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Math (Trig) a.s.a.p, thanks!

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verify the identities:

1.) 2tanx-(1+tanx)^2 = -secx

2.) cos(x-330degrees)=1/2(ã3cosx-sinx)
*the ãis only over the 3*

3.)
1+cos2x/sinx=cotx

any help is greatly appreciated!!

  • Math (Trig) typo edit -

    ã is supposed to represent square root.
    so for number 2
    it's 1/2 (square root 3 cosx-sinx)

  • Math (Trig) a.s.a.p, thanks! -

    1.
    LS = 2sinx/cosx - (1+sinx/cosx)^2
    = 2sinx/cosx - ((cosx + sinx)/cosx)
    = 2sinx/cosx - (cos^2 x + 2sinxcosx + sin^2 x)/cos^2 x

    = (2sinxcosx -(1 + 2sinxcosx))/cos^2 x
    = -1/cos^2 x
    = -sec^2 x

    you had -secx on the right side
    I think you made a typo

    2. LS = cosxcos330 + sinxsin330

    but sin 330 = -sin30 = -1/2
    and cos 330 = cos30 = √3/2

    so above
    = (√3/2)cosx - 1/2 sinx
    = 1/2(√3 cosx - sinx)
    = RS

    Why don't you try the third one.
    Change everything to sines and cosines
    let me know how you did.

  • Math (Trig) a.s.a.p, thanks! -

    for the third one you had

    1+cos2x/sinx=cotx

    Your lack of brackets made this ambiguous.

    I tried (1+cos2x)/sinx=cotx and tested with 40º, did not work

    I tried 1+cos^2 x/sinx=cotx and tested with 40º, did not work

    I tried 1+(cos2x/sinx)=cotx and tested with 40º, did not work

    check your typing of the question.

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