# Math Help - Sin 2A=...

1. ## Sin 2A=...

If I know the value and quadrant of Sin (A+B) and Cos (A-B), how would I find Sin2A and Cos2A?

Btw, 2A would equal (A+B)+(A-B).

Thanks!

2. Originally Posted by JasonW
If I know the value and quadrant of Sin (A+B) and Cos (A-B), how would I find Sin2A and Cos2A?

Btw, 2A would equal (A+B)+(A-B).

Thanks!
nope, think of 2A as A + A

so sin(2A) = sin(A + A) and cos(2A) = cos(A + A)

and you can find each of those using the addition formulas

3. Maybe I should have explained that better. I know the formula for Sin2A and Cos2A, I'm asked for the exact value of them, given that sin (A+B)=3/5 and is in Q1, and that cos (A-B)=12/13 and is in Q4.

4. Originally Posted by JasonW
If I know the value and quadrant of Sin (A+B) and Cos (A-B), how would I find Sin2A and Cos2A?

Btw, 2A would equal (A+B)+(A-B).

Thanks!
Originally Posted by JasonW
Maybe I should have explained that better. I know the formula for Sin2A and Cos2A, I'm asked for the exact value of them, given that sin (A+B)=3/5 and is in Q1, and that cos (A-B)=12/13 and is in Q4.
Let's try this:

We know that $sin(A + B) = \frac{3}{5}$ and A + B is in QI. Thus $cos(A + B) = \frac{4}{5}$. Similarly $sin(A - B) = -\frac{5}{13}$.

So
$sin(2A) = sin([A + B] + [A - B]) = sin(A + B)cos(A - B) + sin(A - B)cos(A + B)$

$= \frac{3}{5} \cdot \frac{12}{13} - \frac{5}{13} \cdot \frac{4}{5}$

etc.

-Dan

5. Ah, ok, that makes sense. Thanks!