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Question: If an inverse trigonometric equation is given by \({{\sin }^{-1}}\left( 1-x \right)-2{{\sin }^{-1}}\...

If an inverse trigonometric equation is given by sin⁡−1(1−x)−2sin⁡−1(x)=π2{{\sin }^{-1}}\left( 1-x \right)-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}, then x is equal to
(a) 0, 12\dfrac{1}{2}
(b) 1, 12\dfrac{1}{2}
(c) 0
(d) 12\dfrac{1}{2}

Explanation

Solution

Hint: We will apply the trigonometric identity given by sin⁡2(p)+cos⁡2(p)=1{{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1and inverse trigonometric identity given by cos⁡−1(p)+sin⁡−1(p)=π2{{\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(1−x)−2sin⁡−1(x)=π2...(i){{\sin }^{-1}}\left( 1-x \right)-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}...(i). Now we will place the term sin⁡−1(1−x){{\sin }^{-1}}\left( 1-x \right) in equation (i) to the right side of the expression. Thus, we get −2sin⁡−1(x)=π2−sin⁡−1(1−x)-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(p)+sin⁡−1(p)=π2{{\cos }^{-1}}\left( p \right)+{{\sin }^{-1}}\left( p \right)=\dfrac{\pi }{2} to the right side of the expression −2sin⁡−1(x)=π2−sin⁡−1(1−x)-2{{\sin }^{-1}}\left( x \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( 1-x \right). Therefore, we have
−2sin⁡−1(x)=π2−(π2−cos⁡−1(1−x)) ⇒−2sin⁡−1(x)=π2−π2+cos⁡−1(1−x) ⇒−2sin⁡−1(x)=cos⁡−1(1−x)...(ii) \begin{aligned} & -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{aligned}
Now we will substitute the term sin⁡−1(x){{\sin }^{-1}}\left( x \right) equal to a. Thus we will have sin⁡−1(x)=a{{\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⁡(a)x=\sin \left( a \right). Now we will use the formula sin⁡2(p)+cos⁡2(p)=1{{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1 or, cos⁡2(p)=1−sin⁡2(p){{\cos }^{2}}\left( p \right)=1-{{\sin }^{2}}\left( p \right) in the expression x=sin⁡(a)x=\sin \left( a \right). After squaring both the sides of this equation we have x2=sin⁡2(a){{x}^{2}}={{\sin }^{2}}\left( a \right). Therefore, we get x2=1−cos⁡2(a){{x}^{2}}=1-{{\cos }^{2}}\left( a \right). Now we will find the value of a from this equation. This can be done as,
x2=1−cos⁡2(a) ⇒cos⁡2(a)=1−x2 ⇒cos⁡(a)=1−x2 ⇒a=cos⁡−11−x2 \begin{aligned} & {{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{aligned}
As we have sin⁡−1(x)=a{{\sin }^{-1}}\left( x \right)=a and a=cos⁡−11−x2a={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}. Thus, we get sin⁡−1(x)=cos⁡−11−x2...(iii){{\sin }^{-1}}\left( x \right)={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}...(iii). After substituting equation (iii) in (ii) we can have −2cos⁡−11−x2=cos⁡−1(1−x)-2{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}={{\cos }^{-1}}\left( 1-x \right). Now we will apply substitution here by substituting x = sin⁡(q)\sin \left( q \right). Thus we get −2cos⁡−11−sin⁡2(q)=cos⁡−1(1−sin⁡(q))-2{{\cos }^{-1}}\sqrt{1-{{\sin }^{2}}\left( q \right)}={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right).
By applying formula sin⁡2(p)+cos⁡2(p)=1{{\sin }^{2}}\left( p \right)+{{\cos }^{2}}\left( p \right)=1 results into a new expression given by −2cos⁡−1cos⁡2(q)=cos⁡−1(1−sin⁡(q))-2{{\cos }^{-1}}\sqrt{{{\cos }^{2}}\left( q \right)}={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right).
Therefore, we have −2cos⁡−1(cos⁡(q))=cos⁡−1(1−sin⁡(q))-2{{\cos }^{-1}}\left( \cos \left( q \right) \right)={{\cos }^{-1}}\left( 1-\sin \left( q \right) \right). By the formula cos⁡−1(cos⁡(x))=x{{\cos }^{-1}}\left( \cos \left( x \right) \right)=x gives us the equation −2q=cos⁡−1(1−sin⁡(q))-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⁡(−2q)=1−sin⁡(q)\cos \left( -2q \right)=1-\sin \left( q \right). As cos⁡(−y)=cos⁡(y)\cos \left( -y \right)=\cos \left( y \right) thus we have cos⁡(2q)=1−sin⁡(q)\cos \left( 2q \right)=1-\sin \left( q \right).
Now we will apply the formula cos⁡(2θ)=1−2sin⁡2(θ)\cos \left( 2\theta \right)=1-2{{\sin }^{2}}\left( \theta \right). This results into 1−2sin⁡2(q)=1−sin⁡(q)1-2{{\sin }^{2}}\left( q \right)=1-\sin \left( q \right) or, 1=2sin⁡2(q)−sin⁡(q)+11=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
2sin⁡2(q)−sin⁡(q)=0 ⇒sin⁡(q)(2sin⁡(q)−1)=0 \begin{aligned} & 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{aligned}
So, we can have either (2sin⁡(q)−1)=0\left( 2\sin \left( q \right)-1 \right)=0 or sin⁡(q)=0\sin \left( q \right)=0. This further results into sin⁡(q)=12\sin \left( q \right)=\dfrac{1}{2} or sin⁡(q)=0\sin \left( q \right)=0. Since, x = sin⁡(q)\sin \left( q \right) therefore, we actually have that x = 0 or 12\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(1−(0))−2sin⁡−1(0){{\sin }^{-1}}\left( 1-\left( 0 \right) \right)-2{{\sin }^{-1}}\left( 0 \right) which is equal to sin⁡−1(1){{\sin }^{-1}}\left( 1 \right). As we know that the value of sin⁡(π2)=1\sin \left( \dfrac{\pi }{2} \right)=1 thus we get sin⁡−1(1)=sin⁡−1(sin⁡(π2)){{\sin }^{-1}}\left( 1 \right)={{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{2} \right) \right) or sin⁡−1(1)=π2{{\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 = 12\dfrac{1}{2}. We will substitute the value to the left hand side of equation (i). Therefore, we get sin⁡−1(1−(12))−2sin⁡−1(12){{\sin }^{-1}}\left( 1-\left( \dfrac{1}{2} \right) \right)-2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) which is equal to sin⁡−1(12)−2sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( \dfrac{1}{2} \right). As we know that the value of sin⁡(π6)=12\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2} thus we get sin⁡−1(sin⁡(π6))−2sin⁡−1(sin⁡(π6)){{\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(sin⁡(x))=x{{\sin }^{-1}}\left( \sin \left( x \right) \right)=x results into
π6−2×π6=π6−π3 ⇒π6−2×π6=π−2π6 ⇒π6−2×π6=−π6 \begin{aligned} & \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{aligned}
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.