How which Trig Identity is that?

Discussion in 'Homework Help' started by k31453, Jun 15, 2013.

  1. k31453

    Thread Starter Member

    May 7, 2013
  2. Georacer


    Nov 25, 2009
    This one pops in mind, but it requires a cosine on the right end. Maybe it's a typo?
    Or it could be that the phase is irrelevant, but the equals sign shouldn't be there anyway.
  3. Shagas

    Active Member

    May 13, 2013
    I think you would first have to expand the sin(1000pit) .
    Because for example Sin(2x) expands into 2sinxcosx so I'm guessing this goes similarly
  4. k31453

    Thread Starter Member

    May 7, 2013

    Can you please show me working out !!
    I tried but it keeps giving me in cos not in sin
  5. amilton542

    Active Member

    Nov 13, 2010
    Just make use of the trig' identities and in your final result observe the relation is a typo.

     cos(\alpha + \beta) \equiv cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)

     cos(2\alpha) \equiv cos^2(\alpha) - sin^2(\alpha)

     cos(2\alpha) \equiv 1 - 2sin^2(\alpha)

     sin^2(\alpha) \equiv \frac{1}{2}[1 - cos(2\alpha)]

     Asin^2(\omega t) \equiv \frac{A}{2}[1 - cos(2\omega t)

     2000sin^2(500\pi t) \equiv \frac{2000}{2}[1 - cos((2)(500\pi t))] \equiv 1000[1 - cos(1000\pi t)]
    Georacer likes this.
  6. DerStrom8

    Well-Known Member

    Feb 20, 2011
    I think Geo is right--it's a typo. I'm pretty sure it should be cos, not sin.

    -1000(-1+cos(1000(pi)t)) is the result, and that is the same as 1000(1-cos(1000(pi)t)).

    EDIT: Amilton beat me to it :p
  7. WBahn


    Mar 31, 2012
    Two observations.

    1) Always, always, ALWAYS ask if the answer makes sense!

    Do the two expressions even agree with each other at the easily evaluated point of t=0?


    So you KNOW it is wrong! No point going any further. It's WRONG!

    2) Had it passed the sanity test (may or may not be right, but at least it would have a fighting change), then you start looking for identities that involve the squares of the trig functions one angle and the first order trig functions of an angle twice as large. In other words, either the "double-angle" formulas or the "half-angle" formulas.

    But since you already know that you CAN"T find an identity that will make this work out, there's no point in trying. What you CAN do is try to write the expression involving squared trig functions in terms of linear terms of the trig functions at twice the frequency, since that appears to be the intent.