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Math Help - Trig Identities

  1. #1
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    Trig Identities

    Establish each identity.

    NOTE: t = theta for short

    (1) csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)

    (2) csc (t) - cot (t) = [sin(t)]/[1 + cot(t)]
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  2. #2
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    For #1, just use the identity csc^{2}(t)=1+cot^{2}(t)

    Sub into the left side:

    (1+cot^{2}(t))^{2}-(1+cot^{2}(t))

    Now, expand that out and I bet it transforms into the right side.
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  3. #3
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    Quote Originally Posted by magentarita View Post
    Establish each identity.

    NOTE: t = theta for short

    (1) csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)

    (2) csc (t) - cot (t) = [sin(t)]/[1 + cot(t)]
    In the trig identity
    sin^2(t) +cos^2(t) = 1 -----------(i)
    divide both sides by sin^2(t),
    1 +cot^2(t) = csc^2(t) -----------(ii)

    The first problem:
    csc^4 (t) - csc^2 (t) = cot^4 (t) + cot^2 (t)
    We play on the LHS to see if it becomes equivalent to the RHS.
    LHS = [1 +cot^2(t)]^2 -[1 +cot^2(t)]
    = [1 +2cot^2(t) +cot^4(t)] -[1 +cot^2(t)]
    = 1 +2cot^2(t) +cot^4(t) -1 -cot^2(t)
    = cot^4(t) +cot^2(t)
    = RHS
    Proven.

    ------------------------------
    The second problem:

    csc (t) - cot (t) = [sin(t)] /[1 + cot(t)]

    From the derived Eq.(ii) above,
    1 = csc^2(t) -cot^2(t) ------------------(iia)

    Using a^2 -b^2 = (a+b)(a-b),
    We multiply, and divide at the same time, the LHS by [csc(t) +cot(t)],
    LHS = csc(t) -cot(t)
    = [csc^2(t) -cot^2(t)] / [csc(t) +cot(t)]
    = [1] / [{1/sin(t)} +{cos(t)/sin(t)}]
    = 1 / [{1 +cos(t)} /sin(t)]
    = 1 * [sin(t) /{1 +cos(t)}
    = sin(t) / [1 +cos(t)]

    Umm, I think there a typo in your posted problem/identity #2. It should be cos(t)....not the posted cot(t)....in the denominator of the RHS.
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  4. #4
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    great...

    Great replies.

    Thanks
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