Question

If an inverse trigonometric equation is given by ${{\sin }^{-1}}\left( 1-x \right)-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}$, then x is equal to
(a) 0, $\dfrac{1}{2}$
(b) 1, $\dfrac{1}{2}$
(c) 0
(d) $\dfrac{1}{2}$

Answer 1

Hint: We will apply the trigonometric identity given by ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$and inverse trigonometric identity given by ${{\cos }^{-1}}\left( p \right)+{{\sin }^{-1}}\left( p \right)=\dfrac{\pi }{2}$.

Complete step-by-step answer:
First we will consider the expression ${{\sin }^{-1}}\left( 1-x \right)-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}...(i)$. Now we will place the term ${{\sin }^{-1}}\left( 1-x \right)$ in equation (i) to the right side of the expression. Thus, we get $-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( 1-x \right)$.
Now we will apply the formula of inverse trigonometric identity which is given by ${{\cos }^{-1}}\left( p \right)+{{\sin }^{-1}}\left( p \right)=\dfrac{\pi }{2}$ to the right side of the expression $-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( 1-x \right)$. Therefore, we have
$\begin{align}
  & -2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}-\left( \dfrac{\pi }{2}-{{\cos }^{-1}}\left( 1-x \right) \right) \\
 & \Rightarrow -2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}-\dfrac{\pi }{2}+{{\cos }^{-1}}\left( 1-x \right) \\
 & \Rightarrow -2{{\sin }^{-1}}\left( x \right)={{\cos }^{-1}}\left( 1-x \right)...(ii) \\
\end{align}$
Now we will substitute the term ${{\sin }^{-1}}\left( x \right)$ equal to a. Thus we will have ${{\sin }^{-1}}\left( x \right)=a$. Now we will take the inverse sine term to the right side of the equation. Therefore, we will have $x=\sin \left( a \right)$. Now we will use the formula ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$ or, ${{\cos }^{2}}\left( p \right)=1-{{\sin }^{2}}\left( p \right)$ in the expression $x=\sin \left( a \right)$. After squaring both the sides of this equation we have ${{x}^{2}}={{\sin }^{2}}\left( a \right)$. Therefore, we get ${{x}^{2}}=1-{{\cos }^{2}}\left( a \right)$. Now we will find the value of a from this equation. This can be done as,
 $\begin{align}
  & {{x}^{2}}=1-{{\cos }^{2}}\left( a \right) \\
 & \Rightarrow {{\cos }^{2}}\left( a \right)=1-{{x}^{2}} \\
 & \Rightarrow \cos \left( a \right)=\sqrt{1-{{x}^{2}}} \\
 & \Rightarrow a={{\cos }^{-1}}\sqrt{1-{{x}^{2}}} \\
\end{align}$
As we have ${{\sin }^{-1}}\left( x \right)=a$ and $a={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}$. Thus, we get ${{\sin }^{-1}}\left( x \right)={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}...(iii)$. After substituting equation (iii) in (ii) we can have $-2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}={{\cos }^{-1}}\left( 1-x \right)$. Now we will apply substitution here by substituting x = $\sin \left( q \right)$. Thus we get $-2{{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\left( q \right)}={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right)$.
By applying formula ${{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1$ results into a new expression given by $-2{{\cos }^{-1}}\sqrt{{{\cos }^{2}}\left( q \right)}={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right)$.
Therefore, we have $-2{{\cos }^{-1}}\left( \cos \left( q \right) \right)={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right)$. By the formula ${{\cos }^{-1}}\left( \cos \left( x \right) \right)=x$ gives us the equation $-2q={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right)$.
Now we will place the inverse cosine term to the left side of the expression we will get $\cos \left( -2q \right)=1-\sin \left( q \right)$. As $\cos \left( -y \right)=\cos \left( y \right)$ thus we have $\cos \left( 2q \right)=1-\sin \left( q \right)$.
Now we will apply the formula $\cos \left( 2\theta \right)=1-2{{\sin }^{2}}\left( \theta \right)$. This results into $1-2{{\sin }^{2}}\left( q \right)=1-\sin \left( q \right)$ or, $1=2{{\sin }^{2}}\left( q \right)-\sin \left( q \right)+1$. At this step we will cancel 1 from both the sides of the expression. Thus we have
$\begin{align}
  & 2{{\sin }^{2}}\left( q \right)-\sin \left( q \right)=0 \\
 & \Rightarrow \sin \left( q \right)\left( 2\sin \left( q \right)-1 \right)=0 \\
\end{align}$
So, we can have either $\left( 2\sin \left( q \right)-1 \right)=0$ or $\sin \left( q \right)=0$. This further results into $\sin \left( q \right)=\dfrac{1}{2}$ or $\sin \left( q \right)=0$. Since, x = $\sin \left( q \right)$ therefore, we actually have that x = 0 or $\dfrac{1}{2}$.
Now we will substitute the value of x = 0 in the left hand side of equation (i). Therefore, we get
${{\sin }^{-1}}\left( 1-\left( 0 \right) \right)-2{{\sin }^{-1}}\left( 0 \right)$ which is equal to ${{\sin }^{-1}}\left( 1 \right)$. As we know that the value of $\sin \left( \dfrac{\pi }{2} \right)=1$ thus we get ${{\sin }^{-1}}\left( 1 \right)={{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right)$ or ${{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2}$. Clearly it is equal to the right side of equation (i). So, x = 0 satisfies the equation.
Now we will check for x = $\dfrac{1}{2}$. We will substitute the value to the left hand side of equation (i). Therefore, we get ${{\sin }^{-1}}\left( 1-\left( \dfrac{1}{2} \right) \right)-2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ which is equal to ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. As we know that the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ thus we get ${{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right) \right)-2{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right) \right)$. After applying the formula ${{\sin }^{-1}}\left( \sin \left( x \right) \right)=x$ results into
$\begin{align}
  & \dfrac{\pi }{6}-2\times \dfrac{\pi }{6}=\dfrac{\pi }{6}-\dfrac{\pi }{3} \\
 & \Rightarrow \dfrac{\pi }{6}-2\times \dfrac{\pi }{6}=\dfrac{\pi -2\pi }{6} \\
 & \Rightarrow \dfrac{\pi }{6}-2\times \dfrac{\pi }{6}=\dfrac{-\pi }{6} \\
\end{align}$
Clearly it is not equal to the right side of equation (i). Therefore, it does not satisfy the equation.
Hence, the correct option is (c).

Note: By placing the inverse trigonometric terms we should be aware that they need to be changed to the simple trigonometric term. As this solution consists of a lot of substitutions, one needs to put equation numbers as per their understanding.