# Homework Help: Trigonometry- addition and factor forumla

1. Nov 23, 2012

### xiphoid

1. The problem statement, all variables and given/known data
If tan(A+B) = 3 and tan(A-B) = 2, find tan2A and tan2B

2. Relevant equations
tan (A - B) = (tan A - tan B)/(1 + (tan A)(tan B) and similar sort of one for tan(A+B)

3. The attempt at a solution
i did some calculation and got tanA= (1-2tanB)/5tanB

after which there seems some thing missing for the proceeding calculations....
wonder what to do next?

2. Nov 23, 2012

### SteamKing

Staff Emeritus
For better service, try posting your math questions in either the Precalculus math or the Calculus sections of the HW forums.

3. Nov 23, 2012

### Staff: Mentor

Topic moved. As SteamKing wrote - this is definitely not "OtherSciences", but Math itself.

4. Nov 23, 2012

### Millennial

The "other identity" of "the sort" you are talking about is given like this:

$$\tan(x+y)=\frac{\sin(x+y)}{\cos(x+y)}=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)-\sin(x)\sin(y)}=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$$

Now, a good start to this question would be to write everything you already have down: (name tan(a)=x and tan(b)=y) $\displaystyle 3=\frac{x+y}{1-xy}$ and $\displaystyle 2=\frac{x-y}{1+xy}$. Then, these are two equations with two unknowns (a and b.) Try to manipulate the expressions and eliminate a term that you would not want in an equation with two unknowns. The rest follows relatively easily.

If you found some value for tan A whose inverse tangent is not very pleasant, you are doing something wrong; the value A comes out very nicely.

Tip: Your expression for tan A is not the simplest one possible. Try to find linear expressions.
Tip-2: There isn't only one solution.

Last edited: Nov 23, 2012