\(\sin^4x+\cos^4x=1-\frac{1}{2}\sin^22x\)
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NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2020
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2020
d/
ĐKXĐ: \(sin2x\ne0\) \(\Leftrightarrow2x\ne k\pi\)
\(\Leftrightarrow1+\frac{cos2x}{sin2x}=\frac{1-cos2x}{sin^22x}\)
\(\Leftrightarrow sin^22x+sin2x.cos2x=1-cos2x\)
\(\Leftrightarrow sin^22x-1+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow-cos^22x+sin2x.cos2x+cos2x=0\)
\(\Leftrightarrow cos2x\left(sin2x-cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x-cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\left(l\right)\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
e/
\(\Leftrightarrow1+cos2x+1+cos4x+1+cos6x=3+3cosx.cos4x\)
\(\Leftrightarrow cos2x+cos6x+cos4x-3cosx.cos4x=0\)
\(\Leftrightarrow2cos4x.cos2x+cos4x-3cosx.cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1-3cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\\2cos2x-3cosx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(2cos^2x-1\right)-3cosx+1=0\)
\(\Leftrightarrow4cos^2x-3cosx-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arccos\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
d/
\(\Leftrightarrow5\left(1+cosx\right)=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow5\left(1+cosx\right)=2+sin^2x-cos^2x\)
\(\Leftrightarrow5+5cosx=2+1-cos^2x-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
AH
Akai Haruma
Giáo viên
28 tháng 11 2019
\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)
là giá trị không phụ thuộc vào biến (đpcm)
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\(B=\sin ^6x+\cos ^6x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=-\sin ^4x-\sin ^2x\cos ^2x+\sin ^2x=-\sin ^2x(\sin ^2x+\cos ^2x)+\sin ^2x\)
\(=-\sin ^2x+\sin ^2x=0\)
là giá trị không phụ thuộc vào biến (đpcm)
AH
Akai Haruma
Giáo viên
28 tháng 11 2019
\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)
\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)
\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)
\(=\frac{(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{1-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{-2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=-2\)
là giá trị không phụ thuộc vào biến $x$
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\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)
\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)
\(=1+\frac{2\sin ^2x}{\cos ^4x}\)
Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 6 2020
\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)
\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)
\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2019
\(A=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)+m\left(sin^4x+cos^4x\right)+\left(m+1\right)sin^22x\)
\(=\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x+m\left[\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right]+\left(m+1\right)sin^22x\)
\(=1-\frac{3}{4}sin^22x+m\left(1-\frac{1}{2}sin^22x\right)+\left(m+1\right)sin^22x\)
\(=m+1+\left(-\frac{3}{4}-\frac{m}{2}+m+1\right)sin^22x\)
\(=m+1+\left(\frac{m}{2}+\frac{1}{4}\right)sin^22x\)
Để biểu thức ko phụ thuộc x \(\Rightarrow\frac{m}{2}+\frac{1}{4}=0\Rightarrow m=-\frac{1}{2}\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)