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)=-(3)/(5) \\ \cos (B)=(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)=(4)/(5)`

Since A is in Quadrant III,

`\sin (A)=-(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)=(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)=(-(3)/(5)\cdot(33)/(65))-(-(4)/(5)\cdot(56)/(65))`

Using a calculator, we have the answer to be:

`\cos (A+B)=(5)/(13)`