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Math Help - Trigonometry

  1. #1
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    Trigonometry

    1. Verify the identity (sinx/1-cosx) -cotx=cscx

    2. Show that the equation sin(x+y)=sin x + sin y is not an identity.

    3. Write the product 11sin (x/2) cos (x/4) as a sum.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    I will give you some hints, try them and see what you come up with

    Quote Originally Posted by johntuan View Post
    1. Verify the identity (sinx/1-cosx) -cotx=cscx
    - start with the left hand side
    - put everything in terms of sine and cosine
    - combine all terms into a single term (simplify)

    2. Show that the equation sin(x+y)=sin x + sin y is not an identity.
    weird question. start on the left hand side, expand it using the addition formula for sine. show that you can't get the right hand side from that

    3. Write the product 11sin (x/2) cos (x/4) as a sum.
    the product to sum formula for sine-cosine is:

    \sin A \cos B = \frac {1}{2} \big[ \sin (A + B) + \sin (A - B) \big]
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  3. #3
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    ok thanks I will try that. Can you tell me what each part does in this function f(x)=2tan(x+(pi/2))
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  4. #4
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    Krizalid's Avatar
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    That's a different question, give it its own thread.
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  5. #5
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    for 1 I am stuck on this L.H.S. sinx/1-cosx - 1/tanx what can I do from here.

    For 2 what would I do from here

    sinxcosy+cosxsiny=sinx+siny
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    for 1 I am stuck on this L.H.S. sinx/1-cosx - 1/tanx what can I do from here.
    i said change everything in terms of sines and cosines did i not? what is tan x doing there. and you still have two terms, combine them into one
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post

    sinxcosy+cosxsiny=sinx+siny
    that's pretty much it, i think

    in general, those two sides are not equal, only for particular values of x and y, which means they can't be identities
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    i said change everything in terms of sines and cosines did i not? what is tan x doing there. and you still have two terms, combine them into one
    What else can cotx equal then if not 1/tanx?
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  9. #9
    MHF Contributor red_dog's Avatar
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    \displaystyle\frac{\sin x}{1-\cos x}-\frac{\cos x}{\sin x}=\frac{\sin^2x-\cos x+\cos^2x}{\sin x(1-\cos x)}=
    \displaystyle =\frac{1-\cos x}{\sin x(1-\cos x)}=\frac{1}{\sin x}=\csc x
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by johntuan View Post
    What else can cotx equal then if not 1/tanx?
    cot x = \frac{cos x}{sin x}

    AND

    tan x = \frac{sin x}{cos x}
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    What else can cotx equal then if not 1/tanx?
    as red_dog pointed out:

    remember that we have a formula for tanx itself in terms of sine and cosine

    \tan x = \frac {\sin x }{\cos x}

    \Rightarrow \cot x = \frac {1}{\tan x} = \frac {1}{ \frac {\sin x }{\cos x}} = \frac {\cos x }{\sin x}
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