KCET · Maths · Trigonometric Ratios & Identities
The value of \(\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}\) is
- A \(\theta\)
- B 1
- C \(\frac{1}{2}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Given, expression is \(\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}\).
We know that,
\[
\begin{aligned}
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \text { Therefore, } \sin \frac{5 \pi}{12} \sin \frac{\pi}{12} \\
& =\frac{1}{2}\left[\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)-\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)\right] \\
& =\frac{1}{2}\left[\cos \left(\frac{4 \pi}{12}\right)-\cos \left(\frac{6 \pi}{12}\right)\right] \\
& =\frac{1}{2}\left[\cos \left(\frac{\pi}{3}\right)-\cos \left(\frac{\pi}{2}\right)\right]=\frac{1}{2}\left[\frac{1}{2}-0\right]=\frac{1}{4} \\
&
\end{aligned}
\]
We know that,
\[
\begin{aligned}
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \text { Therefore, } \sin \frac{5 \pi}{12} \sin \frac{\pi}{12} \\
& =\frac{1}{2}\left[\cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)-\cos \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)\right] \\
& =\frac{1}{2}\left[\cos \left(\frac{4 \pi}{12}\right)-\cos \left(\frac{6 \pi}{12}\right)\right] \\
& =\frac{1}{2}\left[\cos \left(\frac{\pi}{3}\right)-\cos \left(\frac{\pi}{2}\right)\right]=\frac{1}{2}\left[\frac{1}{2}-0\right]=\frac{1}{4} \\
&
\end{aligned}
\]
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