B tan2 x . (cot2x + cos2x + sin2x – 1 ) + 10 B sin2 230 + + sin2670 – cos600
Đọc tiếp
B = tan2 x . (cot2x + cos2x + sin2x – 1 ) + 10
B = sin2 230 + 
+ sin2670 – cos600
Những câu hỏi liên quan
B sin2 230 + + sin2670 – cos600
Đọc tiếp
B = sin2 230 + + sin2670 – cos600
\(B=1-\dfrac{1}{2}+\dfrac{3}{2}=1+1=2\)
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chứng minh
a> cot2x/1+cot2x . 1+tan2x/tan2x = tan2x+cot2x/1=tan4x
b>tan2x-cos2x/sin2x + cot2x-sin2x/cos2x = 2
a: \(VT=\dfrac{cot^2x}{1+cot^2x}\cdot\dfrac{1+tan^2x}{tan^2x}\)
\(=\dfrac{cot^2x}{\dfrac{1}{sin^2x}}\cdot\dfrac{\dfrac{1}{cos^2x}}{tan^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{1}{cos^2x}:\dfrac{1}{sin^2x}\)
\(=\dfrac{cot^2x}{tan^2x}\cdot\dfrac{sin^2x}{cos^2x}\)
\(=cot^2x\)
\(VP=\dfrac{tan^2x+cot^2x}{1+tan^4x}=\dfrac{\dfrac{sin^2x}{cos^2x}+\dfrac{cos^2x}{sin^2x}}{1+\dfrac{sin^4x}{cos^4x}}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}:\dfrac{cos^4x+sin^4x}{cos^4x}\)
\(=\dfrac{sin^4x+cos^4x}{sin^2x\cdot cos^2x}\cdot\dfrac{cos^4x}{cos^4x+sin^4x}=\dfrac{cos^2x}{sin^2x}=cot^2x\)
=>VT=VP
b:
\(\dfrac{tan^2x-cos^2x}{sin^2x}+\dfrac{cot^2x-sin^2x}{cos^2x}\)
\(=\dfrac{\left(\dfrac{sinx}{cosx}\right)^2-cos^2x}{sin^2x}+\dfrac{\left(\dfrac{cosx}{sinx}\right)^2-sin^2x}{cos^2x}\)
\(=\dfrac{sin^2x-cos^4x}{cos^2x\cdot sin^2x}+\dfrac{cos^2x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x+cos^2x-cos^4x-sin^4x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{1-\left(cos^2x+sin^2x\right)^2+2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}\)
\(=\dfrac{2\cdot cos^2x\cdot sin^2x}{cos^2x\cdot sin^2x}=2\)
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Giúp mình với mn...
1)cos2x+cos22x+cos23x+cos24x=2
2) (1-tanx) (1+sin2x)=1+tanx
3) tan2x=sin3x.cosx
4) tanx +cot2x=2cot4x
5) sinx+sin2x+sin3x=cosx+cos2x+cos3x
6)sinx=√2 sin5x-cosx
7) 1/sin2x + 1/cos2x =2/sin4x
8) sinx+cosx=cos2x/1-sin2x
9)1+cos2x/cosx= sin2x/1-cos2x
10)sin3x+cos3x/2cosx-sinx=cos2x
A, sin2 x- 4sinx +3=0
B, 2cos2x- cosx-1=0
C, 3sin2x- 2cosx +2=0
D, 3cosx+ cos2x -cos3x +1=2sinx.sin2x
E, tan2 x+(\(\sqrt{3}\) +1)tanx-\(\sqrt{3}\)=0
F, \(\dfrac{\sqrt{3}}{sin^2x}\)=3cotx + \(\sqrt{3}\)
a, \(sin^2x-4sinx+3=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
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b, \(2cos^2-cosx-1=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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c, \(3sin^2x-2cosx+2=0\)
\(\Leftrightarrow3-3sin^2x+2cosx-5=0\)
\(\Leftrightarrow3cos^2x+2cosx-5=0\)
\(\Leftrightarrow\left(cosx-1\right)\left(3cosx+5\right)=0\)
\(\Leftrightarrow cosx=1\)
\(\Leftrightarrow x=k2\pi\)
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1 tháng 10 2016 lúc 12:00
giải các phương trình sau :
a) \(\tan x+\cot2x=2\cot4x\) ; b) \(\sin x+\sin^2\frac{x}{2}=0,5\) ; c) \(\sin x=\sqrt{2}\sin5x-\cos x\) ; d) \(\frac{1}{\sin2x}+\frac{1}{\cos2x}=\frac{2}{\sin4x}\) ; e) \(\sin x+\cos x=\frac{\cos2x}{1-\sin2x}\)
1> 1 + sinx + cosx + sin2x + cos2x = 0
2> cos2x + 3sin2x + 5 sinx - 3cosx = 3
3> \(\dfrac{\sqrt{2}*(cosx - sinx)}{cotx - 1}\) = \(\dfrac{1}{tanx + cot2x}\)
4> (2cosx - 1)*(2sinx + cosx) = sin2x - sinx
a) Biết sin2=\(\dfrac{9}{15}tính\cos2,\tan2,\cot,biết\cos2=\dfrac{3}{5}tính\sin2,\tan2,\cot2\)
1. Chứng minh các đẳng thức :
a) tan^{2}x - sin^{2}x = tan^{2}x . sin^{2}x
b) \frac{sin4x}{1+cos4x} . \frac{cos2x}{1+cos2x} = tanx 2. Rút gọn
a) A = \frac{sin^{2}x-tan^{2}x}{cos^{2}x - cot^{2}x}
b) B = \frac{1+cosx +cos2x + cos3x}{2cos^{2}x+ cosx - 1}
c) C = \frac{1+sin2x+ cos2x}{1+sin2x- cos2x}
Xem chi tiết
a) tan^{2}x - sin^{2}x = tan^{2}x . sin^{2}x
b) \frac{sin4x}{1+cos4x} . \frac{cos2x}{1+cos2x} = tanx 2. Rút gọn
a) A = \frac{sin^{2}x-tan^{2}x}{cos^{2}x - cot^{2}x}
b) B = \frac{1+cosx +cos2x + cos3x}{2cos^{2}x+ cosx - 1}
c) C = \frac{1+sin2x+ cos2x}{1+sin2x- cos2x}
25 tháng 8 2020 lúc 19:12
giải pt
a) \(sin^2x+2sin^22x+sin^23x-2=0\)
b) \(2cosx.cos\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
c) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)
d) \(1+cot2x=\frac{1-cos2x}{sin^22x}\)
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
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b/
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)
\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=k\pi\end{matrix}\right.\)
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c/
\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)
\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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