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Thread: 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 $\displaystyle \alpha = \arcsin \left(- \frac{4}{5} \right) \Rightarrow \sin \alpha = - \frac{4}{5}$. Note that $\displaystyle \alpha$ is in the fourth quadrant (why?).

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

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

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

    Not sure why.

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

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

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

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

    or would it be


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

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


    Because :

    $\displaystyle 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

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

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

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

    $\displaystyle 2(-12/25) = -24/25$

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

    then

    $\displaystyle \sin (\arcsin(-24/25)$

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

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

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

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

    I'll say it again: this is not the same as this:
    $\displaystyle 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
    $\displaystyle \sin (2 \alpha) = 2 \sin \alpha \cos \alpha$
    you get this:
    $\displaystyle \sin (2 \alpha) = 2 \sin \left(\arcsin \left(-\frac{4}{5} \right)\right) \cos \left(\arcsin \left(-\frac{4}{5} \right)\right)$

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

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

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

    I'll say it again: this is not the same as this:
    $\displaystyle 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 :

    $\displaystyle 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

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

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

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

    $\displaystyle 2(-12/25) = -24/25$

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

    then

    $\displaystyle \sin (\arcsin(-24/25)$

    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 $\displaystyle \sin2\alpha = 2\sin\alpha\cos\alpha$, and the fact that $\displaystyle \alpha$ is in the fourth quadrant (your reason for this is correct, by the way) and you've correctly worked out that $\displaystyle \cos\alpha = \frac35$.

    So you can simply write:

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

    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 $\displaystyle \sin2\alpha = 2\sin\alpha\cos\alpha$, and the fact that $\displaystyle \alpha$ is in the fourth quadrant (your reason for this is correct, by the way) and you've correctly worked out that $\displaystyle \cos\alpha = \frac35$.

    So you can simply write:

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

    Grandad
    I really appreciate your help!

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

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

    Replacing the $\displaystyle \arcsin -4/9 $

    with $\displaystyle \alpha$ 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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