# Thread: How to solve a trigometric function...

1. ## How to solve a trigometric function...

Hey everybody.
I got this math assignment and I've been trying to work this problem but I have NO idea how.

Can anyone help?:
[tex]squareroot[1-cos2theta] = 2sinsqaredtheta/MATH]

I apologize in advance if the math button didn't work...
I'm still new to this forum!

Any help is appreciated,
Stealth

2. Hello, Stealth!

I'll make up my own instructions . . .

Solve for $\displaystyle 0 \leq x < 2\pi\!:\quad \sqrt{1-\cos2\theta} \:=\:2\sin^2\theta$
We're expected to know the identity: .$\displaystyle \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1-\cos2\theta \:=\:2\sin^2\!\theta$

So the left side is: .$\displaystyle \sqrt{1-\cos2\theta} \:=\:\sqrt{2\sin^2\!\theta} \:=\:\sqrt{2}\,\sin\theta$

The equation becomes: .$\displaystyle \sqrt{2}\,\sin\theta \:=\:2\sin^2\!\theta \quad\Rightarrow\quad 2\sin^2\!\theta - \sqrt{2}\,\sin\theta \:=\:0$

Factor: .$\displaystyle \sin\theta\,(2\sin\theta - \sqrt{2}) \:=\:0$

And we have: .$\displaystyle \begin{array}{ccccccc}\sin\theta \:=\:0 & \Rightarrow & \boxed{\theta \:=\:0,\pi} \\ \\[-4mm] 2\sin\theta - \sqrt{2}\:=\:0 & \Rightarrow & \sin\theta \:=\:\dfrac{\sqrt{2}}{2} & \Rightarrow &\boxed{ \theta \:=\:\dfrac{\pi}{4},\:\dfrac{3\pi}{4}} \end{array}$

3. Wow! Thanks! I was looking at the sinsquared formula but I didn't know that I could manipulate the 2 to make it 2sinsquaredtheta = 1-cos2theta!

Thank you sooo much!

Stealth

4. ## Edit

Hi! I noticed something when I worked it out another way!:

We're expected to know the identity: .$\displaystyle \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1-\cos2\theta \:=\:2\sin^2\!\theta$

So the left side is: .$\displaystyle \sqrt{1-\cos2\theta} \:=\:\sqrt{2\sin^2\!\theta} \:=\:\sqrt{2}\,\sin\theta$

On the squareroot(2) part, would that be a plus/minus squareroot(2) because then wouldn't we also have the solutions of 5pi/4, and 7pi/4?

The equation becomes: .$\displaystyle \sqrt{2}\,\sin\theta \:=\:2\sin^2\!\theta \quad\Rightarrow\quad 2\sin^2\!\theta - \sqrt{2}\,\sin\theta \:=\:0$
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