Answer
Equation of the tangent:
$y -1 = 2(x + 1)$
or
$y = 2x + 3$
Graph: (The green curve represents the tangent line, and the blue one represents the parametric equation.)
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Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{d(sin(t) + sin(2t))}{dt} = cos(t) + 2cos(2t)$
$\frac{d{(cos(t) + cos(2t))}}{dt} = -sin(t) -2sin(2t)$
Therefore: $\frac{dy}{dx} = \frac{cos(t)+2cos(2t)}{-sin(t)-2sin(2t)}$
2. At which point $x = -1$ and $y = 1$?
$-1 = cos(t) + cos(2t)$
$-1 = cos(t) + (cos^2(t) - sin^2(t))$
$-1 = cos(t) + cos^2(t) - (1 - cos^2(t) )$
$-1 = cos(t) + 2cos^2(t) - 1 $
$0 = cos(t) + 2cos^2(t)$
$0 = cos(t)(1 + 2cos(t))$
So: $cos(t) = 0$ or $1 + 2cos(t) = 0$
Possible values (between $-\pi$ and $\pi$):
$cos(t) = 0 \longrightarrow t = \frac {\pi} 2 \space and \space t =- \frac{\pi} 2$
$1 + 2cos(t) = 0$
$cos(t) = -\frac{1}{2} \longrightarrow t = \frac{2\pi}{3}$ and $t = -\frac{2\pi} 3$
Now, check these for values on the $y = sin(t) + sin(2t)$ equation:
$y = sin(\frac{\pi}{2}) + sin(2 \frac {\pi} 2) = 1 + 0 = 1$: $y = 1$
$y = sin(\frac{-\pi}{2}) + sin(2 (-\frac {\pi} 2)) = -1 + 0 = -1$: $y \neq 1$
$y = sin(\frac{2\pi}{3}) + sin(2 (\frac {2\pi} 3)) = \frac{\sqrt 3}{2} - \frac{\sqrt 3}{2} = 0$: $y \neq 1$
$y = sin(-\frac{2\pi}{3}) + sin(2 (-\frac {2\pi} 3)) = - \frac{\sqrt 3}{2} + \frac{\sqrt 3}{2} = 0$: $y \neq 1$
- The only answer that satisfies both equations is $t = \frac {\pi} 2$
3. Calculate $\frac{dy}{dx}$ for t = $\frac {\pi} 2$:
$\frac{cos(\frac {\pi} 2) + 2cos(2(\frac {\pi} 2))}{-sin(\frac {\pi} 2) -2sin(2(\frac {\pi} 2))} = \frac {0 + 2(-1)}{-(1) - (2(0))} = 2$
4. Determine the tangent equation:
$y-y_0 = \frac{dy}{dx}(x-x_0)$
$y - 1 = 2 (x - (-1))$
$y -1 = 2(x + 1)$
or
$y = 2x + 3$
