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UnkleRhaukus

  • one year ago

Show that: \[\cos^2 (nx) = \frac{1 + \cos (2nx)}2\]

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  1. Astrophysics
    • one year ago
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    Does 1+cos(2x) = 2cos^2(x) not work here?

  2. ganeshie8
    • one year ago
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    I would start with \(\cos^2nx+\sin^2nx=1\)

  3. UnkleRhaukus
    • one year ago
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    \[LHS=1-\sin^2nx\]

  4. UnkleRhaukus
    • one year ago
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    @Astrophysics how can i show: \[1+\cos(2x) = 2\cos^2(x)\]

  5. amilapsn
    • one year ago
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    @UnkleRhaukus use the identity for \(\cos(A+B)\)

  6. ganeshie8
    • one year ago
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    \[1+\cos(2x) = \mathcal{R}(1+e^{i(2x)}) = \mathcal{R}(1+e^{ix}e^{ix}) \\~\\= \mathcal{R}(1+(\cos x+i\sin x)(\cos x+i\sin x))\\~\\ =\cdots \]

  7. Astrophysics
    • one year ago
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    You can use cos(A+B) = cosAsinB+sinAcosB cos(2x) = cos(x+x) = .... you see :P

  8. UnkleRhaukus
    • one year ago
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    \(\mathcal R\) is \(\Re\), the real component ?

  9. ganeshie8
    • one year ago
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    Yes

  10. anonymous
    • one year ago
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    |dw:1440057813193:dw|

  11. UnkleRhaukus
    • one year ago
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    \[RHS= \frac{1 + \cos nx\sin nx+\sin nx\cos nx}2\\ =\frac{1 + 2\cos nx\sin nx}2\]

  12. anonymous
    • one year ago
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    |dw:1440057916809:dw|

  13. ganeshie8
    • one year ago
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    That will do |dw:1440058025993:dw|

  14. amilapsn
    • one year ago
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    correction @Astrophysics \[\huge\color{red}{ \cos(A+B)=\cos A\cos B-\sin A\sin B}\]

  15. UnkleRhaukus
    • one year ago
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    \[RHS= \frac{1 + \cos nx\cos nx-\sin nx\sin nx}2\\ =\frac{1 + \cos^2nx-\sin^2nx}2\]

  16. anonymous
    • one year ago
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    |dw:1440058094301:dw|

  17. amilapsn
    • one year ago
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    Now use \(\sin^2x+\cos^2x=1\)

  18. UnkleRhaukus
    • one year ago
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    \[\vdots\\=\frac{\cos^2nx+\cos^2nx}2\\ =\cos^2nx\\ =LHS\]

  19. UnkleRhaukus
    • one year ago
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    got it now, not too hard (when using the right formulae)

  20. Astrophysics
    • one year ago
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    Thanks @amilapsn I just realized the mistake lol late night...

  21. amilapsn
    • one year ago
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    Yw! lol :D

  22. UnkleRhaukus
    • one year ago
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    [ no need to over complexify the problem ]

  23. anonymous
    • one year ago
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    something is wrong

  24. anonymous
    • one year ago
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    theres some mistake in there i cant put my finger on it yet

  25. UnkleRhaukus
    • one year ago
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    i suppose \[\cos^2nx+\sin^2nx=1\]follows directly from \[\cos^2\theta +\sin^2\theta=1\] with \(nx=\theta\)

  26. anonymous
    • one year ago
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    ohh i got it xD

  27. anonymous
    • one year ago
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    |dw:1440058675021:dw|

  28. ganeshie8
    • one year ago
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    |dw:1440058596216:dw|

  29. ganeshie8
    • one year ago
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    try a formula for \(\cos^n(x)\) using the trig

  30. ganeshie8
    • one year ago
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    then you will appreciate how complexifying really simplfies things ;p

  31. anonymous
    • one year ago
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    even for this problem i think its still faster than just trig identities

  32. amilapsn
    • one year ago
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    lol...

  33. ganeshie8
    • one year ago
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    Yes it is cute only if you know complex numbers.. otherwise it is complex

  34. Astrophysics
    • one year ago
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    Nice pun ganeshie

  35. amilapsn
    • one year ago
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    complexity to simplicity..... I like that...

  36. ganeshie8
    • one year ago
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    show that \[\sin(nx) = \binom{n}{1}\sin x\cos^{n-1}x-\binom{n}{3}\sin^3 x\cos^{n-3}x+\binom{n}{5}\sin^5 x\cos^{n-5}x-\cdots\] where \(n\in \mathbb{Z^{+}}\)

  37. ganeshie8
    • one year ago
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    try proving that using trig, then il show a one line proof by complexifying :)

  38. UnkleRhaukus
    • one year ago
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    if you change the question, certain methods will be more or less appropriate

  39. ganeshie8
    • one year ago
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    sure, the beauty lies in the eyes of beholder :) for cos(nx) we have \[\cos(nx) = \binom{n}{0}\sin x\cos^{n}x-\binom{n}{2}\sin^2 x\cos^{n-2}x+\binom{n}{4}\sin^4 x\cos^{n-4}x-\cdots\] where \(n\in \mathbb{Z^{+}}\)

  40. ganeshie8
    • one year ago
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    i know at least 3 different methods to prove those identities : 1) trig 2) vectors 3) complex numbers all 3 methods provide different ways to look at the problem, personally i like complex numbers because they give the answer in fewer steps compared to other methods that i am familiar with

  41. anonymous
    • one year ago
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    this is the same proof as for \(\cos(x)^2=\frac12(1+\cos(2x))\) i.e. without the \(n\)

  42. Astrophysics
    • one year ago
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    Vectors...X_X that sounds the most complicated how do you prove it with vectors?

  43. anonymous
    • one year ago
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    consider: $$\cos(2x)=\cos^2(x)-\sin^2(x)\\1=\cos^2(x)+\sin^2(x)$$so it follows adding gives $$\cos(2x)+1=2\cos^2(x)\\\cos^2(x)=\frac12(1+\cos(2x))$$

  44. ganeshie8
    • one year ago
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    I think complex numbers are nothing but vectors in 2 dimensions with some attitude

  45. UnkleRhaukus
    • one year ago
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    i can't remember this formula \[\cos(2x)=\cos^2(x)-\sin^2(x)\]

  46. anonymous
    • one year ago
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    well, \(\cos(x+y)=\cos x\cos y-\sin x\sin y\)

  47. UnkleRhaukus
    • one year ago
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    yeah, i have to derive it form the sum/difference formula \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]

  48. UnkleRhaukus
    • one year ago
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    But how i am supposed to remember this one \[\cos(a\pm b)=\cos a\cos b\mp \sin a\sin b\]

  49. ganeshie8
    • one year ago
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    \[\cos(a\pm b)=\Re e^{i(a\pm b)}\]

  50. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Cosine

  51. UnkleRhaukus
    • one year ago
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    hmm.

  52. anonymous
    • one year ago
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    for \(\cos nx, \sin nx\) we can use recursion usign the angle sum/difference identities since \(\cos((n+1)x)=\cos(nx+x)=\dots\) and similarly in the other case

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