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Homework Help: Proving Identities and Double angles problems

  1. Oct 30, 2005 #1
    I've been working at proving these identites and I just cant seem to figure them out, some of them just come to me others I work on them for 10 mintues or more and just get more and more bogged down.:frown:

    1) csc2x + cot2x = cotx so far i have
    1/sin2x + cos2x/sin2x = cosx/sinx
    1+cos2x/sinx = cosx/sinx

    2) cos^2x + 4cosx + 3 = cosx + 3 so far i have
    4cos^2x + 3 = cosx + 3

    3)1+cosx/1-cosx - 1-cosx/1+cosx = 4cotxcscx so far i have
    1+cosx/sin^2x - sin^2x/1+cosx = 4cosx/4sinx*1/sinx

    4)cos^2x-csc^2x/cot^2x = sin^2x - sec^2x so far i have
    cos^2x - 1/sin^2x/ cos^2X/sin^2x
    cos^2x - 1/sin^2x * sin^2X/cos^2x
    cos^2x - 1/cos^2x
    cos^2x - sec^2x

    5) 1 + sec^2x + cot^2x/ csc^2x = sec^2x

    1+ 1/cos^2x + cos^2x/sin^2x/ 1/sin^2x = 1/cos^2x

    6) sin(x+y) + sin(x-y)/cos(x+y) + cos(x-y) = tanx
    I have absolutely no clue what to do with this one, i've tried changing the left side to cosxsiny+sinxcosy and stuff like that but it just doesnt come to me

    Thanks for any help given to me, I really need some and greatly appreciate it!:smile:
  2. jcsd
  3. Oct 30, 2005 #2


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    Staff Emeritus
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    Gold Member

    First off, you need to learn how to use parentheses correctly:

    a + b / c + d = a + \frac{b}{c} + d


    (a + b) / (c + d) = \frac{a + b}{c + d}

    Secondly, I see you're converting everything into sines and cosines: that's usually a good thing.

    But you're missing two other easy steps:

    (1) Clearing denominators.

    For example, in order to prove

    [tex]\frac{a}{b} = \frac{c}{d}[/tex]

    whenever it is defined all you have to do is to prove

    [tex]ad = bc[/tex]

    whenever both b and d are nonzero. (it's okay to prove this holds even when b or d are zero)

    (2) Working with only one angle.

    In a lot of these, I see that you have trig functions with different arguments. E.G. in the first one, cos x and cos 2x both appear. You can often fix that.
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