JEE PYQ: Trigonometric Ratios And Identities Question 6
Question 6 - 2020 (09 Jan Shift 1)
The value of $\cos^3\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^3\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$ is:
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(1) $\frac{1}{\sqrt{2}}$
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(2) $\frac{1}{2\sqrt{2}}$
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(3) $\frac{1}{2}$
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(4) $\frac{1}{4}$
Show Answer
Answer: (2) $\frac{1}{2\sqrt{2}}$
Solution
Expanding and simplifying yields $\frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{2}}$.
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