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Math Help - A simple "Verify the identity" problem that is confusing me.

  1. #1
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    A simple "Verify the identity" problem that is confusing me.

    Well, the problem is:

    sin4t = 4 sint cost(1-2sin^2t)

    For some reason I always get confused in verifying the identity ><;. Now, I tried various forms of doing it like changing sin4t to sin(2t+2t) but it just ends up getting jumbled up. Any help on how to not get so confused with "Verify the identity" problems or just pushing me in the right direction to solve this one would be appreciated.
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  2. #2
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    Quote Originally Posted by Megaman View Post
    Well, the problem is:

    sin4t = 4 sint cost(1-2sin^2t)

    For some reason I always get confused in verifying the identity ><;. Now, I tried various forms of doing it like changing sin4t to sin(2t+2t) but it just ends up getting jumbled up. Any help on how to not get so confused with "Verify the identity" problems or just pushing me in the right direction to solve this one would be appreciated.
    You're on the right track.

    Let u = 2t such that your equation becomes sin(2u) which is 2sin(u)cos(u). However u=2t so you end up with:

    <br />
sin(4t) = 2sin(2t)cos(2t)

    then plug in what sin(2t) and cos(2t) are equal to in order to get the RHS (but don't forget that 2 at the front)
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    Quote Originally Posted by e^(i*pi) View Post
    You're on the right track.

    Let u = 2t such that your equation becomes sin(2u) which is 2sin(u)cos(u). However u=2t so you end up with:

    <br />
sin(4t) = 2sin(2t)cos(2t)

    then plug in what sin(2t) and cos(2t) are equal to in order to get the RHS (but don't forget that 2 at the front)
    Okay, I understand just about all of it now but I just have one question. I'm a little confused on the whole u = 2t. I know you're changing it to that to use the Double-Angle formula for sin, but isn't it 4t and not 2t?
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    Quote Originally Posted by Megaman View Post
    Okay, I understand just about all of it now but I just have one question. I'm a little confused on the whole u = 2t. I know you're changing it to that to use the Double-Angle formula for sin, but isn't it 4t and not 2t?
    In my substitution I let u=2t and because 4t = 2*2t it would be 2u. This can be expanded out to 2sin(u)cos(u).

    However, because u=2t it must be put back in on that line so we get 2sin(2t)cos(2t). The u was to make the first step appear easier, you don't need to use it. The key to the question is knowing that sin(4t) = sin(2(2t)) so you can use the double angle formula with 2t +2t to simplify the 4t.

    The next step is because the equation still needs to be simplified into terms of t so the double angle formulae can be used again
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  5. #5
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    One step at a time! 4t= 2(2t) so use sin(2a)= 2sin(a)cos(a) with a= 2t. Then repeat with cos(2t) and sin(2t).
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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post
    In my substitution I let u=2t and because 4t = 2*2t it would be 2u. This can be expanded out to 2sin(u)cos(u).

    However, because u=2t it must be put back in on that line so we get 2sin(2t)cos(2t). The u was to make the first step appear easier, you don't need to use it. The key to the question is knowing that sin(4t) = sin(2(2t)) so you can use the double angle formula with 2t +2t to simplify the 4t.

    The next step is because the equation still needs to be simplified into terms of t so the double angle formulae can be used again
    Ah, okay now I understand completely. Thanks a lot. Hopefully by doing about 8-10 more of these I'll get the hang of it.
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