Answer
Tangents: $1 \space and \space -1$
Their equations: $y = x$ and $y = -x$
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Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{d(sin(t)cos(t))}{dt} = (sin(t))'(cos(t)) + (sin(t))(cos(t))'$
$\frac{dy}{dt} = cos(t)cos(t) + sin(t) (-sin(t))$
$\frac{dy}{dt} = cos^2(t) - sin^2(t) = cos(2t)$
$\frac{dx}{dt} = \frac{d(cos(t))}{dt} = -sin(t)$
Thus, $\frac{dy}{dx} = \frac{cos(2t)}{-sin(t)}$
2. Find the tangents at $(0,0)$
$x = cos(t)$
$0 = cos(t)$
Thus, $t = \frac{\pi}{2}$ or $t = \frac{3 \pi}{2}$
- If cos(t) is equal to 0, $y$ is also equal to 0. (y = sin(t) * 0 = 0)
$\frac{dy}{dx} = \frac{cos(2(\frac{\pi}{2}))}{-sin(\frac{\pi}{2})} = \frac{cos(\pi)}{-1} = \frac 1 {-1} = -1$
$\frac{dy}{dx} = \frac{cos(2(\frac{3\pi}{2}))}{-sin(\frac{3\pi}{2})} = \frac{cos(3\pi)}{-(-1)} = \frac 1 {1} = 1$
3. Find their equations:
$y - y_0 = (1)(x - x_0)$
$y = (x)$
$y = x$
and
$y - y_0 = (-1)(x - x_0)$
$y = -1(x)$
$y = -x$
