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Solve the equation \( sin^{-1}6x+sin^{-1}6\sqrt 3 x = -\frac{\pi}{2}. \)

1 Answer

  • \( sin^{-1}x+sin^{-1}y=sin^{-1}\: x\sqrt{1-y^2}+y\sqrt{1-x^2}\)
  • \(sin(-\frac{\pi}{2})=-1\)
By taking 6x in the place of x and \(6\sqrt3x\) in the place of y in the formula of \(sin^{-1}x+sin^{-1}y\) we get
\( sin^{-1} \bigg[ 6x \sqrt{1-108x^2}+6\sqrt 3 x\sqrt{1-36x^2} \bigg]\)\( = -\frac{\pi}{2}\)
Taking sin on both the sides
\(\Rightarrow\: 6x\sqrt{1-108x^2}+6\sqrt 3 x\sqrt{1-36x^2}=sin(-\frac{\pi}{2})=-1\)
We know that \( sin(-\frac{\pi}{2})=-1\)
\( \Rightarrow 6\sqrt 3 x \sqrt{1-36x^2}=-1-6x\sqrt{1-108x^2}\)
Squaring on both the sides we get
\( \Rightarrow 108x^2(1-36x^2)=1+36x^2(1-108x^2)+12x\sqrt{1-108x^2}\)
\(\Rightarrow\: 72x^2-1=12x\sqrt{1-108x^2}\)
Again squaring both the sides we get
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 21, 2013 by rvidyagovindarajan_1

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