Find the point of intersection of f(x):cos(x) and g(x):sin(2x) in the intervall 0;Pi/2

is there no way to find the point but by approximation?

Thanks

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- Apr 29th 2010, 09:00 AM #1

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- Apr 29th 2010, 09:04 AM #2
You can find the exact solutions! Recall that $\displaystyle \sin(2x)=2\sin x\cos x$.

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

So either $\displaystyle \sin x=\tfrac{1}{2}$ or $\displaystyle \cos x=0$. Now you can find exact solutions, keeping in mind they must fall in the interval $\displaystyle [0,\pi/2]$.

Can you take it from here?

- Apr 29th 2010, 09:04 AM #3

- Apr 29th 2010, 09:08 AM #4

- Apr 29th 2010, 09:10 AM #5

- Apr 29th 2010, 11:48 AM #6

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- Apr 29th 2010, 12:51 PM #7

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List of trigonometric identities - Wikipedia, the free encyclopedia contains probably anything you will ever need.

- Apr 29th 2010, 01:03 PM #8

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This is way to complicated, I dont even see which formula are relevant for my level of mathematics. I'd need a site on which the basic rules are listed, which i need for basic calculus exercises. Unfortunately i either found way to complex or way to simple sites...

- Apr 29th 2010, 09:33 PM #9
1. It's your responsibility to use an appropriate serach engine (such as Google) to find and review sites in order to find a site that suits your purpose.

2. $\displaystyle \sin(2x) = \cos(x) \Rightarrow \sin(2x) = \sin\left( \frac{\pi}{2} - x\right)$.

**Case 1:**$\displaystyle 2x = \frac{\pi}{2} - x$.

**Case 2:**$\displaystyle 2x = \pi - \left( \frac{\pi}{2} - x\right)$.

In each case, solve for x.

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