I've been asked to find the coordinates of the points on the curve:
$r = acos2\theta, -\dfrac{\pi}{4} \leq \theta \leq \dfrac{\pi}{4} $
where the tangents are parallel to the initial line.
The first thing I did was say:
$y = rsin\theta \space$ but $\space r = acos2\theta $ $\space \therefore \space y=acos2\theta sin\theta $
After rearranging, I got that:
$y = asin\theta (1 - 2sin^2\theta)$
$\Rightarrow \dfrac{dy}{d\theta} = acos\theta(1-2sin^2\theta) + asin\theta(-4sin\theta cos\theta)$
After rearranging this I found that:
$\dfrac{dy}{d\theta} = acos\theta(1-6sin^2\theta)$
So if I let:
$\dfrac{dy}{d\theta} = 0$
$\Rightarrow acos\theta(1-6sin^2\theta) = 0$
$\Rightarrow cos\theta = 0 \space$ or $\space sin\theta = \dfrac{1}{\sqrt{6}}$
If:
$cos\theta = 0 \Rightarrow \theta = \pm \dfrac{\pi}{2}$
This is outside my range so I can ignore these values.
If:
$sin\theta = \dfrac{1}{\sqrt{6}} \Rightarrow \theta = 0.421 \space (3s.f.)$
I haven't chosen any other values of $\theta$ here as they would be outside my range. However the book I am working from says that:
$\theta = \pm 0.421$
I don't see how this can be true as sin is an even function. Can anyone see where I've gone wrong or what I'm misunderstanding?
Thank you :D
As David Mitra pointed out in the comment below the question:
"$1−6sin^2 θ=0⟺|sinθ|=\dfrac{1}{\sqrt{6}} \space$ (More generally, $\sqrt{x^2}=|x|$)"