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Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$

I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.

So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.

Ducky
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4 Answers4

7

$\cos^4x+\sin^4 x\\=(\cos^2x+\sin^2x)^2-2(\sin x\cos x)^2\\=1-2(\frac{1}{2}\sin(2x))^2\\=1-\frac{1}{2}\sin^2(2x)\\=1-\frac{1}{2}\frac{1-\cos(4x)}{2}\\=\frac{1}{4}\cos(4x)+\frac{3}{4}$

Uncountable
  • 3,520
  • Bravo! Just brings me back to my high school age when I'm having a tough time of those terrible sines and cosines XXD – Vim Feb 17 '15 at 15:49
4

$$\cos^4x+\sin^4x\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-\frac12\sin^22x=1-\frac12\left(\frac12(1-\cos4x)\right)=\frac34+\frac14\cos4x$$ As: $$\sin^2x+\cos^2x=1;\\(a+b)^2=a^2+b^2+2ab;\\\sin2x=2\sin x\cos x;\\\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$$

Also: $$\cos^4x+\sin^4x=\left(\frac{e^{ix}+e^{-ix}}2\right)^4+\left(\frac{e^{ix}-e^{-ix}}2\right)^4\\=\frac1{16}({\small e^{-4 ix}+4 e^{-2 ix}+4 e^{2 ix}+e^{4 ix}+6+e^{-4 ix}-4 e^{-2 ix}-4 e^{2 ix}+e^{4i x}+6})\\=\frac34+\frac14\left(\frac{e^{4ix}+e^{-4ix}}2\right)=\frac34+\frac14\cos4x$$

RE60K
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2

$$\begin{align} \cos^4x + \sin^4 x &= \left( \cos^2 x \right)^2 + \left( \sin^2 x \right)^2 \\[4pt] &= \left( \frac{1 + \cos 2 x}{2}\right)^2 + \left( \frac{1-\cos 2x}{2} \right)^2 \\[4pt] &= \frac{1}{2}\left( 1 + \cos^2 2 x \right) \\[4pt] &= \frac{1}{2}\left( 1 + \frac{1+\cos 4 x}{2} \right) \\[4pt] &= \frac{1}{4}\left(\; 3 + \cos 4 x \;\right) \end{align}$$

From your particular stopping place, you could proceed thusly: $$\begin{align} \cos 4 x + 6 \sin^2 x \cos^2 x &= \cos 4 x + \frac32\cdot(2\sin x \cos x)^2 \\[4pt] &= \cos 4 x + \frac32\cdot \sin^2 2 x \\[4pt] &= \cos 4 x + \frac32\cdot \frac{1 - \cos 4 x}{2} \\[4pt] &= \frac14\left(\;3 + \cos 4 x\;\right) \end{align}$$

Blue
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1

$\cos^2 x \sin^2 x = (1 - \sin^2 x) \sin^ 2 x = \sin^2 x - \sin^ 4 x$ and by the same argument $\cos^2 x \sin^2 x = \cos^2 x - \cos^ 4 x$. Take the average of these two identities to obtain $$ \cos^2 x \sin^2 x = \frac{1}{2} - \frac{\cos^4x + \sin^4 x}{2} \, . $$ Now substitute this and simplify.

Hans Engler
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