# Thread: Differentiation and trigonometry question

1. ## Differentiation and trigonometry question

A curve has parametric equations

$\displaystyle x=\theta + sin\theta , y = 1 + cos\theta, (0 \leq \theta \leq \pi)$.

Show that

$\displaystyle (\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 4cos^2(\frac{\theta}{2})$

I differentiate the parametric equations, square them and get

$\displaystyle cos^2\theta +2cos\theta + 1 + sin^2\theta$, and I know that somehow I need to show that it's equal to $\displaystyle 4cos^2(\frac{\theta}{2}$ but I have no idea - I've looked at some basic trigonometric identities to simplify it but I get nowhere. Is there anyone who can help me with this?

Thanks if you have the time

2. Originally Posted by db5vry
A curve has parametric equations

$\displaystyle x=\theta + sin\theta , y = 1 + cos\theta, (0 \leq \theta \leq \pi)$.

Show that

$\displaystyle (\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2 = 4cos^2(\frac{\theta}{2})$

I differentiate the parametric equations, square them and get

$\displaystyle cos^2\theta +2cos\theta + 1 + sin^2\theta$, and I know that somehow I need to show that it's equal to $\displaystyle 4cos^2(\frac{\theta}{2}$ but I have no idea - I've looked at some basic trigonometric identities to simplify it but I get nowhere. Is there anyone who can help me with this?

Thanks if you have the time
Since $\displaystyle sin^2(\theta)+cos^2(\theta)=1$, then :
$\displaystyle cos^2(\theta) +2cos(\theta) + 1 + sin^2(\theta)=2cos(\theta)+2=2( \, cos(\theta) + 1 \, )$.

Do you think you can do something if you know that $\displaystyle cos(\theta)=cos( \, 2 \, \frac{\theta}{2} \, )$ ?

3. Originally Posted by Miss
Since $\displaystyle sin^2(\theta)+cos^2(\theta)=1$, then :
$\displaystyle cos^2(\theta) +2cos(\theta) + 1 + sin^2(\theta)=2cos(\theta)+2=2( \, cos(\theta) + 1 \, )$.

Do you think you can do something if you know that $\displaystyle cos(\theta)=cos( \, 2 \, \frac{\theta}{2} \, )$ ?
I didn't think of writing it in that form! Well following from your instructions I got

$\displaystyle 2(cos(\frac{2\theta}{2}) + 1)$

$\displaystyle cos^2\theta = \frac{1}{2} + cos2\theta$

$\displaystyle 2cos^2\theta = 1 + 2cos2\theta$

but from this I end up with $\displaystyle 4cos^2\theta$ which isn't the same as $\displaystyle 4cos^2(\frac{\theta}{2})$. Any ideas?

4. Originally Posted by db5vry
I didn't think of writing it in that form! Well following from your instructions I got

$\displaystyle 2(cos(\frac{2\theta}{2}) + 1)$

$\displaystyle cos^2\theta = \frac{1}{2} + cos2\theta$

$\displaystyle 2cos^2\theta = 1 + 2cos2\theta$

but from this I end up with $\displaystyle 4cos^2\theta$ which isn't the same as $\displaystyle 4cos^2(\frac{\theta}{2})$. Any ideas?
Nope.
Replace $\displaystyle cos( \, 2 \, \frac{\theta}{2} \, )$ by $\displaystyle 2cos^2( \, \frac{\theta}{2} \, ) - 1$.
This comes from the identity $\displaystyle cos(2x)=2cos^2(x)-1$

5. Originally Posted by Miss
Nope.
Replace $\displaystyle cos( \, 2 \, \frac{\theta}{2} \, )$ by $\displaystyle 2cos^2( \, \frac{\theta}{2} \, ) - 1$.
This comes from the identity $\displaystyle cos(2x)=2cos^2(x)-1$
I see! I was trying to derive it from going another way around which didn't work!

So with $\displaystyle 2(cos\theta + 1)$

and $\displaystyle cos2\theta = 2cos^2\theta - 1$

it gives $\displaystyle 2(2cos^2(\frac{\theta}{2}))$ where the -1 and 1 cancel, therefore, we get

$\displaystyle 4cos^2(\frac{\theta}{2}))$ which is what I was looking for!

Thanks so much. That was really useful