Toán Chứng minh rằng $sin^{4}x$ + $cos^{4}x$ = $\frac{3 + cos4x}{4}$ 13/10/2021 By Arianna Chứng minh rằng $sin^{4}x$ + $cos^{4}x$ = $\frac{3 + cos4x}{4}$
$VT=\sin^4x+\cos^4x$ $=(\sin^2x+\cos^2x)^2-2\sin^2x.\cos^2x$ $= 1-2(\sin x.\cos x)^2$ $=1-2(\dfrac{1}{2}\sin 2x)^2$ $=1-\dfrac{1}{2}sin^22x$ $=1-\dfrac{1}{2}.\dfrac{1-cos 4x}{2}$ $=1-\dfrac{1}{4}(1-\cos4x)$ $= \dfrac{3}{4}+\dfrac{1}{4}\cos 4x$ $=\dfrac{3+\cos 4x}{4}$ $= VP$ Trả lời
Đáp án: Giải thích các bước giải: $ sin^{4}x + cos^{4}x = (sin²x + cos²x)² – 2sin²xcos²x$ $= 1²- \frac{2}{4}(2sinxcosx)² = 1 – \frac{2}{4}sin²2x $ $= \frac{4 – 2sin²2x}{4} = \frac{3 + (1 – 2sin²2x)}{4} = \frac{3 + cos4x}{4}$ Trả lời
$VT=\sin^4x+\cos^4x$
$=(\sin^2x+\cos^2x)^2-2\sin^2x.\cos^2x$
$= 1-2(\sin x.\cos x)^2$
$=1-2(\dfrac{1}{2}\sin 2x)^2$
$=1-\dfrac{1}{2}sin^22x$
$=1-\dfrac{1}{2}.\dfrac{1-cos 4x}{2}$
$=1-\dfrac{1}{4}(1-\cos4x)$
$= \dfrac{3}{4}+\dfrac{1}{4}\cos 4x$
$=\dfrac{3+\cos 4x}{4}$
$= VP$
Đáp án:
Giải thích các bước giải:
$ sin^{4}x + cos^{4}x = (sin²x + cos²x)² – 2sin²xcos²x$
$= 1²- \frac{2}{4}(2sinxcosx)² = 1 – \frac{2}{4}sin²2x $
$= \frac{4 – 2sin²2x}{4} = \frac{3 + (1 – 2sin²2x)}{4} = \frac{3 + cos4x}{4}$