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Math Help - Trig Review

  1. #1
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    Trig Review

    we got a bunch of review problems for the test thats coming up in a few days, I got most of them figured out but am stuck on these 3 if anyone could give me a few pointers of where to head

    1. Solved
    Code:
      
    TANx + TAN^3x + SINx
    ---------------------- =   COSx + SEC^2x
                TANx
    
    
    TANx(1 + TAN^2x) + SINx
    -------------------------- =
                TANx
    
    TANx(SEC^2x) + SINx
    --------------------- =
            TANx
    
    (SEC^2x) + SINx =
    2. Solved
    Code:
     
    COS(x+y)
    --------- = 1 - TANxTANy
    COSxCOSy
    
    COSxCOSy - SINxSINy
    --------------------- =
                COSxCOSy
    
    COSxCOSy - TANxTANy =
    
    not sure how to turn COSxCOSy into 1
    3.
    Code:
    SIN3x    COS3x
    ------ - ------ = 2
    SINx      COSx
    
    SIN(x+2x)   COS(x+2x)
    --------- - ---------- =
    SINx           COSx
    
    SINxCOS2x + COSxSIN2x      COSxCOS2x - SINxSIN2x
    -----------------------  -  ----------------------- =
    SINx                               COSx
    
    COSxSINxCOS2x + COS^2xSIN^2x - SINxCOSxCOS2x - SIN^2xSIN^2x
    --------------------------------------------------------------- =
                                        SINxCOSx
    
    COS^2xSIN2x -SIN^2xSIN2x
    ---------------------------- =
                  SINxCOSx
    COS^2x(2SINxCOSx) - SIN^2x(2SINxCOSx)
    ------------------------------------ =
                  SINxCOSx
    
    2SINxCOSx(COS^2x - SIN^2x)
    ----------------------------- =
           SINxCOSx
    
    2(COS^2x - SIN^2x) =
    if anyone could give me suggestions of where to go for the next steps in these, that would help me out alot, I'll post back if I figure anything out in the mean time
    Last edited by drewms64; April 4th 2006 at 04:55 PM.
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  2. #2
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    In
    Code:
    =C++
    
    Tanx(Sec^2 x)+Sin x
    -----------------------
             Tan x
    When you divide by tangent you get,
    <br />
\sec^2x+\frac{\sin x}{\tan x}
    But,
    \frac{\sin x}{\tan x}=\cos x
    Because, \tan x=\frac{\sin x}{\cos x}
    Thus, you are left with,
    \sec^2x+\cos x and the proof is complete.
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  3. #3
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    ahh, thanks for the help, didnt notice I had to divide by both sides there and that I could use COS = SIN/TAN


    for 2. , I figured out what I was doing wrong, just gotta finish the 3rd one now

    Code:
    COSxCOSy - SINxSINy
    --------------------- =
                COSxCOSy
    
    COSxCOSy - TANxTANy =
    
    should of been
    
    COSxCOSy     SINxSINy
    ---------- - -------- =
    COSxCOSy      COSxCOSy
    
    1 - TANxTANy = 1 - TANxTANy
    Last edited by drewms64; April 4th 2006 at 02:09 PM.
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  4. #4
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    well I got the 3rd one down to
    2(COS^2x - SIN^2x) =
    but not sure how to make (COS^2x - SIN^2x) into 1, I know that COS^2x + SIN^2x = 1 though

    edit: if I did
    Code:
    2(1-SIN^2x - 1-COS^2x) 
    2(-SIN^2x - COS^2x)
    does -SIN^2x - COS^2x = 1? or could I just multiply the part inside parenthesis by -1 to become (SIN^2x + COS^2x) so it = 1
    Last edited by drewms64; April 4th 2006 at 07:26 PM.
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  5. #5
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    Quote Originally Posted by drewms64
    well I got the 3rd one down to
    2(COS^2x - SIN^2x) =
    but not sure how to make (COS^2x - SIN^2x) into 1, I know that COS^2x + SIN^2x = 1 though
    For #3
    Code:
    =C++
    
    SIN3x    COS3x
    ------ - ------ = 2
    SINx      COSx
    You are doing this the primitive way. Rather add the fractions immediately to get,
    \frac{\sin 3x\cos x-\cos 3x\sin x}{\sin x\cos x}
    Recognizing the sine for angle difference we have,
    \frac{\sin 2x}{\sin x\cos x}
    Using the double angle formula we get,
    \frac{2\sin x\cos x}{\sin x\cos x}
    Thus, =2

    \mathbb{Q}.\mathbb{E}.\mathbb{D}
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  6. #6
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    wow, that was alot easier then what I was trying, thanks for the help
    I'll post back with my results on the test tomorrow
    Last edited by drewms64; April 4th 2006 at 07:43 PM.
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