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anonymous

  • one year ago

hey guys, need help is solving cos(2tan^-1x)

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  1. anonymous
    • one year ago
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    what do you have to do with that function...

  2. zepdrix
    • one year ago
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    simplify probably :)

  3. anonymous
    • one year ago
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    yea fair call, probs a good idea to apply what arctanx means

  4. zepdrix
    • one year ago
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    \[\large\rm \cos(2\color{orangered}{\arctan x})\] If \(\large\rm arctan x=\theta\) then \(\large\rm \tan\theta=x\) This arctangent is just some angle.\[\large\rm \cos(2\color{orangered}{\arctan x})=\cos(2\color{orangered}{\theta})\]So we need to apply our Double Angle Formula for Cosine. It shows up in three different forms, any of them will do.\[\large\rm \cos(2\theta)=2\cos^2\theta-1\]

  5. zepdrix
    • one year ago
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    We should draw a triangle to show what is going on with this tangent function.

  6. zepdrix
    • one year ago
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    \[\large\rm \tan \theta=x\qquad\to\qquad \tan\theta=\frac{x}{1}=\frac{opposite}{adjacent}\]

  7. zepdrix
    • one year ago
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    |dw:1444291200132:dw|

  8. anonymous
    • one year ago
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    |dw:1444291207356:dw|

  9. zepdrix
    • one year ago
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    Then use Pythagorean Theorem to find the missing hypotenuse.

  10. anonymous
    • one year ago
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    ye lel i'll delete mine

  11. zepdrix
    • one year ago
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    And use that information to finish it off

  12. anonymous
    • one year ago
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    probs a good idea to realise \[\theta=\tan ^{-1}x\] and you therefore wants to find \[\cos(2\theta)\]

  13. anonymous
    • one year ago
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    then use trig identities

  14. anonymous
    • one year ago
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    all g?

  15. anonymous
    • one year ago
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    thanks guys.

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