# Thread: Compound Angle Formulas - Confused over evaluating

1. ## Compound Angle Formulas - Confused over evaluating

I have done questions like these before, but the ones I did were only in quadrant I. I really wish I knew how to draw the triangles on here for you, so you see everything that I am working with.

Question:
If x is in the interval $(\frac{\pi}{2}, \pi)$ and y is in the interval $(\pi, \frac {3\pi}{2})$ and cosx = $- \frac {5}{13}$ and tany = $\frac {4} {3}$, evaluate the following.

*I will put the thing to evaluate, then the answer (according to the sheet) under it, then put my attempt after*

a) sin (x + y)
Answer: $- \frac{16}{65}$

b) cos (x - y)
Answer: $- \frac {33}{65}$

c) tan (x - y)
Answer: $\frac {56}{33}$

Okay, now I will show my attempts (for some reason I got the answer for b, as the sheet answer for a. And the answer for b as the sheet answer for a.)

a) $\sin (x + y)$
= $\sin x \cos y + \cos x \sin y$
= $((\frac {12}{13}) (\frac {-4}{-5})) + ((\frac {-5}{13}) (\frac {-3}{-5})$
= $(\frac {48}{65}) + (\frac{-3}{13})$
= $\frac {33}{65}$

b) $cos (x - y)$
= $cosx cosy + sinx siny$
= $((\frac{-5}{13})(\frac{-4}{5}) + ((\frac {12}{13})(\frac{-3}{5}))$
= $\frac {4}{13} + \frac {-36}{65}$
= $\frac {-16}{65}$

c) $tan (x - y)$
= $\frac {tanx - tany} {1 + tanxtany}$
= $\frac {(\frac {12}{13}) - \frac {-3}{-4}} {1 + (\frac {12}{13}) (\frac {-3}{-4}) }$
= $\frac { \frac {9}{52} } {1+ \frac {9}{13} \frac {22}{13}}$
= $\frac {9} {88}$

2. Hello, Xenophobe!

If $x$ is in the interval $\left(\tfrac{\pi}{2}, \pi\right)$ and $y$ is in the interval $\left(\pi, \tfrac{3\pi}{2}\right)$

and $\cos x = -\tfrac{5}{13}$ and $\tan y = \tfrac{4}{3}$, evaluate the following:

$(a)\;\sin(x + y) \qquad (b)\;\cos(x - y) \qquad (c)\;\tan (x - y)$

$x$ is in Quadrant 2, and $\cos x \:=\: -\frac{5}{13} \:=\:\frac{adj}{hyp}$
We have: . $adj = -5,\;hyp = 13$
. . Using Pythagorus: . $opp \:=\:12$

$y$ is in Quadrant 3, and $\tan y \:=\:\frac{4}{3} \:=\:\frac{opp}{adj}$

In Quadrant 3: . $opp = -4,\; adj = -3,\; hyp = 5$

We have these values: . $\begin{array}{cccccccc}
\sin x &=& \dfrac{12}{13} & & \sin y &=& \text{-}\dfrac{4}{5} \\ \\[-3mm]
\cos x &=& \text{-}\dfrac{5}{13} & & \cos y &=& \text{-}\dfrac{3}{5} \\ \\[-3mm]
\tan x &=& \text{-}\dfrac{12}{5} & & \tan y &=& \dfrac{4}{3} \end{array}$

Try those in your compound angle formulas . . .