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Homework Help: Proove trig identity help

  1. Feb 25, 2005 #1
    Proove trig identity help plz

    I think this problem involves the double angle formula which i am not quite familiar with.


    I know the cot=1/tan(x) or cos(x)/sin(x)


    cos2(x)=1-2sin^2 (x)

    But Im not sure what to do can u please help me? :rolleyes:
  2. jcsd
  3. Feb 25, 2005 #2
    here is what im getting so far

    1+cos(2x)=cot(x) sin2(x)

    1+cos(x+x)=[cos(x)/sin(x)] multiplied by sin(x+x)

    1+cos^2(x)=cos(x)/sin(x) multiplied by sin^2(x)

    Im doing something totally wrong please HELP ME
  4. Feb 25, 2005 #3
    [tex] cos2x [/tex] is equal to [tex] cos^2x-sin^2x [/tex] therefore for the left side:

    [tex] 1-sin^2x+cos^2x [/tex]
    [tex] cos^2x+cos^2x [/tex]
    [tex] 2cos^2x [/tex]

    for the right side:
    [tex] \frac{1}{tanx} (2sinxcosx) [/tex]

    [tex] \frac {2sinxcosx}{\frac{sinx}{cosx}} [/tex]

    [tex] 2sinxcosx (\frac {cosx}{sinx}) [/tex]

    [tex] 2cos^2x [/tex]
  5. Feb 26, 2005 #4
    Hi thanks I didnt know [tex] cos2x [/tex] is equal to [tex] cos^2x-sin^2x [/tex] Im just trying to understand what you did on the right side can u explain why you multiplied 1/tan(x) by (2sinxcosx) thanks soo much :smile:
  6. Feb 26, 2005 #5


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    Okay.Initially,the RHS was:

    [tex] \cot x \sin 2x [/tex](1)


    [tex] \cot x=\frac{1}{\tan x} [/tex](2)

    [tex] \sin 2x=2\sin x\cos x [/tex] (3)

    Multiply (2) and (3) and find:

    [tex]\cot x\sin 2x=\frac{2\sin x\cos x}{\tan x} [/tex] (4)

    ,which is just what u asked to prove.

  7. Feb 26, 2005 #6
    ok i totally understand that I think Im over looking one of the trig identities because I didnt know that cos2x=cos^2(x)-sin^2(x) and I also didnt know sin2x=2sin(x)cos(x)

    Hey how do u know that? :redface:
  8. Feb 26, 2005 #7


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    Simple,apply the formulas for addition (both for "sine" and for "cosine") and choose equal arguments
    [tex] \cos 2x=\cos (x+x)=...? [/tex]

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