I've tried to use the fact that $\tanθ = \frac{\sinθ}{\cosθ}.$
and the property $\sin^2θ + \cos^2θ = 1$ and trying to solve for sinθ but i'm running around in circles.
Also tried using the special triangles but the values i'm getting are way bigger than 1
On
The ratio of $\sin x $ over $\cos x$ is $7/4$
$$7^2+4^2=65$$
Thus you have $\sin x =\pm 7/ {\sqrt {65}} $ and $\cos x =\mp 4/{\sqrt {65}}$
With the condition of being reflex you have positive $\cos x$ and negative $\sin x $
On
I just figured it out. The notation was obscuring simple algebra for me.
Using: $$tanθ = sinθ/cosθ$$ and $$sin^2θ+cos^θ=1$$
We know $$tanθ = -7/24$$
therefore, $$sinθ/cosθ = -7/24$$
square the equation to get, $$sin^2θ/cos^2θ=49/576$$
use identity to get, $$sin^2θ/1-sin^2θ=49/576$$
rearrange to get, $$576sin^2θ=49(1-sin^2θ)$$
keep simplifying until we get $$sin^2θ=49/625$$
and therefore $$sinθ=-7/25$$ (negative because it's reflex and thus in 3rd quadrant.)
Can find cosθ using similar steps to give $24/25$
I would have spotted the solution quicker if I just replaced sinθ and cosθ with y for a simpler notation!
$\sin\theta=-\dfrac74\cos\theta$. So,
$$\dfrac{49}{16}\cos^2\theta+\cos^2\theta=1$$
$$\cos^2\theta=\dfrac{16}{65}$$
As $\theta$ is reflex and $\tan\theta<0$, we have $\cos\theta>0$ and $\sin\theta<0$.
$\cos\theta=\dfrac4{\sqrt{65}}$ and $\sin\theta=-\dfrac7{\sqrt{65}}$.