# Trigonometry Addition and Subtraction Formula

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?

• In this case, he is using $A = \theta$ and $B = X$. – Jack Moody Nov 14 '18 at 0:36

## 3 Answers

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).