1. ## Trig problem

Alright, so I've got sin(5/13) and I've tan(-sqrt3)
I'm instructed to use tan(a+b)=tanx-tany/1+tanx(tany), and utilising properties of triangles I can resolve that sin(5/13) has a tan of (5/12)
/I plug it in to my equation to get [(5/13)+(-sqrt3)]/[1+(5/12)(-sqrt3)].. man I'm lost

2. ## Re: Trig problem

it would be easier to answer this if it made a lick of sense...

wth are alpha and beta?

3. ## Re: Trig problem

sin is alpha, tan is beta

4. ## Re: Trig problem

Originally Posted by EngineeringGuy
sin is alpha, tan is beta
Do you mean $\sin{\alpha} = \dfrac{5}{13}$ and $\tan{\beta} =-\sqrt{3}$. ?

If so, is $\alpha$ in quadrant I or II ? Is $\beta$ in quadrant II or IV ?

5. ## Re: Trig problem

bingo, and sine alpha is such that -3pi/2 <a<-pi, and tan beta is pi/2<B<pi

6. ## Re: Trig problem

Originally Posted by EngineeringGuy
bingo, and sine alpha is such that -3pi/2 <a<-pi, and tan beta is pi/2<B<pi
That means both $\alpha$ and $\beta$ are quadrant II angles, which also indicates $\tan{\alpha} < 0$

use the difference identity for tangent ...

$\tan(\alpha - \beta) = \dfrac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}$

In future, please provide the problem statement with all given information in your initial post. Thank you for your cooperation.

7. ## Re: Trig problem

yes thanks and will do but that isn't answering my question really, I'm getting stuck with the operations

8. ## Re: Trig problem

Originally Posted by EngineeringGuy
yes thanks and will do but that isn't answering my question really, I'm getting stuck with the operations
in quadrant II, $\sin{\alpha} = \dfrac{5}{13} \implies \tan{\alpha} = -\dfrac{5}{12}$

you now have values for both $\tan{\alpha}$ and $\tan{\beta}$ ... use the given identity and calculate the value of $\tan(\alpha-\beta)$

9. ## Re: Trig problem

yeah... the answer in the back of the book is -240+169sqrt3/69, I get to -5+12sqrt3/12+5sqrt3 and don't know what to do

10. ## Re: Trig problem

Originally Posted by EngineeringGuy
yeah... the answer in the back of the book is -240+169sqrt3/69, I get to -5+12sqrt3/12+5sqrt3 and don't know what to do
rationalize the denominator ...

$\dfrac{12\sqrt{3}-5}{5\sqrt{3}+12} \cdot \dfrac{5\sqrt{3}-12}{5\sqrt{3}-12}$

also ...
(-5+12sqrt3)/(12+5sqrt3)
... parentheses around the terms in the numerator and denominator in your rational expressions

11. ## Re: Trig problem

that damn conjugate! thank you so much