It may be a long way to find it by mine, but it works. As $$\sin\theta+\sin\phi=a,~~~\cos\theta+ \cos\phi=b$$ then
$$\cos(\theta-\phi)=\cos\phi\cos\theta+\sin\phi\sin\theta=\frac{a^2+b^2-2}2$$ Now use: $$\cos x=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$
You found the proper value of $\cos \theta$ for the reference angle for $\theta$, not for $\theta$ itself. The reference angle is the angle in the first quadrant that has the same trig values in absolute value. Your final step is to find the proper sign of your trig value, based on the quadrant your $\theta$ is in.
Your question tells you that the tangent is negative, which is true for quadrants II and IV. It also tells you that the sine is positive, which is true for quadrants I and II. The only quadrant that fits both requirements is the second quadrant, so that's where $\theta$ lies. In that quadrant, cosine is negative, so you must change your answer from $\frac{15}{17}$ to $-\frac{15}{17}$.
One way to remember the signs for each quadrant is the mnemonic All Students Take Calculus. This tells you which trig ratios are positive in each quadrant: all, sine, tangent, and cosine for quadrants I, II, III, and IV respectively.
There are other ways, such as trig identities, to solve your problem that give the correct sign automatically, but it is good to have several ways to solve your problem.
Best Answer
Edit: Personally I prefer all the other answers, here's another way to think about it if you want more ideas
If $x^2 + y^2 = 1$, then $2x + 2y \frac{dy}{dx} = 0$
Re-arranging, $\frac{dy}{dx} = \frac{-x}{y}$. As you say, $\cos \theta = x$ and $\sin \theta = y$, giving you the slope of a tangent line as $- \cot \theta$. You can then use the equation: $$y - \sin \theta = - \cot \theta (x - \cos \theta)$$