THE MAXIMUM VALUE OF `SIN^4X + COS^4X ` IS

     
$sin^4x+cos^4x$I should rewrite this expression into a new form to plot the function.

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eginalign& = (sin^2x)(sin^2x) - (cos^2x)(cos^2x) \& = (sin^2x)^2 - (cos^2x)^2 \& = (sin^2x - cos^2x)(sin^2x + cos^2x) \& = (sin^2x - cos^2x)(1) longrightarrow,= sin^2x - cos^2xendalign

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eginalignsin^4 x +cos^4 x&=sin^4 x +2sin^2xcos^2 x+cos^4 x - 2sin^2xcos^2 x\&=(sin^2x+cos^2 x)^2-2sin^2xcos^2 x\&=1^2-frac12(2sin xcos x)^2\&=1-frac12sin^2 (2x)\&=1-frac12left(frac1-cos 4x2 ight)\&=frac34+frac14cos 4xendalign


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Let $$displaystyle y=sin^4 x+cos^4 x = left(sin^2 x+cos^2 x ight)^2-2sin^2 xcdot cos^2 x = 1-frac12left(2sin xcdot cos x ight)^2$$

Now using $$ sin 2A = 2sin Acos A$$

So, we get $$displaystyle y=1-frac12sin^2 2x$$


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Note that $a^2 + b^2 = (a+b)^2 - 2ab$

$$(sin^2 x)^2 + (cos^2 x)^2 = (sin^2 x + cos^2 x)^2 - 2sin^2 xcos^2 x =(sin^2 x + cos^2 x)^2 - 2(sin xcos x)^2 = \ 1 -frac sin^2 2x2$$

Note the following results:

$$ sin^2 x + cos^2 x = 1$$

$$ sin x cos x = fracsin 2x2$$


Expand in terms of complex exponentials.

$$sin^4 x + cos^4 x = left( frace^ix - e^-ix2i ight)^4 + left( frace^ix + e^-ix2 ight)^4$$

Notice that $i^4 = +1$, so we get

$$sin^4 x + cos^4 x = frac116 left( 2e^4ix + 2 e^-4ix + 12 ight)$$

where we use the relation $(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 ab^3 + b^4$. The terms of the form $a^3 b$ and $ab^3$ all cancel by addition.

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This leaves us with a final result:

$$sin^4 x + cos^4 x = frac416 left(frace^4ix + e^-4ix2 ight) + frac1216 = frac34 + frac14 cos 4x$$


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answered Sep 30, năm ngoái at 17:14
MuphridMuphrid
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If you want khổng lồ express in functions of higher frequencies like this $$sum_k=0^N sin(kx) + cos(kx)$$ Then you can use the Fourier transform together with convolution theorem. This will work out for any sum of powers of cos & sin, even $sin^666(x)$.


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answered Sep 30, năm ngoái at 17:09
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