## anonymous 5 years ago 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?

1. Owlfred

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2. anonymous

you need the "double angle" formula $cos(2x)=cos^2(x)-sin^2(x)$

3. anonymous

perhaps i should write $cos(2\theta)=cos^2(\theta)-sin^2(\theta)$ and here $\theta=sin^{-1}(x)$

4. anonymous

in order to use the formula you need to know $cos^2(sin^{-1}(x))$ and $sin^2(sin^{-1}(x))$

5. anonymous

$sin(sin^{-1}(x))=x$ so $sin^2(sin^{-1}(x))=x^2$

6. anonymous

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 $cos(sin^{-1}(x))=\sqrt{1-x^2}$

7. anonymous

another easy method is just remembering it because it is always the same. in any case you get $cos^2(sin^{-1}(x))-sin^2(sin^{-1}(x))=1-x^2-x^2=1-2x^2$

8. anonymous

A condition so that our answer is not complex is that x is less than or equal to 1 as well.