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Math Help - Urgent Question

  1. #1
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    Urgent Question

    Hi guys,

    How do I do arithematic operations with functions like:

    2cos(2t-45(degrees)) - 3cos(4t-45) + 5sin(3t+45)

    Any help would be greatly appreciated.

    Thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thebearot
    Hi guys,

    How do I do arithematic operations with functions like:

    2cos(2t-45(degrees)) - 3cos(4t-45) + 5sin(3t+45)

    Any help would be greatly appreciated.

    Thanks
    The basic rules are:
    sin(A \pm B) = sin(A)cos(B) \pm sin(B)cos(A)
    cos(A \pm B) = cos(A)cos(B) \mp sin(A)sin(B)

    We get the "double angle" formulae from these:
    sin(2A) = 2sin(A)cos(A)
    cos(2A) = cos^2(A) - sin^2(B) = 2cos^2(A) - 1 = 1 - 2sin^2(A)

    Hint: It takes a bit of work, but you can simplify sin(3t) as sin(2t+t) and cos(4t) as cos(2t+2t).

    -Dan
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  3. #3
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    Thanks for the quick reply. I'll try to solve it with the formulas.
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  4. #4
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    Quote Originally Posted by topsquark
    Hint: It takes a bit of work, but you can simplify sin(3t) as sin(2t+t) and cos(4t) as cos(2t+2t).
    There is an elegant way to derive a general formula for,
    \sin nx \mbox{ and }\cos nx
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  5. #5
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    Hello, ThePerfectHacker!

    There is an elegant way to derive a general formula for: \sin nx \mbox{ and }\cos nx

    I "discovered" a method while in college . . . with my very own ritual.


    Suppose we want the trig formulas for 6\theta

    (1) Expand (x + 1)^6:\;\;x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1


    (2) Construct a fraction with these terms,
    Write the terms alternately in the denominator and numerator.
    Start with the denominator.

    . . \frac{6x^5\qquad 20x^3\qquad 6x}{x^6\qquad 15x^4\qquad 15x^2\qquad 1}


    (3) Insert alternating signs in the numerator and the denominator.

    . . \frac{6x^5 -20x^3 + 6x}{x^6 - 15x^4 + 15x^2 -1}


    (4) Replace x with \tan\theta and we have:

    . . \boxed{\tan6\theta \;= \;\frac{6\tan^2\!\theta -20\tan^3\!\theta + 6\tan\theta}{\tan^6\!\theta - 15\tan^4\!\theta + 15\tan^2\!\theta -1}} \bf{(1)}


    (5) Multiply top and bottom by \cos^6\!\theta and we get two formulas:

    . . \tan6\theta \;=  \;\frac{\boxed{\sin6\theta}}{\boxed{\cos6\theta}} \;= \;\frac{\boxed{6\sin^2\!\theta\cos^4\!\theta - 20\sin^3\!\theta\cos^3\!\theta + 6\sin\theta\cos^5\!\theta}}{\boxed{\sin^6\!\theta - 15\sin^4\!\theta\cos^2\!\theta + 15\sin^2\!\theta\cos^4\!\theta - \cos^6\!\theta}} \begin{array}{ccc}\bf{(2)} \\ \\ \bf{(3)}\end{array}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This is all possible because of DeMoivre's Theorem.

    Since \cos6\theta + i\sin6\theta\;= \;(\cos\theta + i\sin\theta)^6, we can expand the right side,

    then equate real and imaginary components, giving us (2) and (3) above.

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  6. #6
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    Quote Originally Posted by Soroban
    Hello, ThePerfectHacker!


    I "discovered" a method while in college . . . with my very own ritual.
    That must have been 80 years ago. Anyway you have a good method. The one I was speaking about also uses de Moiver's theorem. Then you just use binomial expansion.
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