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Cos^4x + sin^4x =

  1. Mar 30, 2015 #1
    I am a bit confused here.

    Cos2x + sin2x = 1

    Thus can I say

    Cos4x + sin4x = 1

    If I just sqroot each term:
    sqroot Cos4x + sqroot sin4x = sqroot (1) = 1

    ?
     
  2. jcsd
  3. Mar 30, 2015 #2
    No. Sqrt(a+b) and sqrt(a)+sqrt(b) is different. If you take sqrt on both sides, it has to be sqrt(a+b)=sqrt(c).
    But you can still write##cos^4x+sin^4x=(cos^2x+sin^2x)-2cos^2xsin^2x##
     
  4. Mar 30, 2015 #3

    PeroK

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    You can write it, but it wouldn't be correct. Instead:

    ##cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x = 1 - 2cos^2xsin^2x##
     
  5. Mar 30, 2015 #4
    I think that the simplest form will be:
    $$...=1-2\cos^2x\sin^2x=1-\frac{1}{2}\sin^2(2x)$$
    because:
    $$2\sin x\cos x=\sin(2x)$$
     
  6. Mar 30, 2015 #5
    it was a typo
     
  7. Mar 30, 2015 #6

    Svein

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    Using complex analysis makes it still more elegant. On one hand: [itex]e^{4ix}=cos(4x)+isin(4x) [/itex]. On the other hand: [itex]e^{4ix}=(cos(x)+isin(x))^{4}= cos^{4}(x) + 4icos^{3}(x)sin(x) + 6i^{2}cos^{2}(x)sin^{2}(x)+4i^{3}cos(x)sin^{3}(x)+i^{4}sin^{4}(x) [/itex]. Multiplying out: [itex]e^{4ix}=cos^{4}(x) + 4icos^{3}(x)sin(x) - 6cos^{2}(x)sin^{2}(x)-4icos(x)sin^{3}(x)+sin^{4}(x)[/itex]. Thus: [itex]cos(4x)+isin(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x) + i(4cos^{3}(x)sin(x) )-4cos(x)sin^{3}(x)) [/itex]. Taking the real part: [itex]cos(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x) [/itex]
     
  8. Apr 6, 2015 #7
    From the definition of trigonometric function
    cos=x/r
    sin=y/r
    From the Pythagorean theorem
    r^2=x^2+y^2

    cos^4(x) + sin^4(x)
    =(x^4+y^4)/r^4
    =((x^2+y^2)^2-2x^2y^2)/(r^4)
    =(r^4-2x^2y^2)/r^4
    !=1
     
    Last edited: Apr 6, 2015
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