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Math Help - Sin 2A=...

  1. #1
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    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!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JasonW View Post
    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
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  3. #3
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    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.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JasonW View Post
    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!
    Quote Originally Posted by JasonW View Post
    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
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  5. #5
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    Ah, ok, that makes sense. Thanks!
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