Finding a tangent and normal line to a curve Given $$\gamma(t)=(2\cos(t)-\cos(2t),2\sin(t)-\sin(2t))$$
at $t=\pi/4$ find the equation for the tangent and normal line
I believe I know how to do this believe I it correctly yet my answer for both the tangent line differ from the text in the same way.  I may be making a stupid mistake but I have checked my work several times and believe it to be correct.
using the point $\langle\sqrt2,\sqrt2-1\rangle$ and tangent vector $\langle\sqrt2-1,1\rangle$
Then using the usual method of $P+tv$ to parametrize the line and solving x and y for t and substituting i obtain  
$$y+1-\sqrt2=-\frac{x-\sqrt2}{\sqrt2-1}$$  
whereas the book has the answer as
$$y-(\frac{1}{\sqrt2}-1)=-\frac{x-\sqrt2}{\sqrt2-1}$$
They only differ by the one term on the left side and my equation for the normal line differs in the same way so if I can figure this out the normal will follow.  
If you can put my mind at ease and tell me the books wrong I would be so grateful. Or I suppose hearing I am wrong could be beneficial too.
 A: Here is another approach, using parametric differentiation: 
First of all we have, $x(\frac{\pi}{4})=\sqrt{2}$, $ \ y(\frac{\pi}{4})=\sqrt{2}-1$. So we can write:
$$\frac{dy}{dx}\bigg|_{(\sqrt{2},\sqrt{2}-1)}=\frac{dy/dt\big|_{\frac{\pi}{4}}}{dx/dt\big|_{\frac{\pi}{4}}}=-\frac{\cos t-\cos2t\big|_{\frac{\pi}{4}}}{\sin t-\sin2t\big|_{\frac{\pi}{4}}}=-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}-1}=-\frac{\sqrt{2}}{\sqrt{2}-2}=\frac{1}{\sqrt{2}-1}$$
thus, we get for the tangent at $(\sqrt{2},\sqrt{2}-1)$:
$$y-y(\frac{\pi}{4})= \frac{dy}{dx}\bigg|_{(\sqrt{2},\sqrt{2}-1)}\cdot\big(x-x(\frac{\pi}{4})\big)$$ 
which yields
$$y-(\sqrt{2}-1)=\frac{x-\sqrt{2}}{\sqrt{2}-1}$$
for the tangent line to the curve $\gamma(t)$, at $t=\frac{\pi}{4}$ i.e. at the point $(\sqrt{2},\sqrt{2}-1)$. 
A: Neither of the stated answers in the original question correspond to a tangent or normal line at $t = \pi/4$.  The slope is negative, when in fact it should be positive as the calculation of the tangent vector so clearly shows.
Second, you can easily verify which of the two equations doesn't even pass through $t = \pi/4$ simply by substituting in $(\sqrt{2}, \sqrt{2}-1)$ into both equations and seeing which one is correct.
