigonometry
JEE Main 2021 — Mathematics Trigonometry
easy
mcq
2021
Official previous-year question
Verified 30 May 2026.
Question
The value of $\cos^2 15° - \cos^2 75°$ is:
Options
- A
$\frac{\sqrt{3}}{2}$
- B
$\frac{1}{2}$
- C
$\frac{1}{\sqrt{2}}$
- D
$\frac{\sqrt{3}}{4}$
Solution
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