MHT CET · Maths · Trigonometric Ratios & Identities
The value of \(\sin ^{2}\left(\frac{\pi}{8}\right)=\)
- A \(\frac{\sqrt{2}+1}{2 \sqrt{2}}\)
- B \(\frac{\sqrt{5}+1}{2 \sqrt{2}}\)
- C \(\frac{\sqrt{5}-1}{2 \sqrt{2}}\)
- D \(\frac{\sqrt{2}-1}{2 \sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{2}-1}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
\sin ^{2}\left(\frac{\pi}{8}\right) &=\frac{1-\cos 2\left(\frac{\pi}{8}\right)}{2} \\
&=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{1-\frac{1}{\sqrt{2}}}{2}=\frac{\sqrt{2}-1}{2 \sqrt{2}}
\end{aligned}
\)
\begin{aligned}
\sin ^{2}\left(\frac{\pi}{8}\right) &=\frac{1-\cos 2\left(\frac{\pi}{8}\right)}{2} \\
&=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{1-\frac{1}{\sqrt{2}}}{2}=\frac{\sqrt{2}-1}{2 \sqrt{2}}
\end{aligned}
\)
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