1. ## Compound Angle Formulas

Can someone describe to me clearly, how you would solve these questions:

1. Angles x and y are located in the first quadrant such that sinx=4/5 and cosy=7/25.
a) determine an exact value for cos x
b) determine an exact value for sin y
c) determine an exact value for sin(x+y)

2. angle x lies in the third quadrant, and tanx=7/24
a) determine an exact value for cos2x
b) determine an exact value for sin2x

for 1 a and b I got it, but for 1 c I am not sure how to approach it

2. Hello, skeske1234!

1. Angles $\displaystyle x$ and $\displaystyle y$ are in the Quadrant 1 so that: $\displaystyle \sin x=\tfrac{4}{5},\;\cos y=\tfrac{7}{25}$

a) Determine an exact value for $\displaystyle \cos x\;\;{\color{blue}\tfrac{3}{5}}$
b) Determine an exact value for $\displaystyle \sin y\;\;{\color{blue}\tfrac{24}{25}}$
c) Determine an exact value for $\displaystyle \sin(x+y)$

We're expected to know this identity: .$\displaystyle \sin(x+y) \:=\:\sin x\cos y + \cos x\sin y$

So we have: .$\displaystyle \sin(x+y) \;=\;\tfrac{4}{5}\!\cdot\!\tfrac{7}{25} + \tfrac{3}{5}\!\cdot\!\tfrac{24}{25} \;=\;\tfrac{28}{125} + \tfrac{72}{125} \;=\;\tfrac{100}{125} \;=\;\frac{4}{5}$

2. Angle $\displaystyle x$ is in the quadrant 3, and $\displaystyle \tan x= \tfrac{7}{24}$

a) Determine an exact value for $\displaystyle \cos2x$
b) Determine an exact value for $\displaystyle \sin2x$
From $\displaystyle \tan x = \tfrac{7}{24}$ in Quadrant 3, we have: .$\displaystyle \begin{array}{c}\sin x \:=\:\text{-}\frac{7}{25} \\ \\[-3mm] \cos x \:=\:\text{-}\frac{24}{25}\end{array}$

a) We're expected to know that: .$\displaystyle \cos2x \:=\:\cos^2\!x - \sin^2\!x$

So we have: .$\displaystyle \cos2x \;=\;\left(\text{-}\tfrac{24}{25}\right)^2 - \left(\text{-}\tfrac{7}{25}\right)^2 \;=\;\tfrac{576}{625} - \tfrac{49}{625} \;=\;\frac{527}{625}$

b) We're expected to know that: .$\displaystyle \sin2x \:=\:2\sin x\cos x$

So we have: .$\displaystyle \sin2x \;=\;2\left(\text{-}\tfrac{7}{25}\right)\left(\text{-}\tfrac{24}{25}\right) \;=\;\frac{336}{625}$