Question

\(\cos \left( \sin^{-1}(\frac{\sqrt{3}}{200})+\cos^{-1}(\frac{\sqrt{3}}{200}) \right)=\)

• A. \(\frac{\pi}{3}\)
• B. \(\frac{\pi}{4}\)
• C. \(\frac{\pi}{6}\)
• D. 1
• E. 0

Answer 1

Answer: (E)

Let \( x = \sin^{-1} \left( \frac{\sqrt{3}}{200} \right) \).
This means that \( \sin(x) = \frac{\sqrt{3}}{200} \).
Similarly, let \( y = \cos^{-1} \left( \frac{\sqrt{3}}{200} \right) \),
so \( \cos(y) = \frac{\sqrt{3}}{200} \).
Now, consider the sum \( x + y \). Using the identity: \[ \sin^{-1}(z) + \cos^{-1}(z) = \frac{\pi}{2} \quad \text{for any } z \text{ in the domain of } \sin^{-1} \text{ and } \cos^{-1} \] we have: \[ x + y = \frac{\pi}{2} \] Now, the original expression becomes: \[ \cos \left( \frac{\pi}{2} \right) = 0 \]

The correct option is (E) : \( 0\)

Answer 2

We are asked to find the value of \(\cos \left( \sin^{-1}(\frac{\sqrt{3}}{200})+\cos^{-1}(\frac{\sqrt{3}}{200}) \right)\).

We know that for any \(x\) in the interval \([-1, 1]\), \(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\).

Since \(\frac{\sqrt{3}}{200}\) is between -1 and 1, we can apply this identity.

Let \(A = \sin^{-1}(\frac{\sqrt{3}}{200})\) and \(B = \cos^{-1}(\frac{\sqrt{3}}{200})\). Then \(A + B = \frac{\pi}{2}\).

Therefore, we have \(\cos(A+B) = \cos\left(\frac{\pi}{2}\right)\).

Since \(\cos\left(\frac{\pi}{2}\right) = 0\), we have \(\cos \left( \sin^{-1}(\frac{\sqrt{3}}{200})+\cos^{-1}(\frac{\sqrt{3}}{200}) \right) = 0\).