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anonymous

  • 5 years ago

cos(2tan-1x) rewrite as an algebraic expression

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  1. anonymous
    • 5 years ago
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    there is something wrong in the expression!!

  2. anonymous
    • 5 years ago
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    the tan-1 is inverse tan not minus 1

  3. anonymous
    • 5 years ago
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    oh I see

  4. anonymous
    • 5 years ago
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    sorry about that

  5. anonymous
    • 5 years ago
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    Well, arctan(x) is just an angle in a right angled triangle, then tan of which is x. So if you draw a triangle where tan(angle) = x, what are the side lengths?

  6. anonymous
    • 5 years ago
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    Maybe it makes more sense like this: let α = arctan(x) <=> tan(α) = x Therefore the opposite side over the ajacent side = .... ,

  7. anonymous
    • 5 years ago
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    OK I'll do one more step but no further without some sign you are thinking about it yourself: In a RA triangle with one angle α, the opposite side over the adjacent side = x, so let the sides be x and 1, respectively. Therefore, by Pythagoras, the hypotenuse = ...., and so cos α = ....

  8. anonymous
    • 5 years ago
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    sorry i was grabbing lunch

  9. anonymous
    • 5 years ago
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    OK then. Do you see where I'm going, though?

  10. anonymous
    • 5 years ago
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    yeah i understand. a^2 +b^2 = c ^2. what i don't understand is where the 2 comes int0 play

  11. anonymous
    • 5 years ago
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    before the tangent

  12. anonymous
    • 5 years ago
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    You can treat tan^-1(x) as a normal angle, so if you know what cos(tan^-1(x)) is just use the formula for cos(2a)

  13. anonymous
    • 5 years ago
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    im sorry but that doesn't make any sense to me. my brain is peicing together what you're saying

  14. anonymous
    • 5 years ago
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    piecing*

  15. anonymous
    • 5 years ago
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  16. anonymous
    • 5 years ago
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    Does that image help explain it better?

  17. anonymous
    • 5 years ago
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    hold on one second

  18. anonymous
    • 5 years ago
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    You could similarly use cos^2(x) + sin^2(x) = 1 -> 1 + tan^2(a) - 1/cos^2(a) -> cos^2(a) = 1/(1+tan^2(a)) And we know tan(a) = x as a = tan^-1(x)

  19. anonymous
    • 5 years ago
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    Let y=arctanx. The original equation then becomes cos2y. Using an identity this becomes cosy^2-siny^2. So your goal here is to obtain cosy^2 and siny^2. So we let y=arctanx. This means tany=x. We square and add 1 to both sides because of the identity tany^2+1=secy^2. So tany^2+1=x^2+1 and using the identity just mentioned, secy^2=x^2+1. Inverse both sides to get cosy^2= 1/ x^2+1. Now to obtain siny, simply multiple cosy (which you get by sqrting the cosy^2 from above) on both sides by tany. And since in the beginning we said tany=x. We get siny=x/sqrt(x^2+1). Subbing everything back into the orignal we get 1-x^2/(x^2+1)

  20. anonymous
    • 5 years ago
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    so if i just take what i find for tan (opposite/adjacent) and distribute a 2 before putting it back into terms of cosine?

  21. anonymous
    • 5 years ago
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    No, you take what is cos (adj/hyp = 1/sqrt(x^2+1) ) And then use cos(2a) = 2cos^2(a) - 1 = 2(1/sqrt(x^2+1) )^2 -1 = 2/(x^2+1) -1

  22. anonymous
    • 5 years ago
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    You know tan(a) = x by definition

  23. anonymous
    • 5 years ago
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    okay so cos(2a) got it. so you're just using tan to find the value's of the triangle?

  24. anonymous
    • 5 years ago
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    Yes, from which you can work out cos (or sin, but you don't need both here)

  25. anonymous
    • 5 years ago
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    ohhhhh okay. tan-1x = a ? so 2tan-1x = 2a right then cos(2a) is what i want?

  26. anonymous
    • 5 years ago
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    Yes, indeed :D

  27. anonymous
    • 5 years ago
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    ahh thanks let me work it out and ill reply with an answer

  28. anonymous
    • 5 years ago
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    cos(2sqrt(1+x^2)/x) ?

  29. anonymous
    • 5 years ago
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    Sorry there is a slight error. You use the triangle (or the equations) to work out cos(a) and THEN you use this value of cos(a) in the formula cos(2a) = 2cos^2(a) - 1. You do not need to work out cos or anything complicated, ad you already have the value of cos(a).

  30. anonymous
    • 5 years ago
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    my book says the answer is (1-x^2)/(1+x^2)

  31. anonymous
    • 5 years ago
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    but idk how to get there

  32. anonymous
    • 5 years ago
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    You could similarly use cos^2(x) + sin^2(x) = 1 -> 1 + tan^2(a) - 1/cos^2(a) -> cos^2(a) = 1/(1+tan^2(a)) And we know tan(a) = x as a = tan^-1(x) So cos^2(a) = 1/(1+x^2) and so cos(2a) is just this minus 1. Both would give the same value of cos(a) (and cos^2(a) ) , it just depends how you prefer to think about it

  33. anonymous
    • 5 years ago
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    OK, I'll help you with the next bit. From the triangle, you know cos(a) = adj/hyp = 1/sqrt(1+x^2) So cos^2(a) = 1/(1+x^2) Cos(2a) = 2cos^2(a) - 1 = 2/(1+x^2) - 1 = (2 - 1 - x^2)/1+x^2 = (1-x^2)/(1+x^2)

  34. anonymous
    • 5 years ago
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    okay yes

  35. anonymous
    • 5 years ago
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    Sorry the simplification/fraction look messy but hopefully the brackets make it clear

  36. anonymous
    • 5 years ago
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    You could also have got sin^2(a) form the triangle (x^2/(1+x^2)) and used cos(2a) = cos^2(a) - sin^2(a) = (1-x^2)/(1+x^2), it's all the same.

  37. anonymous
    • 5 years ago
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    those are pythag identities ?

  38. anonymous
    • 5 years ago
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    thats what i was missing. i didin't know i was supposed to be using those to get my answer.

  39. anonymous
    • 5 years ago
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    cos(2a) = cos^2(a) - sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a) This is a specfic version of the cos formulae: cos(a+b) = cos(a)cos(b) - sin(a)sin(b) It is not a pythag identity, and is pretty hard to prove

  40. anonymous
    • 5 years ago
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    no no i meant...

  41. anonymous
    • 5 years ago
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    cos^2 - 1

  42. anonymous
    • 5 years ago
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    that's part of cos^2 + sin^2 = 1 right?

  43. anonymous
    • 5 years ago
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    you used that to get the answer?

  44. anonymous
    • 5 years ago
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    It depends, you either use that, or the triangle to get the first part - I showed both methods - but you still need cos(2a) later.

  45. anonymous
    • 5 years ago
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    okay thanks! off to take my test.

  46. anonymous
    • 5 years ago
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    Have fun!

  47. anonymous
    • 5 years ago
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    lol sure

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