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Thread: Help

  1. #1 Help 
    Forum Freshman daniel's Avatar
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    There's a task in my math book that says:
    Write the expression

    \/2*(cos(3x+ π/4))

    using cos3x and sin3x. (\/2 means the square root of 2, I couldn't locate the square root sign on my keyboard).

    I ended up with \/2*(cos3x-sin3x), but the answer is supposed to be cos3x-sin3x. What did I do wrong?

    Oh, and, sorry if my translation makes no sense. I'm Norwegian and not very used to talk about mathematics in any other language than Norwegian


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  3. #2 Re: Help 
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by daniel
    There's a task in my math book that says:
    Write the expression

    \/2*(cos(3x+ ?/4))

    using cos3x and sin3x. (\/2 means the square root of 2, I couldn't locate the suare root sign on my keyboard).

    I ended up with \/2*(cos3x-sin3x), but the answer is supposed to be cos3x-sin3x. What did I do wrong?

    Oh, and, sorry if my translation makes no sense. I'm Norwegian and not very used to talk about mathematics in any other language than Norwegian

    Am I to assume your "n" is supposed to be pi?

    If so, then that's the problem.

    cheers
    william


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  4. #3  
    Forum Freshman daniel's Avatar
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    Yeah, I just copied the pi sign from a website and it showed up mysteriously similar to an "n" here. How is that the problem?
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  5. #4  
    Forum Ph.D. william's Avatar
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    Okay, not really the problem so much as the "solution".

    cos(pi/4) = sqrt(2)/2 = sin(pi/4)

    sqrt(2)*sqrt(2)/2 = 1

    and your troublesome coefficient disappears.

    Hope that helps,
    william
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  6. #5  
    Forum Ph.D. william's Avatar
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    I assume you're using
    cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
    by the way....

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  7. #6  
    Forum Freshman daniel's Avatar
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    Well, it says cos(3x+pi/4), not cos(pi/4)
    Here's what I did to reach the answer I got:
    sqrt2*(cos(3x+ pi/4))
    =
    sqrt2*(cos3x*sin(pi/4) - cos(pi/4)*sin3x)
    =
    sqrt2*(cos3x - sin3x)
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  8. #7  
    Forum Freshman daniel's Avatar
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    Oh, snap, I just got understood why the sqrt2 goes away. Thanks for your excellent assistance to my enlightenment, dear William.
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  9. #8  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by daniel
    Well, it says cos(3x+pi/4), not cos(pi/4)
    Here's what I did to reach the answer I got:
    sqrt2*(cos(3x+ pi/4))
    =
    sqrt2*(cos3x*sin(pi/4) - cos(pi/4)*sin3x)
    =
    sqrt2*(cos3x - sin3x)
    You're confusing cos(a+b) with sin(a+b)...
    and what you wrote indeed says "cos(pi/4)" (and sin(pi/4)) no?
    (But you still need to use the correct trig identity....)

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  10. #9  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by daniel
    Oh, snap, I just got understood why the sqrt2 goes away. Thanks for your excellent assistance to my enlightenment, dear William.
    No problem.
    But make sure you use the correct trig identity. It looks like you are using the one for sin(a-b)....

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  11. #10  
    Forum Freshman daniel's Avatar
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    Quote Originally Posted by william
    You're confusing cos(a+b) with sin(a+b)...
    Ah, yes, I didn't see that until now ^_^
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