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Math Help - Sine of Arcsine

  1. #1
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    Sine of Arcsine

    Was hoping to get some help on this. Not sure how to approach.


    Sin[2Arcsin(-4/5)]

    Thank you!
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  2. #2
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    Quote Originally Posted by Impalord View Post
    Was hoping to get some help on this. Not sure how to approach.


    Sin[2Arcsin(-4/5)]

    Thank you!
    Let \alpha = \arcsin \left(- \frac{4}{5} \right) \Rightarrow \sin \alpha = - \frac{4}{5}. Note that \alpha is in the fourth quadrant (why?).

    Now use \sin (2 \alpha) = 2 \sin \alpha \cos \alpha.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Let \alpha = \arcsin \left(- \frac{4}{5} \right) \Rightarrow \sin \alpha = - \frac{4}{5}. Note that \alpha is in the fourth quadrant (why?).

    Now use \sin (2 \alpha) = 2 \sin \alpha \cos \alpha.

    For some reason I didn't think \sin(2 \alpha) = 2 \sin (\alpha)

    Not sure why.

    So it's in quadrant 4 because the domain of the arcsine function is

     (\pi /2) -> -(\pi /2) right?

    Now if I plug -4/5 into the equation I have

     2 \sin(-4/5) \cos(-4/5) ?

    or would it be


     2 \sin(-4/5) \cos(3/5) ?

    Thanks!
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  4. #4
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    All right, I think I've got it.


    Because :

     2 \sin x = 2 \sin x \cos x

    and if then after using the pythagorean theorem

    we get 4^2 + 3^2 = 5^2

    so

     2( \sin -4/5 \cos 3/5)

    Luckily, in this case, I know the answer is -24/25

    So I can see how the (-4/5)(3/5) multiply to make

    2(-12/25) = -24/25

    and then we can make it back into \arcsin (-24/25)

    then

    \sin (\arcsin(-24/25)

    And because of the identity property the answer is -24/25.

    Does that seem right?

    Thank you
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  5. #5
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    Quote Originally Posted by Impalord View Post
     2( \sin -4/5 \cos 3/5)
    No, no, no. Watch your notation! -4/5 and 3/5 are not angles! They are the ratios.

    Look at what Mr. fantastic wrote:
    Quote Originally Posted by mr fantastic View Post
    Let \alpha = \arcsin \left(-\frac{4}{5} \right)
    When plugging into the double angle formula
    \sin (2 \alpha) = 2 \sin \alpha \cos \alpha
    you get this:
    \sin (2 \alpha) = 2 \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) \cos \left(\arcsin \left(-\frac{4}{5} \right)\right)

    This is not the same as this:
     2( \sin -4/5 \cos 3/5)

    We are saying that \alpha is in Quadrant IV, so
    \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) = -\frac{4}{5}.
    You figured out correctly that
    \cos \left(\arcsin \left(-\frac{4}{5} \right)\right) = \frac{3}{5}.

    So we substitute and get
    \sin (2 \alpha) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right)

    I'll say it again: this is not the same as this:
     2( \sin -4/5 \cos 3/5)

    I'm sorry if I sound irritated -- I can't tell you how many times I see errors in notation like this. If this was an assignment and I was grading it I would only give partial credit.


    01
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  6. #6
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    Quote Originally Posted by yeongil View Post
    No, no, no. Watch your notation! -4/5 and 3/5 are not angles! They are the ratios.

    Look at what Mr. fantastic wrote:


    When plugging into the double angle formula
    \sin (2 \alpha) = 2 \sin \alpha \cos \alpha
    you get this:
    \sin (2 \alpha) = 2 \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) \cos \left(\arcsin \left(-\frac{4}{5} \right)\right)

    This is not the same as this:
     2( \sin -4/5 \cos 3/5)

    We are saying that \alpha is in Quadrant IV, so
    \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) = -\frac{4}{5}.
    You figured out correctly that
    \cos \left(\arcsin \left(-\frac{4}{5} \right)\right) = \frac{3}{5}.

    So we substitute and get
    \sin (2 \alpha) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right)

    I'll say it again: this is not the same as this:
     2( \sin -4/5 \cos 3/5)

    I'm sorry if I sound irritated -- I can't tell you how many times I see errors in notation like this. If this was an assignment and I was grading it I would only give partial credit.


    01

    Great! thanks a bunch for clearing that up. As I was working through that, It seemed wrong. Now I know why, I really appreciate it.
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  7. #7
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    Hello Impalord
    Quote Originally Posted by Impalord View Post
    All right, I think I've got it.


    Because :

     2 \sin x = 2 \sin x \cos x You mean sin2x. Grandad.

    and if then after using the pythagorean theorem

    we get 4^2 + 3^2 = 5^2

    so

     2( \sin -4/5 \cos 3/5)

    Luckily, in this case, I know the answer is -24/25

    So I can see how the (-4/5)(3/5) multiply to make

    2(-12/25) = -24/25

    and then we can make it back into \arcsin (-24/25)

    then

    \sin (\arcsin(-24/25)

    And because of the identity property the answer is -24/25.

    Does that seem right?

    Thank you
    Your answer is more or less correct, but rather confused. Mr F has given you the formula \sin2\alpha = 2\sin\alpha\cos\alpha, and the fact that \alpha is in the fourth quadrant (your reason for this is correct, by the way) and you've correctly worked out that \cos\alpha = \frac35.

    So you can simply write:

    \sin\left(2\arcsin(-\frac45)\right)=\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times(-\tfrac45)\times\tfrac35 = -\tfrac{24}{25}

    Grandad
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  8. #8
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    Quote Originally Posted by Grandad View Post
    Hello ImpalordYour answer is more or less correct, but rather confused. Mr F has given you the formula \sin2\alpha = 2\sin\alpha\cos\alpha, and the fact that \alpha is in the fourth quadrant (your reason for this is correct, by the way) and you've correctly worked out that \cos\alpha = \frac35.

    So you can simply write:

    \sin\left(2\arcsin(-\frac45)\right)=\sin2\alpha = 2\sin\alpha\cos\alpha = 2\times(-\tfrac45)\times\tfrac35 = -\tfrac{24}{25}

    Grandad
    I really appreciate your help!

    For some reason I wasn't seeing the identity because I wasn't seeing

     \sin[2 \arcsin (-4/9)] = \sin(2 \alpha)

    Replacing the  \arcsin -4/9

    with  \alpha the way Mr. Fantastic pointed out. Now that I see that, the problem is pretty straight forward. Thank you all again!
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