Solution Given that we only have the Tangent information, we cannot narrow the Quadrant to only one quadrant. In fact, Tangent is positive in the first and third quadrants, so we will have two possible pairs of solutions for the Sine and Cosine functions. As before, I will sketch a "generic" right triangle, treating \(4\) as the fraction \(\frac{{4}}{{1}}\) to construct the opposite, adjacent, and the angle in question:

Next we calculate the hypotenuse:

\[ \solve{ 1^2+4^2 &=&c^2\\ 1+16&=&c^2\\ 17&=&c^2\\ \sqrt{{17}}&=&c } \]

We only take the positive root since this is a distance. Now that we have the hypotenuse, we can write down the two possible solutions for the Sine and Cosine. Either both are positive, or both are negative:

\[ \solve{ \sin(\theta)=\dfrac{{4}}{\sqrt{{17}}}&&\cos=\dfrac{{1}}{\sqrt{{17}}}\\ &\text{ OR }&\\ \sin(\theta)=-\dfrac{{4}}{\sqrt{{17}}}&&\cos=-\dfrac{{1}}{\sqrt{ 17 } } } \]

Of course, we should rationalize those fractions to \[ \solve{ \sin(\theta) = \dfrac{4\sqrt{{17}} }{{17}} &&\cos=\dfrac{\sqrt{{17}} }{{17}}\\ &\text{ OR }&\\ \sin(\theta) = -\dfrac{4\sqrt{{17}} }{{17}} &&\cos=-\dfrac{\sqrt{{17}} }{{17}} } \]

Here we can see the two solutions as we might find them using the Unit Circle (at least to identify that there ARE two solutions):