# Thread: A Trigonometry III Math Question

1. ## A Trigonometry III Math Question

Given that sin A = 4/5 and cos B = -1/3, and that A and B are in the same quadrant, without finding the angles A and B, find the value of sin (A+B). Thanks a lot

2. Hello, Leo!

Given that: . $\sin A = \tfrac{4}{5},\;\cos B = \text{-}\tfrac{1}{3}$ and that $A$ and $B$ are in the same quadrant,
without finding the angles $A$ and $B$, find the value of $\sin(A+B)$
Sine is positive in Quadrants 1 and 2.
Cosine is negative in Quadrants 2 and 3.
. . Hence, angles $A$ and $B$ are in Quadrant 2.

We have: . ${\color{blue}\sin A = \frac{4}{5}}$ in Quadrant 2.

Since $\sin^2\!A + \cos^2\!A \:=\:1$, we have: . $\left(\tfrac{4}{5}\right)^2 + \cos^2\!A \:=\:1 \quad\Rightarrow\quad \cos^2\!A \:=\:1-\tfrac{16}{25}$
. . $\cos^2\!A \:=\:\tfrac{9}{25} \quad\Rightarrow\quad \cos A \:=\:\pm\tfrac{3}{5}$
Since $A$ is in Quadrant 2: . ${\color{blue}\cos A \:=\:-\frac{3}{5}}$

We have: . ${\color{blue}\cos B = -\frac{1}{3}}$ in Quadrant 2.

Since $\sin^2\!B + \cos^2\!B \:=\:1$, we have: . $\sin^2\!B + \left(\text{-}\tfrac{1}{3}\right)^2 \:=\:1 \quad\Rightarrow\quad \sin^2\!B \:=\:1 - \tfrac{1}{9}$
. . $sin^2\!B \:=\:\tfrac{8}{9}\quad\Rightarrow\quad \sin B \:=\:\pm\tfrac{2\sqrt{2}}{3}$
Since $B$ is in Quadrant 2: . ${\color{blue}\sin B \:=\:\frac{2\sqrt{2}}{3}}$

We know that: . $\sin(A + B) \;=\;\sin A\cos B + \sin B\cos A$

Substitute: . $\sin(A + B) \;=\;\left(\frac{4}{5}\right)\left(-\frac{1}{3}\right) + \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{3}{5}\right) \;=\;\frac{-4 - 6\sqrt{2}}{15}$