Find for all 0<=x<=90 , the least and most value of

Am i supposed to do it like this :

For most value :

and just substitute the minimum value of cos and sin which is -1 into the equation ?

or the question means something else . Thanks .

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- Aug 13th 2009, 04:20 AMthereddevilstrigo(2)
Find for all 0<=x<=90 , the least and most value of

Am i supposed to do it like this :

For most value :

and just substitute the minimum value of cos and sin which is -1 into the equation ?

or the question means something else . Thanks . - Aug 13th 2009, 04:57 AMsongoku
- Aug 13th 2009, 04:58 AMmr fantastic
No. For starters it's impossible for sin x and cos x to both equal 1 at the same time (that is, for the same value of x). There are other reasons but one is enough.

For the given domain, you will find that the maximum and minimum values occur at the endpoints, that is, at x = 0 degrees and x = 90 degrees respectively. - Aug 13th 2009, 06:23 AMthereddevils
- Aug 13th 2009, 09:35 AMsongoku
Hi thereddevils

Your answer is maximum value. You optimize the value of the numerator and minimize the value of denominator. That will give you the maximum value :) - Aug 14th 2009, 01:59 AMthereddevils
- Aug 14th 2009, 05:28 AMmr fantastic
- Aug 14th 2009, 05:36 AMthereddevils
- Aug 15th 2009, 04:34 AMsongoku
- Aug 15th 2009, 06:09 AMluobo
- Aug 15th 2009, 08:50 AMthereddevils
- Aug 15th 2009, 09:08 AMsongoku
Hi thereddevils

As Mr. F has said : subs. the endpoints of the domain

When x = 0 --------------------> cos x = 1 and sin x = 0

When x = 90 degrees -------> cos x = 0 and sin x = 1

So, in this case, the maximum value of sin and cos is 1 and minimum value is 0.

When you want to find the maximum value of , you make the value of the numerator as big as possible and the denominator as small as possible.

Because the max. value of cos is 1, then you take x = 0 to optimize the numerator. For x = 0, the value of sin = 0, so you'll get :

To find the min. value, you make the numerator as small as possible and the denominator as big as possible. Give it a try (Sun) - Aug 15th 2009, 09:20 AMthereddevils
- Aug 15th 2009, 09:28 AMsongoku
You're welcome :)