# [SOLVED] Cosine of the Sum of Two Angles

• Mar 17th 2010, 06:40 PM
beardedoneder
[SOLVED] Cosine of the Sum of Two Angles
Given sin $\displaystyle \theta$ = $\displaystyle \frac{-5}{13}$ in quadrant III, tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$ in quadrant II; find the value of sin($\displaystyle \theta$ + $\displaystyle \phi$):

I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

Please tell me what I'm doing wrong.
• Mar 17th 2010, 06:58 PM
bigwave
Quote:

Originally Posted by beardedoneder
Given sin $\displaystyle \theta$ = $\displaystyle \frac{-5}{13}$ in quadrant III, tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$ in quadrant II; find the value of sin($\displaystyle \theta$ + $\displaystyle \phi$):

I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

Please tell me what I'm doing wrong.

first if tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$is in Q2
or actually should be $\displaystyle \tan\phi= \frac{8}{-15}$ if it is in Q2
then $\displaystyle \sin\phi = \frac{8}{17}$

therefore

$\displaystyle \sin(\theta + \phi)=\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \Rightarrow \frac{75}{221} - \frac{96}{221} \Rightarrow -\frac{21}{221}$

just got too late with latex
• Mar 17th 2010, 07:01 PM
skeeter
Quote:

Originally Posted by beardedoneder
Given sin $\displaystyle \theta$ = $\displaystyle \frac{-5}{13}$ in quadrant III, tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$ in quadrant II; find the value of sin($\displaystyle \theta$ + $\displaystyle \phi$):

I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

Please tell me what I'm doing wrong.

• Mar 17th 2010, 07:08 PM
pickslides
I would do it like this

Quote:

Originally Posted by beardedoneder
Given sin $\displaystyle \theta$ = $\displaystyle \frac{-5}{13}$ in quadrant III,

use $\displaystyle \sin^2\theta+\cos^2\theta = 1$ to find $\displaystyle \cos\theta$

Quote:

Originally Posted by beardedoneder
tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$ in quadrant II;

use $\displaystyle 1+\tan^2\phi = \sec^2\phi$ to find $\displaystyle \cos\phi$

and then $\displaystyle \sin^2\phi+\cos^2\phi = 1$ to find $\displaystyle \sin\phi$
• Mar 17th 2010, 07:22 PM
Soroban
Hello, beardedoneder!

Quote:

Given: .$\displaystyle \begin{array}{ccc}\sin\theta \:=\:\text{-}\frac{5}{13}& \text{in quadrant III} \\ \\ [-3mm] \tan\phi \:=\:\text{-}\frac{8}{15} & \text{in quadrant II}\end{array}$

find the value of: .$\displaystyle \sin(\theta + \phi)$

We have: .$\displaystyle \sin\theta \:=\:-\frac{5}{13} \:=\:\frac{opp}{hyp}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = -5,\;hyp = 13$

Pythagorus says: .$\displaystyle adj = \pm12$

. . In Quadrant III: $\displaystyle adj = -12$

Hence: .$\displaystyle \begin{Bmatrix} \sin\theta &=& \text{-}\dfrac{5}{13} \\ \\[-3mm] \cos\theta &=& \text{-}\dfrac{12}{13} \end{Bmatrix}$ .[1]

We have: .$\displaystyle \tan \phi \:=\:-\frac{8}{15} \:=\:\frac{opp}{adj}$

$\displaystyle \phi$ is in Quadrant II with: $\displaystyle opp = 8,\;adj = -15$

Pythagorus says: .$\displaystyle hyp = 17$

Hence: .$\displaystyle \begin{Bmatrix}\sin\phi &=& \dfrac{8}{17} \\ \\[-3mm] \cos\phi &=& \text{-}\dfrac{15}{17} \end{Bmatrix}$ .[2]

We want: .$\displaystyle \sin(\theta + \phi) \;=\;\sin\theta\cos\phi + \cos\theta\sin\phi$

Substitute [1] and [2]:

$\displaystyle \sin(\theta + \phi) \;=\;\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \;=\;\frac{75}{221} - \frac{96}{221} \;=\;-\frac{21}{221}$

• Apr 7th 2010, 04:02 PM
beardedoneder
Quote:

Originally Posted by beardedoneder
Given sin $\displaystyle \theta$ = $\displaystyle \frac{-5}{13}$ in quadrant III, tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$ in quadrant II; find the value of sin($\displaystyle \theta$ + $\displaystyle \phi$):

I've posted my solution -- which the answer key says is wrong -- here, http://imagebin.ca/img/tnINM0J.png .

Please tell me what I'm doing wrong.

I got a copy of the Solution Key. It solves the problem the same way I do, but it gives a product of $\displaystyle \frac{225}{221}$ for $\displaystyle -\frac{5}{13} \cdot -\frac{15}{17}$ instead of $\displaystyle \frac{75}{221}$. Which is why the answer key gives an answer of $\displaystyle \frac{129}{221}$, instead of $\displaystyle -\frac{21}{221}$. My solution and answer were correct. (Clapping)
• Apr 7th 2010, 04:11 PM
beardedoneder
Did you mean to say, ...
Quote:

Originally Posted by bigwave
first if tan $\displaystyle \phi$ = $\displaystyle \frac{-8}{15}$is in Q2
or actually should be $\displaystyle \tan\phi= \frac{8}{-15}$ if it is in Q2
then $\displaystyle \sin\phi = \frac{8}{17}$

therefore

$\displaystyle \sin(\theta + \phi)=\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \Rightarrow \frac{75}{221} - \frac{96}{221} \Rightarrow -\frac{21}{221}$

just got too late with latex

Did you mean to say, $\displaystyle -\frac{12}{13} \cdot \frac{8}{17}$ ?
• Apr 7th 2010, 04:14 PM
beardedoneder
Did you mean to say, ...
Quote:

Originally Posted by Soroban
Hello, beardedoneder!

Substitute [1] and [2]:

$\displaystyle \sin(\theta + \phi) \;=\;\left(\text{-}\frac{5}{13}\right)\left(\text{-}\frac{15}{17}\right) + \left(\text{-}\frac{12}{13}\right)\left(\frac{8}{15}\right) \;=\;\frac{75}{221} - \frac{96}{221} \;=\;-\frac{21}{221}$

Did you mean to say, $\displaystyle -\frac{12}{13} \cdot \frac{8}{17}$ ?