# Thread: Find the extrema of a trigonometric function

1. ## 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?

2. 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 $\displaystyle 2cos(x)=0$ and $\displaystyle 2sin(2x)=0$. 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:

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

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

This means that:

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

$\displaystyle cos(x)=2cos(x)sin(x)$
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$\displaystyle 1=2sin(x)$ Edit, MY logic was wrong on this step. See the post below.
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You're logic is all wrong in you set $\displaystyle 2cos(x)=0$ and $\displaystyle 2sin(2x)=0$. 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:

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

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

This means that:

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

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

$\displaystyle 1=2sin(x)$
You're potentially dividing by 0 here, it would be better to factorise

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

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

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

Either $\displaystyle cos(x) = 0$ or $\displaystyle sin(x) = \frac{1}{2}$

4. Oh yes, thank you. Now MY logic is wrong when I cancel the cosines.

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

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### how to find relative extrema in trig functions

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