pt lượng giác

Question

1) $sin^6x$ - $cos^6x$ + 2cos2x - $cos^24x$ =1

2) $sin^6x$ - $cos^6x$ = 13/12

tran85295 · Answer 1 · 7/19/2016 4:24:16 PM

$A=\sin^6x-\cos^6x=(\sin^2x-\cos^2x)(\sin^4x+\sin ^2x \cos^2 x+\cos^4x)$

$=-\cos 2x.\Big[1-\sin^2x.\cos^2x \Big] $

Ta luôn có $\begin{cases}-\cos2x \le 1 \\ 0\le1-\sin^2x\cos^2x \le 1\end{cases}$

$\Rightarrow A \le 1$

Trả lời 19-07-16 04:24 PM

tran85295
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=> pt VN – tran85295 19-07-16 04:25 PM

score 5 · Answer 2

2) $(sin^2x)^3-(cos^2x)^3=\frac{13}{12}$

$3sin^2x-3cos^2x-sin^23x-cos^23x=\frac{13}{3}$

$-3cos2x-1=\frac{13}{3}$

$cos2x=\frac{16}{9}$

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