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shun Group Title

how can we derive thew formula for half angle for tangent.. pls help..

  • one year ago
  • one year ago

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  1. IsTim Group Title
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    Is this Advanced Functions?

    • one year ago
  2. IsTim Group Title
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    The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?

    • one year ago
  3. eyust707 Group Title
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    \[\tan \theta = {\sin \theta \over \cos \theta}\]

    • one year ago
  4. eyust707 Group Title
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    The derivative of tangent is: \[\frac{ d }{ d \theta } \tan \theta = \sec^2\theta \]

    • one year ago
  5. qpHalcy0n Group Title
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    I think he's looking for a proof of the tangent half angle formula, not an actual derivative.

    • one year ago
  6. ash2326 Group Title
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    @shun you want to derive \(\tan \frac {\theta}{2}\)??

    • one year ago
  7. ash2326 Group Title
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    We know that \[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\times \tan B}\] Put \[A=B=\frac {\theta}{2}\] \(\large \text{you'd get a relation of}\ \tan \theta \ \text{in terms of} \tan \frac{\theta}2\)

    • one year ago
  8. qpHalcy0n Group Title
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    Here, to prove it, use the half angle formulae for sine and cosine. \[\sin(x/2) = \pm \sqrt{ (1-cosx / 2}\] \[\cos(x/2) = \pm \sqrt{ (1+cosx / 2) }\]

    • one year ago
  9. shun Group Title
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    @ash yes,, actually,, it is,

    • one year ago
  10. ash2326 Group Title
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    you got it?

    • one year ago
  11. shun Group Title
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    yes i got it,, tnx,,

    • one year ago
  12. satellite73 Group Title
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    \[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]

    • one year ago
  13. shun Group Title
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    @qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation.. ..

    • one year ago
  14. qpHalcy0n Group Title
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    The derivation is based on those formulae. The half angle tangent is half angle sine / half angle cosine. You can prove half angle sine and cosine as well, if needs be.

    • one year ago
  15. shun Group Title
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    @sat.. yes that the beginning.. did you know some restrictions?

    • one year ago
  16. shun Group Title
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    @qp ok ok.. i get it.. tnx,,,

    • one year ago
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