ช่วยพิสูจน์หน่อยครับ(ตรีโกณ)

[MIN+]

...ช่วยพิสูจน์หน่อยครับ(ตรีโกณ)...

[กิตติ]

$\sin^6\theta +\cos^6\theta=1-\frac{3}{4}\sin^22\theta$
$\sin^6\theta +\cos^6\theta=(\sin^2\theta +\cos^2\theta)(\sin^4\theta-\sin^2\theta\cos^2\theta +\cos^4\theta)$
$=\sin^4\theta-\sin^2\theta(1-\sin^2\theta )+\cos^4\theta$
$=\sin^4\theta-\sin^2\theta+\sin^4\theta+\cos^4\theta$
$=2\sin^4\theta-\sin^2\theta+\cos^4\theta$
$=1-2\sin^2\theta\cos^2\theta+\sin^4\theta-\sin^2\theta$
$=1-2\sin^2\theta\cos^2\theta-\sin^2\theta(1-\sin^2\theta)$
$=1-2\sin^2\theta\cos^2\theta-\sin^2\theta\cos^2\theta$
$=1-3\sin^2\theta\cos^2\theta$
$=1-\frac{3}{4}\sin^22\theta $

$(\sin^2\theta +\cos^2\theta)^2=1$
$=\sin^4\theta +\cos^4\theta+2\sin^2\theta\cos^2\theta$
$\sin^4\theta +\cos^4\theta=1-2\sin^2\theta\cos^2\theta$

[กิตติ]

$\sin^8\theta +\cos^8\theta=\frac{1}{8}\sin^42\theta-\sin^22\theta+1$
$(\sin^2\theta +\cos^2\theta)^2=1$
$\sin^4\theta +\cos^4\theta=1-2\sin^2\theta\cos^2\theta$
$(\sin^4\theta +\cos^4\theta)^2=\sin^8\theta +\cos^8\theta+2\sin^4\theta \cos^4\theta$
$(1-2\sin^2\theta\cos^2\theta)^2=1-4\sin^2\theta\cos^2\theta+4\sin^4\theta\cos^4\theta$
$\sin^8\theta +\cos^8\theta+2\sin^4\theta \cos^4\theta=1-4\sin^2\theta\cos^2\theta+4\sin^4\theta\cos^4\theta$
$\sin^8\theta +\cos^8\theta=1-4\sin^2\theta\cos^2\theta+2\sin^4\theta\cos^4\theta$
$=1-\sin^22\theta+(2)(\frac{\sin2\theta}{2})^4 $
$=1-\sin^22\theta+\frac{1}{8}(\sin^42\theta)$

[MIN+]

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