Find the exact value of cos(2sin^-1 (x)), where x>0 The 2 suggests that I should use the double angle formula, what then?

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Find the exact value of cos(2sin^-1 (x)), where x>0 The 2 suggests that I should use the double angle formula, what then?

Mathematics
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you need the "double angle" formula \[cos(2x)=cos^2(x)-sin^2(x)\]
perhaps i should write \[cos(2\theta)=cos^2(\theta)-sin^2(\theta)\] and here \[\theta=sin^{-1}(x)\]

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in order to use the formula you need to know \[cos^2(sin^{-1}(x))\] and \[sin^2(sin^{-1}(x))\]
\[sin(sin^{-1}(x))=x\] so \[sin^2(sin^{-1}(x))=x^2\]
only challenge is finding \[cos(sin^{-1}(x))\] easy method is to draw a right triangle and label one angle \[sin^{-1}(x)\] the angle whose sine is x. then make the "opposite side" x and the hypotenuse 1 so that opposite/hypotenuse = x. by pythagoras the other side is \[\sqrt{1-x^2}\] and therefore \[cos(sin^{-1}(x))=\sqrt{1-x^2}\]
another easy method is just remembering it because it is always the same. in any case you get \[cos^2(sin^{-1}(x))-sin^2(sin^{-1}(x))=1-x^2-x^2=1-2x^2\]
A condition so that our answer is not complex is that x is less than or equal to 1 as well.

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