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