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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

When I do with trigometry I get problem...

And I need help to understand this. I know what vertical and horizontal means.

exemple this one i am working with

\(\displaystyle x=2\cos\theta\), \(\displaystyle y=sin2\theta\)

Vertical tangent:

We know that vertical tangent is when \(\displaystyle \frac{dx}{d\theta}=0\) and \(\displaystyle \frac{dy}{d\theta} \neq 0\)

and we got \(\displaystyle \frac{dx}{d\theta}= -2\sin\theta\) so we got \(\displaystyle 0=-2\sin\theta\) and when I solve it I get \(\displaystyle \theta=0\)

Horizontal tangent

We know that horizontal tangent is when \(\displaystyle \frac{dy}{d\theta}=0\) and \(\displaystyle \frac{dx}{d\theta} \neq 0\)

so we got \(\displaystyle \frac{dy}{d\theta}=2\cos2\theta\) and then we will get \(\displaystyle 0=2\cos2\theta\) and i get \(\displaystyle \theta=\frac{1}{2}\)

I understand after we solved for \(\displaystyle \theta\) we put it in the x and y to get the point. I got problem solving those equation with trigometry.

Regards,