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Math Help - Single trigonometric functions...

  1. #1
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    Single trigonometric functions...

    If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

    Thanks,


    -Tom
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TomCat View Post
    If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

    Thanks,


    -Tom
    You mean a single trig function to represent that? I don't think it can be done. However, to answer your comment about the double angle formulas this is what you would get:

    f(x) = 3cos(2x) + 4sin(2x) = 3(cos^2(x) - sin^2(x)) + 4*2sin(x)cos(x)

    f(x) = 3cos^2(x) + 8sin(x)cos(x) - 3sin^2(x)

    As it happens you can factor this:
    f(x) = [3cos(x) - sin(x)][cos(x) - 3sin(x)]

    But that doesn't give you the form you want.

    If you mean can you get it just in terms of sine or cosine, THAT we can do:

    f(x) = 3cos(2x) + 4sin(2x) = 3(1 - 2sin^2(x)) + 8sin(x)cos(x)

    f(x) = 3 - 6sin^2(x) + 8 sin(x) \sqrt{1 - sin^2(x)}

    -Dan
    Last edited by topsquark; October 31st 2006 at 07:23 PM. Reason: Missed a factor of 2!
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  3. #3
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    But,
    \cos x\not = \sqrt{1-\sin^2 x}

    Maybe in physics okay, but not heir.
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  4. #4
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    I thought cos^2(x) + sin^2(x) = 1

    Rearranging gives cos(x) = root(1 - sin^2(x))?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    But,
    \cos x\not = \sqrt{1-\sin^2 x}

    Maybe in physics okay, but not heir.
    (Sigh) Okay.

    f(x) = 3 - 6sin^2(x) \pm 8sin(x) \sqrt{1 - sin^2(x)} with the sign of the \pm chosen depending on which quadrant x is in.

    -Dan
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  6. #6
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    Quote Originally Posted by classicstrings View Post
    I thought cos^2(x) + sin^2(x) = 1

    Rearranging gives cos(x) = root(1 - sin^2(x))?
    These are not equivalent statements.
    ---
    Basically, look at the graphs show below.


    ~~~
    Topsquark, you should define a function,
    \mbox{TPH}(x) that is 1 if x\in [-\pi/2 +2\pi k,\pi/2+2\pi k] otherwise -1
    Then you can say,
    \cos x=\mbox{TPH}(x)\sqrt{1-\sin^2 x}

    I actually think my function can be useful. Look at the Heaviside function.

    Can you express my function as a fourier series or will it lead to Gibbs phenomena?
    Attached Thumbnails Attached Thumbnails Single trigonometric functions...-picture6.gif  
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  7. #7
    Member classicstrings's Avatar
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    How do I type sin^2 on my calculator? When I type sin it automatically gives me sin( .
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  8. #8
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    Quote Originally Posted by classicstrings View Post
    How do I type sin^2 on my calculator? When I type sin it automatically gives me sin( .
    Hi,

    you have to use brackets
    sin^2(x) has to be typed: (sin(x))^2

    EB
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  9. #9
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    Okay... I need to get just one type (sin or cos).
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  10. #10
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    Quote Originally Posted by TomCat View Post
    If I were to turn f(x) = 3 cos(2x) + 4 sin(2x) into a single function... Where would I want to start? Would I just use the double angle formulas and go from there?

    Thanks,


    -Tom
    The idea with this type of question is to reduce

    <br />
f(x)=3 \cos(2x) + 4 \sin(2x)<br />

    to some thing like:

    <br />
f(x)=r [\sin(\theta) \cos(2x) + \cos(\theta) \sin(2x)]<br />

    Then to use the addition formula for \sin to reduce this to:

    <br />
f(x)=r \sin(\theta+2x)<br />

    For this to work we need:

    r \sin(\theta)=3 and r \cos(\theta)=4,

    so \theta = \arctan(3/4) and r=\sqrt{3^2+4^2}=5.

    RonL
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  11. #11
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    I think I get what you're saying. I was looking over this in class with a friend, and she was just confusing me more. I don't think she was doing it right either.

    I would just do something similar for f(x)= Acos(bx) + Bsin(bx) correct?

    Thank you!


    -Tom
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by TomCat View Post
    I think I get what you're saying. I was looking over this in class with a friend, and she was just confusing me more. I don't think she was doing it right either.

    I would just do something similar for f(x)= Acos(bx) + Bsin(bx) correct?

    Thank you!


    -Tom
    Yes, exactly.

    RonL
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  13. #13
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    Hmm... That was over the heads of everyone I showed that to. Are there any other ways to do it that are more Pre-Calcish?
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  14. #14
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    Quote Originally Posted by TomCat View Post
    Hmm... That was over the heads of everyone I showed that to. Are there any other ways to do it that are more Pre-Calcish?
    It was pure basic trigonometry, which should be pre-calculus (not that
    I would know we don't have the stereotyped math sequence that you
    seem to have in the US).

    The only things you need to know to do it is the sin addition formula and
    the trig form of Pythagoras's theorem, which are:

    <br />
\sin(A+B)=\sin(A)\cos(B) + \cos(A)\sin(B)<br />

    and:

    <br />
\sin^2(\theta)+\cos^2(\theta)=1<br />

    and the rest is just basic algebra.

    RonL
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    It was pure basic trigonometry, which should be pre-calculus (not that
    I would know we don't have the stereotyped math sequence that you
    seem to have in the US).

    The only things you need to know to do it is the sin addition formula and
    the trig form of Pythagoras's theorem, which are:

    <br />
\sin(A+B)=\sin(A)\cos(B) + \cos(A)\sin(B)<br />

    and:

    <br /> <br /> <br />
\sin^2(\theta)+\cos^2(\theta)=1<br />

    and the rest is just basic algebra.

    RonL
    You might possibly call it "Advanced Trigonometry" but it's definately Pre-Calculus level.

    -Dan
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