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

Toolbox:
• $sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$
• $cos^{-1}x=sin^{-1}\sqrt{1-x^2}$
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 $sin^{-1}x+sin^{-1}(1-x)=$
$sin^{-1}\bigg[x\sqrt{1-(1-x)^2}+(1-x)\sqrt{1-x^2}\bigg]=cos^{-1}x$
$\Rightarrow\: sin^{-1} \bigg[ x \sqrt{2x-x^2}+(1-x)\sqrt{1-x^2} \bigg] = cos^{-1}x$
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
$x^2(2x-x^2)=x^2(1-x^2)$
$\Rightarrow\:x^2(2x-x^2-1+x^2)=0$
$x=0\:or\:x=\frac{1}{2}$
edited Mar 20, 2013