tan (α-β)

Printable View

April 2nd 2012, 02:15 PM
Ashz

1 Attachment(s)

tan (α-β)

finding the exact value of tan (α-β) when:
tan α = -(5/12) in quad II and cos β = 3/4 and lies in quad I
Triangle α
x = -12
y = 5
r = 13

Triangle β
x = 3
y = √7 (suppose to be square root of 7)
r = 4

tan (α-β) = (tan α - tan ß) / 1 + tan (α) tan (ß)
tan (α-β) = ((-5/12) - (√7/4)) / 1 + (-5/12)(√7/4)
tan (α-β) = ((-5/12) - (3√7/12)) / (48/48) + ((-5√7)/48)
tan (α-β) = ((-5-3√7)/12)((48)/(48-5√7)
tan (α-β) = -1/4 (is what I eventually came up with but it's wrong :( and I am unsure of what I'm doing wrong...)

I've also added a screen shot of the actual equation just in case the above is too confusing...
April 2nd 2012, 05:30 PM
Soroban

Re: tan (α-β)

Hello, Ashz!

Quote:

$\text{Find the exact value of }\tan(\alpha - \beta)\text{ when: }\:\begin{Bmatrix}\tan\alpha \,=\, \text{-}\frac{5}{12} & \text{in quad II} \\ \cos\beta \,=\, \frac{3}{4} & \text{in quad I} \end{Bmatrix}$

We are given: . $\tan\alpha = \text{-}{\tfrac{5}{12}$

We have: . $\cos\beta = \tfrac{3}{4} = \tfrac{adj}{hyp} \quad\Rightarrow\quad opp = \sqrt{7} \quad\Rightarrow\quad \tan\beta = \tfrac{\sqrt{7}}{3}$

$\tan(\alpha-\beta) \;=\;\frac{\tan\alpha - \tan\beta}{1 + \tan\slpha\tan\beta} \;=\; \dfrac{\text{-}\frac{5}{12} - \frac{\sqrt{7}}{3}}{1 + \left(\text{-}\frac{5}{12}\right)\left(\frac{\sqrt{7}}{3}\right )} \;=\;\frac{\text{-}15 - 12\sqrt{7}}{36 - 5\sqrt{7}}$

Rationalize: . $\frac{\text{-}15 - 12\sqrt{7}}{36-5\sqrt{7}} \cdot\frac{36 + 5\sqrt{7}}{36 + 5\sqrt{7}}$

. . . . $=\;\frac{\text{-}540 - 75\sqrt{7} - 432\sqrt{7} - 420}{1296 - 175} \;=\; -\frac{960 + 507\sqrt{7}}{1121}$
April 2nd 2012, 06:30 PM
Ashz

Re: tan (α-β)

Thanks for replying back ~ but I have 2 additional questions.

1. before we rationalize, how did you get from the tan (α-β) equation to the (-15-12√7)/(36-5√7) ~ you multiplied it the whole equation by 36/36 right?

2. then when you rationalize the equation with 36+5√7 ~ this might be an algebra question, and I've probably forgotten, but why can't you just multiply by 5√7 instead of what you have?
April 3rd 2012, 03:19 AM
biffboy

Re: tan (α-β)

2 If you just multiply by 5root7 there will still be a root in the denominator so you wont have rationalized it.
April 3rd 2012, 12:39 PM
Ashz

Re: tan (α-β)

Sorry :( but now Im confused ~ bc I thought that you would have to rationalize the denominator if there is a square root there... b/c if I didn't have to rationalize it in the first place, why even bother multiplying it by the 5√7 in the first place... or even the 36+5√7...
April 3rd 2012, 11:22 PM
biffboy

Re: tan (α-β)

The question didn't say we had to rationalize so we could have left the answer as we had it earlier
April 4th 2012, 09:39 AM
Ashz

Re: tan (α-β)

I'm sorry, but I think your causing me more confusion on this question :( Sorban rationalized the answer and I know I didn't post to rationalize the answer, but the attachment I posted showed that it needed to be rationalized. I am prob the one that caused the confusion on this question, but I tried to minimize it with posting the attachment. I think I will just PM Sorban for more information, thanks for your help anyway biffboy. I do appreciate it.