Answer
$t=\frac{(2k+1)\pi}{2}$, where $k$ is an integer
$y=-x$ when k is odd
$y=x$ when k is even
Work Step by Step
Remember that the slope of a tangent line is equal to $y-y_{1}=\frac{dy}{dx}(x-x_{1})$
$x=cos(t)$
$\frac{dx}{dt}=-sin(t)$
$y= sin(t)cos(t)$
$\frac{dy}{dt}= cos ^{2}(t)-sin ^{2}(t)$
We are given the point $(0,0)$ on the graph to find the tangents, so we plug in the coordinates for our x and y functions.
$0=cos(t)$
$t=\frac{(2k+1)\pi}{2}$, where $k$ is an integer
$0=sin(t)cos(t)$
$t=\frac{k\pi}{2}$, where $k$ is an integer
We now look for the values of $t$ where $x$ and $y$ are equal.
$t=\frac{k\pi}{2}$ has some values where t is not equal to zero, such as $k=0$, but for $t=\frac{(2k+1)\pi}{2}$, there is overlap in values, such as $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, so we'll use that function to evaluate
$\frac{dy}{dx}|_{T}= \frac{cos^{2}(\frac{(2k+1)\pi}{2})-sin^{2}(\frac{(2k+1)\pi}{2})}{sin(\frac{(2k+1)\pi}{2})}$
We now plug in two points and see what the values are
With $k=0$
$\frac{dy}{dx}= \frac{cos^{2}(\frac{(2(0)+1)\pi}{2})-sin^{2}(\frac{(2(0)+1)\pi}{2})}{sin(\frac{(2(0)+1)\pi}{2})} =\frac{cos^{2}(\frac{\pi}{2})-sin^{2}(\frac{\pi}{2})}{sin(\frac{\pi}{2})}$ = $ \frac{0-1}{-1}$ = $1$
This remains true whenever $K$ is even.
with $k=1$
$\frac{dy}{dx}= \frac{cos^{2}(\frac{(2(1)+1)\pi}{2})-sin^{2}(\frac{(2(1)+1)\pi}{2})}{sin(\frac{(2(1)+1)\pi}{2})} =\frac{cos^{2}(\frac{3\pi}{2})-sin^{2}(\frac{3\pi}{2})}{sin(\frac{3\pi}{2})}$ = $ \frac{0-1}{1}$ = $-1$
This remains true when $K$ is odd.
We now plug in these results into our tangent slope
When $K$ is even:
$y-0=1(x-0), y=x$
When $K$ is odd:
$y-0=-1(x-0), y=-x$
