# Thread: [SOLVED] Need help with Trig Sum Formulas Question

1. ## [SOLVED] Need help with Trig Sum Formulas Question

Can someone help me solve this:

tan(a-b)

cosa = -4/5 Lies in Quadrant 3

sinb = 1/3 Lies in Quadrant 1

2. Originally Posted by gmorrow2
Can someone help me solve this:

tan(a-b)

cosa = -4/5 Lies in Quadrant 3

sinb = 1/3 Lies in Quadrant 1

for angle $a$ , sketch a reference triangle in quad III

adjacent side = -4 , hypotenuse = 5 , opposite side = ?

$\tan{a} = \frac{opp}{adj}$

for angle b , sketch a reference triangle in quad I

opposite side = 1 , hypotenuse = 3 , adjacent side = ?

$\tan{b} = \frac{opp}{adj}$

lastly, calculate $\tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$

3. I did all the math already, I guess I should of posted the answer I got (which is why I'm asking for help) to see if I got it right

Here is what I got:

$

$
$\tan(a-b)=\frac{3-8\sqrt{2}}{4-6\sqrt{2}}

$

4. Originally Posted by gmorrow2
I did all the math already, I guess I should of posted the answer I got (which is why I'm asking for help) to see if I got it right

Here is what I got:

$

$
$\tan(a-b)=\frac{3-8\sqrt{2}}{4-6\sqrt{2}}

$

Show your calculations of tan(a), tan(b) and tan(a-b)

5. $
\tan{a} = \frac{3}{4}
$

$
\tan{b} = \frac{\sqrt{8}}{8}
$

The answer I posted above is the overall calculations. The radicals are messing with me I just need to know if I did them right.

6. note that $\frac{\sqrt{8}}{8} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}$

$\tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$

$\tan(a-b)=\frac{\frac{3}{4}-\frac{\sqrt{2}}{4}}{1+\frac{3\sqrt{2}}{16}}$

multiply by $\frac{16}{16}$ to clear the fractions ...

$\tan(a-b) = \frac{4(3-\sqrt{2})}{16+3\sqrt{2}}$

7. AHH that's what I was missing the radicals always get me.

Thanks, I'll redo it and see if I get the same thing, Thanks!