# Thread: Finding Equation of Tangent Line (Parametric) - # 2

1. ## Finding Equation of Tangent Line (Parametric) - # 2

$\displaystyle x = \cos t + \cos 2t$

$\displaystyle y = \sin t + \sin 2t$

$\displaystyle (-1,1)$

$\displaystyle (y - 1) = m(x + 1)$

$\displaystyle \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t$ ???

Find t

$\displaystyle -1 = \cos t + \cos 2t$

$\displaystyle -1 = \cos 3t$

$\displaystyle \arccos(-1) = 3t$

$\displaystyle 180 = 3t$

$\displaystyle t = 30$

2. ## Re: Finding Equation of Tangent Line (Parametric) - # 2

Originally Posted by Jason76
$\displaystyle x = \cos t + \cos 2t$

$\displaystyle y = \sin t + \sin 2t$

$\displaystyle (-1,1)$

$\displaystyle (y - 1) = m(x + 1)$

$\displaystyle \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t$ ???
I presume the "???" is because you just can't believe that is correct! You are right $\displaystyle \frac{a+ b}{c+ d}$ is NOT equal to $\displaystyle \frac{a}{c}+ \frac{b}{d}$

Find t

$\displaystyle -1 = \cos t + \cos 2t$

$\displaystyle -1 = \cos 3t$
Oh! Ouch! Surely you know that cos(t)+ cos(2t) is NOT equal to cos(3t)?
[/quote]

$\displaystyle \arccos(-1) = 3t$

$\displaystyle 180 = 3t$

$\displaystyle t = 30$[/QUOTE]

3. ## Re: Finding Equation of Tangent Line (Parametric) - # 2

How can we solve for t, since we know that this cos stuff cannot be added?

Originally Posted by HallsofIvy
I presume the "???" is because you just can't believe that is correct! You are right $\displaystyle \frac{a+ b}{c+ d}$ is NOT equal to $\displaystyle \frac{a}{c}+ \frac{b}{d}$

Oh! Ouch! Surely you know that cos(t)+ cos(2t) is NOT equal to cos(3t)?
$\displaystyle \arccos(-1) = 3t$

$\displaystyle 180 = 3t$

$\displaystyle t = 30$[/QUOTE][/QUOTE]

4. ## Re: Finding Equation of Tangent Line (Parametric) - # 2

Originally Posted by Jason76
$\displaystyle x = \cos t + \cos 2t$

$\displaystyle y = \sin t + \sin 2t$

$\displaystyle (-1,1)$

$\displaystyle (y - 1) = m(x + 1)$

$\displaystyle \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t$ ???

Find t

$\displaystyle -1 = \cos t + \cos 2t$ <snip>
$-1=\cos(t)+\cos^2(t)-\sin^2(t)=\cos(t)+2\cos^2(t)-1$

$0=\cos(t)+2\cos^2(t)$

Let $u=\cos(t)$

$0=2u^2+u =u(2u+1)$

$u=0 \vee u=-\dfrac 1 2$

$\cos(t)=0 \Rightarrow t=\dfrac {\pi}{2} + \pi k, ~k \in \mathbb{Z}$

$\cos(t)=-\dfrac 1 2 \Rightarrow t = \dfrac {2\pi}{3}+2\pi k \vee t=\dfrac{4\pi}{3} + 2\pi k, ~k\in \mathbb{Z}$