Question

\( -\frac{2\pi}{5} \) is the principal value of

• A. \( \sin^{-1}\left[\sin\left(\frac{7\pi}{5}\right)\right] \)
• B. \( \tan^{-1}\left[\tan\left(\frac{7\pi}{5}\right)\right] \)
• C. \( \cos^{-1}\left[\cos\left(\frac{7\pi}{5}\right)\right] \)
• D. \( \sec^{-1}\left[\sec\left(\frac{7\pi}{5}\right)\right] \)

Hint:

For inverse trigonometric functions, always check the principal value range first. The value of \( \sin^{-1}x \) always lies in \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \).

Answer 1

The Correct Option is A

Step 1: Understand the required principal value.
We have to identify which inverse trigonometric expression has principal value:
\[ -\frac{2\pi}{5}. \]
The angle \( -\frac{2\pi}{5} \) lies in the interval:
\[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \]
This is the principal value range of \( \sin^{-1}x \).

Step 2: Simplify the given angle.
The angle inside the trigonometric functions is:
\[ \frac{7\pi}{5}. \]
Since:
\[ \frac{7\pi}{5}=2\pi-\frac{3\pi}{5}, \]
it lies in the third quadrant because:
\[ \pi<\frac{7\pi}{5}<\frac{3\pi}{2}. \]

Step 3: Evaluate the sine expression.
Using the periodicity of sine:
\[ \sin\left(\frac{7\pi}{5}\right)=\sin\left(2\pi-\frac{3\pi}{5}\right). \]
Since \( \sin(2\pi-\theta)=-\sin\theta \), we get:
\[ \sin\left(\frac{7\pi}{5}\right)=-\sin\left(\frac{3\pi}{5}\right). \]
Also,
\[ \sin\left(\frac{3\pi}{5}\right)=\sin\left(\pi-\frac{2\pi}{5}\right)=\sin\left(\frac{2\pi}{5}\right). \]
Therefore:
\[ \sin\left(\frac{7\pi}{5}\right)=-\sin\left(\frac{2\pi}{5}\right)=\sin\left(-\frac{2\pi}{5}\right). \]

Step 4: Apply principal value range of inverse sine.
The principal value range of \( \sin^{-1}x \) is:
\[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \]
Since \( -\frac{2\pi}{5} \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \), we get:
\[ \sin^{-1}\left[\sin\left(\frac{7\pi}{5}\right)\right] = -\frac{2\pi}{5}. \]

Step 5: Check the tangent option.
The principal value range of \( \tan^{-1}x \) is:
\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). \]
Now:
\[ \tan\left(\frac{7\pi}{5}\right)=\tan\left(\frac{7\pi}{5}-\pi\right)=\tan\left(\frac{2\pi}{5}\right). \]
So:
\[ \tan^{-1}\left[\tan\left(\frac{7\pi}{5}\right)\right]=\frac{2\pi}{5}, \]
not \( -\frac{2\pi}{5} \).

Step 6: Check the cosine and secant options.
The principal value range of \( \cos^{-1}x \) is:
\[ [0,\pi]. \]
So \( \cos^{-1}x \) cannot have a negative principal value. Hence option (C) is not correct.
Similarly, the principal value range of \( \sec^{-1}x \) is generally:
\[ [0,\pi] - \left\{\frac{\pi}{2}\right\}. \]
So option (D) also cannot give \( -\frac{2\pi}{5} \).

Step 7: Final conclusion.
Only option (A) gives the required principal value:
\[ -\frac{2\pi}{5}. \]
Final Answer:
\[ \boxed{\sin^{-1}\left[\sin\left(\frac{7\pi}{5}\right)\right]}. \]