# I need help evaluating an expression under specific conditions

• Apr 15th 2013, 06:01 PM
AKPunk49
I need help evaluating an expression under specific conditions

Evaluate the expression under the given conditions.tan(θ + http://www.webassign.net/images/phi.gif); cos θ = −1/3, θ in Quadrant III, sin http://www.webassign.net/images/phi.gif= 1/4, http://www.webassign.net/images/phi.gif in Quadrant 4

Copied from my math homework, I'm not getting the right answer.

• Apr 15th 2013, 06:51 PM
chiro
Re: I need help evaluating an expression under specific conditions
Hey AKPunk49.

Can you show us what you tried: (Hint: did you try using the angle addition formulae?)?
• Apr 15th 2013, 07:01 PM
Prove It
Re: I need help evaluating an expression under specific conditions
$\displaystyle \sin{(\phi)}$ can not possibly be $\displaystyle \frac{1}{4}$ in the fourth quadrant. Why?
• Apr 15th 2013, 07:30 PM
AKPunk49
Re: I need help evaluating an expression under specific conditions
Fixed it, sorry.
• Apr 15th 2013, 07:41 PM
Prove It
Re: I need help evaluating an expression under specific conditions
It doesn't change the fact that $\displaystyle \sin{(\phi)}$ can not possibly be positive in the fourth quadrant. In fact, I don't see anything different at all...
• Apr 16th 2013, 02:58 PM
AKPunk49
Re: I need help evaluating an expression under specific conditions
Oh my-lanta...its in Quad. 2. I'm having an odd week, sorry for that.

The book tells me to make two right triangles in the corresponding quads and then find the identities, which I did. Then I plugged it all into the addition formula for tangent. I got -sr8/3+1/4 all divided by 1-(-sr8/3)(1/4)

Sorry if that's confusing...
• Apr 16th 2013, 03:01 PM
AKPunk49
Re: I need help evaluating an expression under specific conditions
Quote:

Originally Posted by chiro
Hey AKPunk49.

Can you show us what you tried: (Hint: did you try using the angle addition formulae?)?

I did, I don't know if I'm not doing it correctly or if I just don't know how to properly reduce.
• Apr 16th 2013, 05:54 PM
Soroban
Re: I need help evaluating an expression under specific conditions
Hello, AKPunk49!

Quote:

$\text{Given: }\:\cos\theta = -\tfrac{1}{3}\,\text{ in Quadrant 3},\,\sin\phi = \tfrac{1}{4}\,\text{ in Quadrant 2}$

$\text{Find: }\:\tan(\theta + \phi)$

We are given: . $\cos\theta \,=\, \frac{\text{-}1}{3}\,=\,\frac{adj}{hyp}$ in Quad 3.
. . Then: . $opp = \text{-}\sqrt{8} \quad\Rightarrow\quad \tan\theta \,=\, \frac{opp}{adj} \,=\,\frac{\text{-}\sqrt{8}}{\text{-}1}$
. . $\tan\theta \,=\,\sqrt{8}$

We are given: . $\sin\phi \,=\,\frac{1}{4} \,=\,\frac{opp}{hyp}$
. . Then: . $adj \,=\,\text{-}\sqrt{15} \quad\Rightarrow\quad \tan\phi \,=\,\frac{opp}{adj} \,=\,\frac{1}{\text{-}\sqrt{15}}$
. . $\tan\phi \,=\,-\frac{1}{\sqrt{15}}$

We have: . $\tan(\theta + \phi) \:=\:\frac{\tan\theta + \tan\phi}{1-\tan\theta\tan\phi}$

. . . . . . . . . . . . . . . . $=\;\frac{\sqrt{8} + \left(\text{-}\frac{1}{\sqrt{15}}\right)}{1 - \sqrt{8}\left(-\tfrac{1}{\sqrt{15}}\right)} \;=\;\frac{\sqrt{8} - \frac{1}{\sqrt{15}}}{1 + \frac{\sqrt{8}}{\sqrt{15}}}$

Multiply by $\frac{\sqrt{15}}{\sqrt{15}}\!:\;\;\frac{\sqrt{120} - 1}{\sqrt{15} + \sqrt{8}}$

Therefore: . $\tan(\theta + \phi) \;=\;\frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}}$
• Apr 16th 2013, 11:31 PM
Prove It
Re: I need help evaluating an expression under specific conditions
Quote:

Originally Posted by Soroban
Hello, AKPunk49!

We are given: . $\cos\theta \,=\, \frac{\text{-}1}{3}\,=\,\frac{adj}{hyp}$ in Quad 3.
. . Then: . $opp = \text{-}\sqrt{8} \quad\Rightarrow\quad \tan\theta \,=\, \frac{opp}{adj} \,=\,\frac{\text{-}\sqrt{8}}{\text{-}1}$
. . $\tan\theta \,=\,\sqrt{8}$

We are given: . $\sin\phi \,=\,\frac{1}{4} \,=\,\frac{opp}{hyp}$
. . Then: . $adj \,=\,\text{-}\sqrt{15} \quad\Rightarrow\quad \tan\phi \,=\,\frac{opp}{adj} \,=\,\frac{1}{\text{-}\sqrt{15}}$
. . $\tan\phi \,=\,-\frac{1}{\sqrt{15}}$

We have: . $\tan(\theta + \phi) \:=\:\frac{\tan\theta + \tan\phi}{1-\tan\theta\tan\phi}$

. . . . . . . . . . . . . . . . $=\;\frac{\sqrt{8} + \left(\text{-}\frac{1}{\sqrt{15}}\right)}{1 - \sqrt{8}\left(-\tfrac{1}{\sqrt{15}}\right)} \;=\;\frac{\sqrt{8} - \frac{1}{\sqrt{15}}}{1 + \frac{\sqrt{8}}{\sqrt{15}}}$

Multiply by $\frac{\sqrt{15}}{\sqrt{15}}\!:\;\;\frac{\sqrt{120} - 1}{\sqrt{15} + \sqrt{8}}$

Therefore: . $\tan(\theta + \phi) \;=\;\frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}}$

which of course you should then simplify...

\displaystyle \begin{align*} \frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}} &= \frac{\left( 2\sqrt{30} - 1 \right) \left( \sqrt{15} + 2\sqrt{2} \right) }{ \left( \sqrt{15} - 2\sqrt{2} \right) \left( \sqrt{15} + 2\sqrt{2} \right) } \\ &= \frac{2\sqrt{15 \cdot 30} + 4\sqrt{ 2 \cdot 30 } - \sqrt{15} - 2\sqrt{2} }{15 - 8} \\ &= \frac{2\sqrt{225 \cdot 2} + 4\sqrt{4 \cdot 15} - \sqrt{15} - 2\sqrt{2}}{7} \\ &= \frac{2 \cdot 15\sqrt{2} + 4\cdot 2 \sqrt{15} - \sqrt{15} - 2\sqrt{2}}{7} \\ &= \frac{28\sqrt{2} + 7\sqrt{15} }{7} \\ &= 4\sqrt{2} + \sqrt{15} \end{align*}