# Thread: I need help evaluating an expression under specific conditions

1. ## I need help evaluating an expression under specific conditions

Evaluate the expression under the given conditions.tan(θ + ); cos θ = −1/3, θ in Quadrant III, sin = 1/4, in Quadrant 4

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

2. ## 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?)?

3. ## Re: I need help evaluating an expression under specific conditions

$\displaystyle \displaystyle \sin{(\phi)}$ can not possibly be $\displaystyle \displaystyle \frac{1}{4}$ in the fourth quadrant. Why?

4. ## Re: I need help evaluating an expression under specific conditions

Fixed it, sorry.

5. ## Re: I need help evaluating an expression under specific conditions

It doesn't change the fact that $\displaystyle \displaystyle \sin{(\phi)}$ can not possibly be positive in the fourth quadrant. In fact, I don't see anything different at all...

6. ## 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...

7. ## Re: I need help evaluating an expression under specific conditions

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.

8. ## Re: I need help evaluating an expression under specific conditions

Hello, AKPunk49!

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

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

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

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

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

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

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

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

9. ## Re: I need help evaluating an expression under specific conditions

Originally Posted by Soroban
Hello, AKPunk49!

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

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

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

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

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

Therefore: .$\displaystyle \tan(\theta + \phi) \;=\;\frac{2\sqrt{30} - 1}{\sqrt{15} - 2\sqrt{2}}$
which of course you should then simplify...

\displaystyle \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*}