Chứng minh rằng $sin^{8}x$ + $cos^{8}x$ = 1 – $4sin^{2}x$.$cos^{2}x$ + 2$sin^{4}x$.$cos^{4}x$ 13/10/2021 Bởi Ariana Chứng minh rằng $sin^{8}x$ + $cos^{8}x$ = 1 – $4sin^{2}x$.$cos^{2}x$ + 2$sin^{4}x$.$cos^{4}x$
Ta có $\sin^8x + \cos^8x = (\sin^4x + \cos^4x)^2 – 2\sin^4x \cos^4x$ $= [(\sin^2x + \cos^2x)^2 – 2\sin^2 \cos^2 ]^2 – 2\sin^4x \cos^4x$ $= (1 – 2\sin^2x \cos^2x)^2 – 2\sin^4x \cos^4x$ $= 1 + 4\sin^4x \cos^4x – 4\sin^2x \cos^2x – 2\sin^4x \cos^4x$ $= 1 + 2\sin^4x \cos^4x – 4\sin^2x \cos^2x= VP$ Bình luận
Ta có
$\sin^8x + \cos^8x = (\sin^4x + \cos^4x)^2 – 2\sin^4x \cos^4x$
$= [(\sin^2x + \cos^2x)^2 – 2\sin^2 \cos^2 ]^2 – 2\sin^4x \cos^4x$
$= (1 – 2\sin^2x \cos^2x)^2 – 2\sin^4x \cos^4x$
$= 1 + 4\sin^4x \cos^4x – 4\sin^2x \cos^2x – 2\sin^4x \cos^4x$
$= 1 + 2\sin^4x \cos^4x – 4\sin^2x \cos^2x= VP$
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