Find the extrema of a trigonometric function

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November 15th 2009, 08:20 AM
bosmith

Find the extrema of a trigonometric function

I am to find the extrema of the following function. f(x) = 2 sin x + cos 2x when 0 < x < 2pi. taking the derivative of the function I come up with 2 cos x -2 sin 2x. At this point I need to find the extrema on the interval 0 < x < 2pi. I think the solution is to evaluate 2 cos x = 0. being cos x = -2 and then, -2 sin 2x = 0. that being sin x = 2/2 = sin 1, being pi/2. OR, am I to use the double angle identity in the function and change the derivative of 2 cos x to, 1- 2sin^2 x - 2 sin 2x = 0. I am confused at this point. Can someone please help?
November 15th 2009, 08:36 AM
adkinsjr

Quote:

Originally Posted by bosmith

I am to find the extrema of the following function. f(x) = 2 sin x + cos 2x when 0 < x < 2pi. taking the derivative of the function I come up with 2 cos x -2 sin 2x. At this point I need to find the extrema on the interval 0 < x < 2pi. I think the solution is to evaluate 2 cos x = 0. being cos x = -2 and then, -2 sin 2x = 0. that being sin x = 2/2 = sin 1, being pi/2. OR, am I to use the double angle identity in the function and change the derivative of 2 cos x to, 1- 2sin^2 x - 2 sin 2x = 0. I am confused at this point. Can someone please help?

You're logic is all wrong when you set $2cos(x)=0$ and $2sin(2x)=0$ . Remember, in algebra, you can only to set factors equal to zero. You can't just set the terms of a sum equal to zero. You could factor the derivative:

$2cos(x)-2sin(2x)=0$

$2(cos(x)-sin(2x))=0$

This means that:

$cos(x)-sin(2x)=0$

$cos(x)=2cos(x)sin(x)$
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$1=2sin(x)$ Edit, MY logic was wrong on this step. See the post below.
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November 15th 2009, 08:39 AM
e^(i*pi)

Quote:

Originally Posted by adkinsjr

You're logic is all wrong in you set $2cos(x)=0$ and $2sin(2x)=0$ . Remember in algebra, you have to set factors equal to zero. You can't just set the terms of a sum equal to zero. You could factor the derivative:

$2cos(x)-2sin(2x)=0$

$2(cos(x)-sin(2x))=0$

This means that:

$cos(x)-sin(2x)=0$

$cos(x)=2cos(x)sin(x)$

$1=2sin(x)$

You're potentially dividing by 0 here, it would be better to factorise

$cos(x) = 2cos(x)sin(x)$

$2cos(x)sin(x)-cos(x) = 0$

$cos(x)(2sin(x)-1) = 0$

Either $cos(x) = 0$ or $sin(x) = \frac{1}{2}$
November 15th 2009, 08:45 AM
adkinsjr

Oh yes, thank you. Now MY logic is wrong when I cancel the cosines.
November 15th 2009, 08:53 AM
bosmith

adkinsjr

Thank you for assisting me. I also want to thank as well as e^(i*pi) . Alright!!!