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!

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- Nov 3rd 2007, 04:29 PM #1

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- Nov 3rd 2007, 04:31 PM #2

- Nov 3rd 2007, 04:51 PM #3

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- Nov 3rd 2007, 05:00 PM #4
Let's try this:

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

So

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

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

etc.

-Dan

- Nov 3rd 2007, 06:59 PM #5

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