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shun

  • 3 years ago

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

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  1. IsTim
    • 3 years ago
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    Is this Advanced Functions?

  2. IsTim
    • 3 years ago
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    The derivaive of TAN is cosx/sinx I think? Or is that just the reciprocal?

  3. eyust707
    • 3 years ago
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    \[\tan \theta = {\sin \theta \over \cos \theta}\]

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

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

  6. ash2326
    • 3 years ago
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    @shun you want to derive \(\tan \frac {\theta}{2}\)??

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

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

  9. shun
    • 3 years ago
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    @ash yes,, actually,, it is,

  10. ash2326
    • 3 years ago
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    you got it?

  11. shun
    • 3 years ago
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    yes i got it,, tnx,,

  12. anonymous
    • 3 years ago
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    \[\frac{\text{half angle formula for sine}}{\text{half angle formula for cosine}}\]

  13. shun
    • 3 years ago
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    @qp tnx.. i get it.. tank you.. but that it is the formulaim asking for the derivation.. ..

  14. qpHalcy0n
    • 3 years ago
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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.

  15. shun
    • 3 years ago
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    @sat.. yes that the beginning.. did you know some restrictions?

  16. shun
    • 3 years ago
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    @qp ok ok.. i get it.. tnx,,,

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