Chứng minh:cos^3x.sinx - sin^3x.cosx = sin4x/4
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NM
1
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)
KK
6 tháng 4 2017
1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)
\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)
\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )
b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)
\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)
\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)
\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)
\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )
c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)
\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)
\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)
\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)
\(VT=\dfrac{1-sin2x}{1+sin2x}\)
\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)
\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)
\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )
d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)
\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)
\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )
A
1
17 tháng 12 2019
\(4\left(\sin^4x+\cos^4x\right)+\sqrt{3}\sin4x=2\)
<=> \(4\left[\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x\right]+\sqrt{3}\sin4x=2\)
<=> \(4\left(1-\frac{1}{2}\sin^22x\right)+\sqrt{3}\sin4x=2\)
<=> \(4-2\sin^22x+\sqrt{3}\sin4x=2\)
<=> \(-2\sin^22x+\sqrt{3}\sin4x=-2\)
<=> \(\cos4x-1+\sqrt{3}\sin4x=-2\)
<=> \(\cos4x+\sqrt{3}\sin4x=-1\)
<=> \(\frac{1}{2}\cos4x+\frac{\sqrt{3}}{2}\sin4x=-\frac{1}{2}\)
<=> \(\cos\frac{\pi}{3}.\cos4x+\sin\frac{\pi}{3}.\sin4x=\cos\frac{2\pi}{3}\)
<=> \(\cos\left(4x-\text{}\text{}\frac{\pi}{3}\right)=\cos\frac{2\pi}{3}\)
Phương trình lượng giác cơ bản. Em làm tiếp nhé!
MA
1
HP
1 tháng 6 2021
\(4\left(sin^4x+cos^4x\right)+sin4x-2=0\)
\(\Leftrightarrow4\left(1-2sin^2x.cos^2x\right)+2sin2x.cos2x-2=0\)
\(\Leftrightarrow2-2sin^22x+2sin2x.cos2x=0\)
\(\Leftrightarrow2\left(1-sin^22x+sin2x.cos2x\right)=0\)
\(\Leftrightarrow2\left(cos^22x+sin2x.cos2x\right)=0\)
\(\Leftrightarrow2cos2x\left(cos2x+sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x+sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2};x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
b/
\(sin^23x-cos^24x=sin^25x-cos^26x\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos6x-\frac{1}{2}-\frac{1}{2}cos8x=\frac{1}{2}-\frac{1}{2}cos10x-\frac{1}{2}-\frac{1}{2}cos12x\)
\(\Leftrightarrow cos6x+cos8x=cos10x+cos12x\)
\(\Leftrightarrow2cos7x.cosx=2cos11x.cosx\)
\(\Leftrightarrow cosx\left(cos11x-cos7x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos11x=cos7x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\11x=7x+k2\pi\\11x=-7x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{9}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
d/
\(\Leftrightarrow2sin8x.cosx=cos\left(\frac{\pi}{2}-2x\right)+1-1-cos\left(\frac{\pi}{2}+4x\right)\) (hạ bậc vế phải)
\(\Leftrightarrow2sin8x.cosx=sin2x+sin4x\)
\(\Leftrightarrow2sin8x.cosx=2sin3x.cosx\)
\(\Leftrightarrow cosx\left(sin8x-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin8x=sin3x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=3x+k2\pi\\8x=\pi-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)
28 tháng 6 2021
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 8 2020
\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)
\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)
\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)
\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)
\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{k\pi}{3}\\x=\frac{k\pi}{5}\end{matrix}\right.\)
TK
1
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 9 2020
\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow4-2sin^22x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)
\(\Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\)
\(\Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=-\frac{1}{2}\)
\(\Leftrightarrow...\)
\(cos^3x.sinx-sin^3x.cosx=sinx.cosx\left(cos^2x-sin^2x\right)\)
\(=\frac{1}{2}sin2x.cos2x=\frac{1}{4}sin4x\)
Sin4x chia 4 á bạn :v