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My professor showed us how to solve $\cos(\theta - X)$ where $\cos(\theta) = \frac{3}{5}$ and is in Quadrant IV, and $\tan(X) = -\sqrt{3}$ and is in Quadrant II.
Since, $\cos(A-B) = \cos(A)\cos(B)+\sin(A)\sin(B)$, he solved it by doing: $$\frac{3}{5} \cdot \frac{-1}{2} + \frac{-4}{5} \cdot \frac{\sqrt{3}}{2}$$ $$\frac{-3-4\sqrt{3}}{10}$$
My question is, how did he get the values for, $\cos(A)$, $\cos(B)$, $\sin(A)$, and $\sin(B)$ in this equation?
Using that difference formula, we just need to deduce the following values $$\cos(\theta) = \frac{3}{5},\quad \cos(X) = -\frac{1}{2},\quad \sin(\theta) = -\frac{4}{5},\quad \text{ and } \quad \sin(X) = \frac{\sqrt{3}}{2}.$$ The first one is given, so we just need to determine the other three.
First let's show that $\sin(\theta) = -4/5$. Recall that $(\cos(\theta),\sin(\theta))$ is the point on the unit circle corresponding to $\theta$ (which is in Quadrant IV). So by the Pythagorean theorem, $$\sin^2(\theta) + \cos^2(\theta) = 1.$$ Since $\cos(\theta) = 3/5$, we get $\sin^2(\theta) = 16/25$. We noted that $\theta$ is in Quadrant IV, so we must have $\sin(\theta) = -4/5$ (as opposed to $+4/5)$.
To determine the remaining two values, note that the equation $\tan(X) = -\sqrt 3$ has a unique solution in Quadrant II (up to added integer multiples of $2\pi$). Namely, $X = \frac{2\pi}{3} + 2\pi k,$ for some integer $k$. Since $\sin$ and $\cos$ are $2\pi$-periodic, we get $$\sin(X) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt 3}{2} \qquad \text{ and } \qquad \cos(X) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}.$$
You have definitely seen trig identities such as $$\sin ^2x +\cos^2 x =1$$ And $$\sec ^2x=1+\tan ^2 x$$
Well you need to apply the identities to get the unknown parts from the given information.
In this case, your professor is using $A = \theta$ and $B = X$. You are given $\cos(A) = \frac{3}{5}$. From this, you know that for a right triangle, the adjacent side is 3 and the hypotenuse is 5. By the Pythagorean Theorem, this means that the opposite side is 4 since $3^{2} + 4^{2} = 5^{2}$. Also, since you are in the 4th quadrant, your opposite side must be negative ($y$ values are negative in the 4th quadrant). Thus, $\sin(A) = \frac{-4}{5}$.
For $\tan(B) = -\sqrt{3}$, you know that $\tan(B) = \frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}} = \frac{\sin(B)}{\cos(B)}$ (you are in the second quadrant, so your cosine is negative and your sine is positive).
