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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?
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?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need the "double angle" formula \[cos(2x)=cos^2(x)sin^2(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0perhaps i should write \[cos(2\theta)=cos^2(\theta)sin^2(\theta)\] and here \[\theta=sin^{1}(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in order to use the formula you need to know \[cos^2(sin^{1}(x))\] and \[sin^2(sin^{1}(x))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[sin(sin^{1}(x))=x\] so \[sin^2(sin^{1}(x))=x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only 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 \[\sqrt{1x^2}\] and therefore \[cos(sin^{1}(x))=\sqrt{1x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0another easy method is just remembering it because it is always the same. in any case you get \[cos^2(sin^{1}(x))sin^2(sin^{1}(x))=1x^2x^2=12x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A condition so that our answer is not complex is that x is less than or equal to 1 as well.
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