# Thread: Help with polar coordinates

1. ## Help with polar coordinates

Consider a curve given in polar coordinates by the equation
r = sec(theta +pi/4).
Find the equation of the tangent line (in Cartesian coordinates) at the point where  theta= 0.
Thanks.

2. Originally Posted by khuezy
Consider a curve given in polar coordinates by the equation
r = sec(theta +pi/4).
Find the equation of the tangent line (in Cartesian coordinates) at the point where  theta= 0.
Thanks.
We have that $\displaystyle r=\frac{1}{\cos\left(\theta+\frac{\pi}{4}\right)}$ Now a little trig will lead you to $\displaystyle r=\frac{\sqrt{2}}{\cos(\theta)-\sin(\theta)}$, or $\displaystyle 1=\frac{\sqrt{2}}{r\cos(\theta)-r\sin(\theta)}\implies{1=\frac{\sqrt{2}}{x-y}}$

I think you can take it from there.

3. ## ?

how did u do that trig?

4. Originally Posted by khuezy
how did u do that trig?
Using the two facts $\displaystyle \cos\left(A+B\right)=\cos\left(A\right)\cos\left(B \right)-\sin\left(A\right)\sin\left(B\right)$ and $\displaystyle \sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

So

\displaystyle \begin{aligned}\cos\left(x+\frac{\pi}{4}\right)&=\ cos(x)\cos\left(\frac{\pi}{4}\right)-\sin(x)\sin\left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2}\cos(x)-\frac{\sqrt{2}}{2}\sin(x)\\ &\implies\frac{1}{\cos\left(x+\frac{\pi}{4}\right) }\\ &=\frac{1}{\frac{\sqrt{2}}{2}\cos(x)-\frac{\sqrt{2}}{2}\sin(x)}\\ &=\frac{\sqrt{2}}{\cos(x)-\sin(x)}\end{aligned}

5. ## awesome

thanks, how did u get the root2/2 to go on top and the /2 to go away?