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Thread: Need help.

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    Default Need help.

    a and b are two acute angles. Prove that if
    sin^2(a) + sin^2(b) = sin(a + b), then
    a + b = pi/2

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    Quote Originally Posted by Zurm View Post
    a and b are two acute angles. Prove that if
    sin^2(a) + sin^2(b) = sin(a + b), then
    a + b = pi/2
    Do your own homework.
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    I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
    sin(pi/2)=1
    Check a=b=pi/4 (just kind of a "duh" example, but one that should definitely work if the condition is true.)
    Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
    Here's some identities you might be able to use.
    sin(a)=cos(pi/2-a)
    cos(a)=sin(pi/2-a)
    cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
    sin(a)=-sin(-a)
    sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
    sin^2(a)=(1-cos(2a))/2
    cos^2(a)=(1+cos(2a))/2)
    sin^2(a)+cos^2(a)=1

    I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.
    Last edited by alfaroverall; 12-16-2008 at 03:03.
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    Quote Originally Posted by alfaroverall View Post
    I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
    sin(pi/2)=1
    Check a=b=pi/4
    Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
    Here's some identities you might be able to use.
    sin(a)=cos(pi/2-a)
    cos(a)=sin(pi/2-a)
    cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
    sin(a)=-sin(-a)
    sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
    sin^2(a)=(1-cos(2a))/2
    cos^2(a)=(1+cos(2a))/2)
    sin^2(a)+cos^2(a)=1

    I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.
    Let him do his own work.
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    Quote Originally Posted by ejnomad07 View Post
    Let him do his own work.
    At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

    Wow this is actually way simpler than I thought, here you go:
    sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
    Hence:
    sin(a)=cos(b)
    cos(a)=sin(b)
    (Look at the equation for a second it should be pretty clear).
    Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

    Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.
    Last edited by alfaroverall; 12-16-2008 at 03:19.
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    Quote Originally Posted by alfaroverall View Post
    I actually put quite a bit of effort into this but couldn't get a rigorous proof of it, which honestly pissed me off. Here's an example that I used to put me on the right track:
    sin(pi/2)=1
    Check a=b=pi/4 (just kind of a "duh" example, but one that should definitely work if the condition is true.)
    Then (sqrt(2)/2)^2+(sqrt(2)/2)^2=1/2+1/2=1, so that example works.
    Here's some identities you might be able to use.
    sin(a)=cos(pi/2-a)
    cos(a)=sin(pi/2-a)
    cos(a)=cos(-a) (and so cos(pi/2-a)=cos(a-pi/2)=sin(a))
    sin(a)=-sin(-a)
    sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
    sin^2(a)=(1-cos(2a))/2
    cos^2(a)=(1+cos(2a))/2)
    sin^2(a)+cos^2(a)=1

    I rather hope this isn't due tomorrow, because I'd like to be able to come back to it once I have some time, if only for the challenge.
    Quote Originally Posted by alfaroverall View Post
    At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

    Wow this is actually way simpler than I thought, here you go:
    sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
    Hence:
    sin(a)=cos(b)
    cos(a)=sin(b)
    (Look at the equation for a second it should be pretty clear).
    Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

    Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.
    Lol, when I first saw the OP I was thinking "alfaroverall is solving that problem right now"

    I wish I liked math

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    Quote Originally Posted by alfaroverall View Post
    At the moment, that's what I'm doing...that's just a bunch of identities that I considered as I tried to approach the problem.

    Wow this is actually way simpler than I thought, here you go:
    sin^2(a)+sin^2(b) = sin(a)cos(b)+cos(a)sin(b) (identity listed above)
    Hence:
    sin(a)=cos(b)
    cos(a)=sin(b)
    (Look at the equation for a second it should be pretty clear).
    Because sin(a)=cos(pi/2-a) (again identity listed above), b=pi/2-a. Add a to both sides and you show that a+b=pi/2. Alternatively, cos(a)=sin(b), and cos(a)=sin(pi/2-a), so then a=pi/2-b, and by the same reasoning a+b=pi/2. Q.E.D.

    Trig really comes down to knowing the identities and not making the problem too complicated. If you know them, you can usually make it through any problem unscathed.
    I haven't done any of this for a while. Will you explain how you know sin a = cos b?
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  10. #10

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    Quote Originally Posted by Mulambo View Post
    I haven't done any of this for a while. Will you explain how you know sin a = cos b?
    In the first term on the left and right side, you have a sin(a) factor. In the second term on the left and right hand side, you have a sin(b) factor. But in the first one on the right hand side, the second sin(a) factor is replaced by a cos(b) factor, while in the second one on the right hand site, the second sin(b) factor is replaced by a cos(a) factor. Those second factors have to be equal to the factors they "replaced" in order for the equation to turn out equal.

    It's a little easier to understand if you think of it this way. If sin^2(a)=sin(a)cos(b) and simultaneously sin^2(b)=cos(a)sin(b), then the relationship we're looking at is true. (In short, if a=c and b=d, then a+b=c+d.) So if you check each relationship, you can do some routine algebra to find what I showed above:
    sin^2(a)=sin(a)cos(b)
    (divide both sides by sin(a))
    sin(a)=cos(b)
    Similarly:
    sin^2(b)=cos(a)sin(b)
    sin(b)=cos(a)
    These are then only BOTH true (with both a and b being acute) if a+b=pi/2, by what I showed above.
    Last edited by alfaroverall; 12-16-2008 at 04:03.
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