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Trignometry

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2 Apr 2008 21:45:57 IST

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The ratio of, the greatest value of 2-cosx+(sinx) to the power 2, to its least value.

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The ratio of, the greatest value of 2-cosx+(sinx) to the power 2, to its least value.

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sandeep ramesh

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2 Apr 2008 21:48:06 IST

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max of sinx-cosx = sqrt(2) and min is -sqrt(2) hence the answer is [2+sqrt(2)/2-sqrt(2)]^2 = [3+2sqrt(2)]^2

= 17+12sqrt(2)

sreeraman nagasubramaniyan

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2 Apr 2008 21:57:10 IST

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$-\sqrt{(a)^2+(-b)^2}<=a\sin(x) - b\cos(x)<=\sqrt{ (a)^2+(-b)^2} \\ \\ \Rightarrow -\sqrt{2}<=\sin(x) - \cos(x) <= \sqrt{2} \\ \\ \Rightarrow 2-\sqrt{2}<=2+\sin(x) - \cos(x) <=2+\sqrt{2} \\ \\ \mbox{Least is} \ 2-\sqrt{2} \ , \mbox{greatest is} \ 2+\sqrt{2} \\ \\ \mbox{Required ratio is} \ \frac{(2+\sqrt{2})^2}{2-\sqrt{2}} = 10+7\sqrt{2}$

sandeep ramesh

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2 Apr 2008 21:59:40 IST

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isnt the denominator square also???

Abram Benny

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4 Apr 2008 21:00:46 IST

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Thanx for the ans. But i have the answer as 13/4. I am not sure that the ans is right. But i really don't know how the ans. was reached. That's why i asked. Can u pls. make another reply?

sandeep ramesh

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4 Apr 2008 21:11:18 IST

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Oh so you meant sin^2x and not the whole expr square? :D

Well try writing sin^2x as 1-cos^2x and then you can easily find the max n min of that expr by differentiation :)

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