Question

You are given that cos(A) =-3/5with A in Quadrant III, and cos(B) = 33/65 with B in Quadrant I. Find cos(A + B). Give your answer as a fraction.

Answer 1

Given

$\begin{gathered} \cos (A)=-\frac{3}{5} \\ \cos (B)=\frac{33}{65} \end{gathered}$

To evaluate:

$\cos (A+B)$

Use the identity:

$\cos (A+B)=\cos (A)\cos (B)-\sin (A)\sin (B)$

To get sin(A):

Using the trigonometric ratios:

$\begin{gathered} \cos \theta=\frac{\text{adj}}{\text{hyp}},\sin \theta=\frac{\text{opp}}{\text{hyp}} \\ \therefore \\ adj=3 \\ hyp=5 \end{gathered}$

Using the Pythagorean Theorem, we have:

$\begin{gathered} opp^2=hyp^2-adj^2 \\ opp^2=5^2-3^2 \\ opp^2=25-9 \\ opp^2=16 \\ opp=\sqrt[]{16}=4 \end{gathered}$

Hence,

$\sin (A)=\frac{4}{5}$

Since A is in Quadrant III,

$\sin (A)=-\frac{4}{5}$

To get sin(B):

Using the trigonometric ratios:

$\begin{gathered} \cos \theta=\frac{\text{adj}}{\text{hyp}},\sin \theta=\frac{\text{opp}}{\text{hyp}} \\ \therefore \\ adj=33 \\ hyp=65 \end{gathered}$

Using the Pythagorean Theorem, we have:

$\begin{gathered} opp^2=hyp^2-adj^2 \\ opp^2=65^2-33^2 \\ opp^2=4225-1089 \\ opp^2=3116 \\ opp=\sqrt[]{3116}=56 \end{gathered}$

Hence,

$\sin (B)=\frac{56}{65}$

To Evaluate cos (A + B):

Recall:

$\cos (A+B)=\cos (A)\cos (B)-\sin (A)\sin (B)$

Inputting all the necessary values, we have:

$\cos (A+B)=(-\frac{3}{5}\cdot\frac{33}{65})-(-\frac{4}{5}\cdot\frac{56}{65})$

Using a calculator, we have the answer to be:

$\cos (A+B)=\frac{5}{13}$