cho mk hỏi câu này với :)
chứng minh rằng: sin5x -2sinx ( cos2x +cos4x)=sinx
Những câu hỏi liên quan
Chung minh. 1-cos2x/1+cos2x=tan^2x
Bien doi thanh tich
a, A= sina +sinb+sin(a+b)
b, B=cosa +cosb +cos(a+b)+1
c, C= 1 + sina + cosa
d. D = sinx + sin3x +sin5x+sin7x
Chứng minh
a, sinx*sin(pi/3-x)*sin(pi/3+x)=1/4sin3x
b, cosx*cos(pi/3-x)*cos(pi/3+x)=1/4cos3x
c, cos5x*cos3x+sin7x*sinx=cos2x *cos4x
d, sin5x -2sinx(cos2x+cos4x)=sinx
cosx-2cos3x1+sqrt{3}sinx
sinx+sinxleft(x+dfrac{pi}{3}right)+sin4xsinleft(2x-dfrac{pi}{3}right)
left(1-dfrac{1}{2sinx}right)cos^22x2sinx-3+dfrac{1}{sinx}
( sinx -2cosx)cos2x + sinx (cos4x - 1)cosx +dfrac{cos2x}{2sinx}
left(dfrac{cos4x+sin2x}{cos3x+sin3x}right)^22sqrt{2}sinleft(x+dfrac{pi}{4}right)+3
Đọc tiếp
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
Giải các phương trình sau
a. Cosx+cos2x+cos3x+cos4x=0
b. Sinx+sin3x+sin5x+sin7x=0
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx.\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=\pi+k2\pi\end{matrix}\right.\)
\(sinx+sin7x+sin3x+sin5x=0\)
\(\Leftrightarrow2sin4x.cos3x+2sin4x.cosx=0\)
\(\Leftrightarrow sin4x\left(cos3x+cosx\right)=0\)
\(\Leftrightarrow sin4x.cos2x.cosx=0\)
\(\Leftrightarrow sin4x=0\)
\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)
Lý do chỉ cần 1 pt sin4x=0 do sin4x bao hàm cả cosx và cos2x ở trong đó
Chứng minh:
\(\sin5x-2\sin x\times\left(\cos4x+\cos2x\right)=\sin x\)
\(sin5x-2sinx\left(cos4x+cos2x\right)=sin5x-2.2sinx.cosx.cos3x\)
\(=sin5x-2sin2x.cos3x\)
\(=sin5x-\left(sin5x+sin\left(-x\right)\right)\)
\(=-sin\left(-x\right)=sinx\)
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sin^3 x +cos^3 x -3sinx cosx+1=0
3 cosx -3sin2x= √3(cos2x+sinx)
4sin^3x +3sin^2x cosx -sinx-cos^3x=0
√3sin4x-cos4x=sinx- √3cosx
m.n giúp mk chứng minh với ạ
\(\dfrac{\sqrt{2}\left(sinx-cox\right)^2\left(1+2sin2x\right)}{sin3x+sin5x}=1-tanx\)
\(sin\left(2x-\dfrac{\pi}{4}\right)cos2x-2\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
(sin2x+cos2x)cosx+2cos2x -sinx=0
sinx + cosxsin2x + \(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)
\(\sqrt{3}cos5x-2sin3xcos2x-sinx=0\)
HELPING NOW!!!
Giair phương trình lượng giác sau:
1) cosx - cos2x +cos3x = 0
2) cos2x - sin2x = sin3x + cos4x
3) cos2x + 2sinx - 1 - 2sinxsosx = 0
4) 1+ sinx - cosx = sin2x - cos2x
5) \(\sqrt{2}\) sin (2x+\(\dfrac{\pi}{4}\)) - sinx - 3cosx +2 =0
6) sin2x + 2cos2x = 1+sinx - 4cosx
Chứng minh rằng
\(\dfrac{sin5x}{sinx}-\dfrac{cos5x}{cosx}=4cos2x\)
a, cos4x + 12sin2x -1 0
b, cos4x - sin4x + cos4x 0
c, 5.(sinx + dfrac{cos3x+sin3x}{1+2sin2x} ) 3 + cos2x với mọi xinleft(0;2piright)
d, dfrac{sin3x}{3}dfrac{sin5x}{5}
e, dfrac{sin5x}{5sinx}1
f, cos23x - cos2x - cos2x 0
g, cos4x + sin4x + cos(x-dfrac{pi}{4} ) . sin(3x-dfrac{pi}{4} ) - dfrac{3}{2} 0
h, sinleft(2x+dfrac{5pi}{2}right) - 3cosleft(x-dfrac{7pi}{2}right) 1 + 2sinx với xinleft(dfrac{pi}{2};2piright)
i, 5sinx - 2 3.( 1- sinx ) . tan3x
k, ( sin2x + sqrt{3}cos2x)2 - 5 cos le...
Đọc tiếp
a, cos4x + 12sin2x -1 = 0
b, cos4x - sin4x + cos4x = 0
c, 5.(sinx + \(\dfrac{cos3x+sin3x}{1+2sin2x}\) ) = 3 + cos2x với mọi x\(\in\left(0;2\pi\right)\)
d, \(\dfrac{sin3x}{3}=\dfrac{sin5x}{5}\)
e, \(\dfrac{sin5x}{5sinx}=1\)
f, cos23x - cos2x - cos2x =0
g, cos4x + sin4x + cos(\(x-\dfrac{\pi}{4}\) ) . sin(\(3x-\dfrac{\pi}{4}\) ) - \(\dfrac{3}{2}\) = 0
h, sin\(\left(2x+\dfrac{5\pi}{2}\right)\) - 3cos\(\left(x-\dfrac{7\pi}{2}\right)\)= 1 + 2sinx với x\(\in\left(\dfrac{\pi}{2};2\pi\right)\)
i, 5sinx - 2 = 3.( 1- sinx ) . tan3x
k, ( sin2x + \(\sqrt{3}cos2x\))2 - 5 = cos \(\left(2x-\dfrac{\pi}{6}\right)\)
l, \(\dfrac{2.\left(cos^6x+sin^6x\right)-sinx.cosx}{\sqrt{2}-2sinx}=0\)
m, \(\dfrac{\left(1+sinx+cos2x\right).sin\left(x+\dfrac{\pi}{4}\right)}{1+tanx}=\dfrac{1}{\sqrt{2}}cosx\)
Mọi người giúp mình nha ! Mình cần gấp cho ngày mai 