KCET · Maths · Inverse Trigonometric Functions
The value of \(\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]\) is
- A \(\frac{\sqrt{5}}{3}\)
- B \(\frac{2 \sqrt{5}}{3}\)
- C \(\frac{4 \sqrt{5}}{9}\)
- D \(\frac{2 \sqrt{5}}{9}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 \sqrt{5}}{9}\)
Step-by-step Solution
Detailed explanation
We have, \(\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]\)
\(=\sin \left[\cos ^{-1}\left(2 \cdot\left(\frac{\sqrt{5}}{3}\right)^{2}-1\right)\right]\)
\(\left[\because 2 \cos ^{-1} x=\cos ^{-1}\left(2 x^{2}-1\right)\right]\)
\(=\sin \left[\cos ^{-1}\left(\frac{1}{9}\right)\right]\)
\(=\sin \left[\sin ^{-1} \sqrt{1-\left(\frac{1}{9}\right)^{2}}\right]\)
\(\left[\because \cos ^{-1} x=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\right]\)
\(=\sin \left[\sin ^{-1} \sqrt{\frac{80}{81}}\right]\)
\[
=\frac{\sqrt{80}}{9}=\frac{4 \sqrt{5}}{9}
\]
\(=\sin \left[\cos ^{-1}\left(2 \cdot\left(\frac{\sqrt{5}}{3}\right)^{2}-1\right)\right]\)
\(\left[\because 2 \cos ^{-1} x=\cos ^{-1}\left(2 x^{2}-1\right)\right]\)
\(=\sin \left[\cos ^{-1}\left(\frac{1}{9}\right)\right]\)
\(=\sin \left[\sin ^{-1} \sqrt{1-\left(\frac{1}{9}\right)^{2}}\right]\)
\(\left[\because \cos ^{-1} x=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\right]\)
\(=\sin \left[\sin ^{-1} \sqrt{\frac{80}{81}}\right]\)
\[
=\frac{\sqrt{80}}{9}=\frac{4 \sqrt{5}}{9}
\]
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