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Choose the correct answer. If \(\sin^{-1}(1-x) -2 \sin^{-1} x=\large\frac {\pi}{2}, \) then \(x\) is equal to

\[(A) \quad 0, \frac {1}{2} \qquad (B) \quad 1, \frac {1} {2} \qquad (C) \quad 0 \qquad (D) \quad \frac {1}{2} \]
Can you answer this question?
 
 

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Toolbox:
  • \( sin\bigg(\large \frac{\pi}{2}+y \bigg) = cosy\)
  • \( cos2\theta = 1-2sin^2\theta\)
Ans: (C) $0$
Given $sin^{-1}(1-x) -2 sin^{-1} x=\large\frac {\pi}{2},$, let's rewrite the given equation as \( sin^{-1}(1-x)=\large\frac{\pi}{2}+2sin^{-1}x\)
 
Take $sin$ on both sides
\(\Rightarrow sin\big(sin^{-1}(1-x)\big)=sin\big(\large\frac{\pi}{2}+2sin^{-1}x\big)\)
 
Taking \(y=2sin^{-1}x,\:we\:get\:sin\big(\large\frac{\pi}{2}+2sin^{-1}x\big)=cos(2sin^{-1}x)\)
\( \Rightarrow 1-x=cos( \: 2sin^{-1}x)\)
 
Put \( sin^{-1}x=\theta \Rightarrow x = sin\theta\)
\( \Rightarrow\:1-sin\theta = cos2\theta\)
\( \Rightarrow 1-sin\theta=1-2sin^2\theta\)
\(\Rightarrow\:2sin^2\theta-sin\theta=0\)
\(\Rightarrow\:sin\theta(2sin\theta-1)=0\)
\( \Rightarrow sin\theta = 0 \: or \: sin\theta=\large\frac{1}{2}\)
\( \Rightarrow x = 0 \: or \: \frac{1}{2}\)
 
But \( x=\large\frac{1}{2}\) doesnot satisfy
\( \Rightarrow x = 0\)

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by rvidyagovindarajan_1
 

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