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Find the values of x which satisfy the equation \[ sin^{-1}x+sin^{-1}(1-x)=cos^{-1}x.\]

1 Answer

  • \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]\)
  • \( cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
By taking 1-x in the place of y in the formula of \(sin^{-1}x+sin^{-1}y\)
the given equation can be written as \(sin^{-1}x+sin^{-1}(1-x)=\)
\(\Rightarrow\: sin^{-1} \bigg[ x \sqrt{2x-x^2}+(1-x)\sqrt{1-x^2} \bigg] = cos^{-1}x\)
But we know that \(cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
\(\Rightarrow\:sin^{-1} \bigg[ x \sqrt{2x-x^2}+(1-x)\sqrt{1-x^2} \bigg] = sin^{-1}[\sqrt{1-x^2} ]\)
\( \Rightarrow x\sqrt{2x-x^2}+(1-x)\sqrt{1-x^2}=\sqrt{1-x^2}\)
\( \Rightarrow x\sqrt{2x-x^2}=\sqrt{1-x^2} (1-(1-x))=\sqrt{1-x^2}.x\)
Squaring both the sides we get
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 20, 2013 by rvidyagovindarajan_1

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