Question

Let $f:[-1,1]\rightarrow[\frac{\pi}{2},\frac{3\pi}{2}], f(x)=sin^{-1}x$ and $g:[-1,1]\rightarrow[0,\pi], g(x)=cos^{-1}x$ be two bijective functions. Inverse of bijective functions $f$ and $g$ are $h:[\frac{\pi}{2},\frac{3\pi}{2}]\rightarrow[-1,1], h(x)=sinx$ and $k:[0,\pi]\rightarrow[-1,1], k(x)=cosx$ respectively. Then $sin(sin^{-1}x-cos^{-1}x)$ is equal to

• A. 1
• B. 0
• C. -1
• D. $cos(2sin^{-1}x)$

Answer 1

Answer: (A)

1

Answer 2

Let $A = \sin^{-1}x$ and $B = \cos^{-1}x$. According to the problem definition:

1. $A \in [\frac{\pi}{2}, \frac{3\pi}{2}]$ such that $\sin A = x$.
2. $B \in [0, \pi]$ such that $\cos B = x$.

From these, we have $\sin A = \cos B$. We know that $\cos B = \sin(\frac{\pi}{2} - B)$. So, $\sin A = \sin(\frac{\pi}{2} - B)$.

This implies two possibilities for the relationship between $A$ and $B$: (i) $A = \frac{\pi}{2} - B + 2n\pi$ for some integer $n$. (ii) $A = \pi - (\frac{\pi}{2} - B) + 2n\pi = \frac{\pi}{2} + B + 2n\pi$ for some integer $n$.

Let's analyze the ranges of $A$ and $B$ for any $x \in [-1, 1]$:

• If $x \in [0, 1]$: $A \in [\frac{\pi}{2}, \pi]$ (since $\sin A \ge 0$ in this range). $B \in [0, \frac{\pi}{2}]$ (since $\cos B \ge 0$ in this range). For case (i): $A = \frac{\pi}{2} - B + 2n\pi$. Since $B \in [0, \frac{\pi}{2}]$, $\frac{\pi}{2} - B \in [0, \frac{\pi}{2}]$. For $A \in [\frac{\pi}{2}, \pi]$, this implies $A = \frac{\pi}{2}$ and $\frac{\pi}{2}-B = \frac{\pi}{2}$, so $B=0$. This occurs when $x=1$. In this specific case, $A-B = \frac{\pi}{2}-0 = \frac{\pi}{2}$. For case (ii): $A = \frac{\pi}{2} + B + 2n\pi$. Since $B \in [0, \frac{\pi}{2}]$, $\frac{\pi}{2} + B \in [\frac{\pi}{2}, \pi]$. This range matches $A \in [\frac{\pi}{2}, \pi]$. So, taking $n=0$, we have $A = \frac{\pi}{2} + B$. This gives $A-B = \frac{\pi}{2}$.

• If $x \in [-1, 0)$: $A \in (\pi, \frac{3\pi}{2}]$ (since $\sin A < 0$ in this range). $B \in (\frac{\pi}{2}, \pi]$ (since $\cos B < 0$ in this range). For case (i): $A = \frac{\pi}{2} - B + 2n\pi$. Since $B \in (\frac{\pi}{2}, \pi]$, $\frac{\pi}{2} - B \in [-\frac{\pi}{2}, 0)$. This range does not match $A \in (\pi, \frac{3\pi}{2}]$ for $n=0$. If $n=1$, $A = \frac{5\pi}{2}-B$, then $A \in [\frac{3\pi}{2}, 2\pi)$. This only matches $A=\frac{3\pi}{2}$ when $B=\pi$, which occurs when $x=-1$. In this specific case, $A-B = \frac{3\pi}{2}-\pi = \frac{\pi}{2}$. For case (ii): $A = \frac{\pi}{2} + B + 2n\pi$. Since $B \in (\frac{\pi}{2}, \pi]$, $\frac{\pi}{2} + B \in (\pi, \frac{3\pi}{2}]$. This range matches $A \in (\pi, \frac{3\pi}{2}]$. So, taking $n=0$, we have $A = \frac{\pi}{2} + B$. This gives $A-B = \frac{\pi}{2}$.

In all valid cases for $x \in [-1, 1]$, we consistently find that $A - B = \frac{\pi}{2}$. Therefore, $\sin(\sin^{-1}x - \cos^{-1}x) = \sin(A-B) = \sin(\frac{\pi}{2}) = 1$.