sin3x+cos3x=2(cos5x+sin5x)
sinx - sin3x + sin5x =0
sin2x + sin22x = sin23x
cos3x - cos5x = sinx
sin3x + sin5x + sin7x = 0
sinx + sin2x + sin3x - cosx - cos2x - cos3x = 0
Chứng minh đẳng thức sau: \(\frac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}=tan3x\)
\(\frac{sinx+sin5x+sin3x}{cosx+cos5x+cos3x}=\frac{2sin3x.cos2x+sin3x}{2cos3x.cos2x+cos3x}=\frac{sin3x\left(2cos2x+1\right)}{cos3x\left(2cos2x+1\right)}=\frac{sin3x}{cos3x}=tan3x\)
sin3x + 1=2sin22x
sin2xcos3x = sin5x
cos5x + cos3x + sin2x =0
sin5x + 1 = 2sin2x
sin3xcosx + 2cos22x = 1 + cos3xsinx
sin3x + 1=2sin22x
<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)
<=> sin3x + 1 = 1 - cos4x
<=> sin3x = -cos4x
<=> sin3x + cos4x = 0
<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).
<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0
<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0
<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0
<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)
<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)
Tìm x để A=1, biết A=\(\frac{cos3x+cos5x+cos7x+cos9x}{sin3x+sin5x+sin7x+sin9x}\)
\(A=\frac{cos3x+cos9x+cos5x+cos7x}{sin3x+sin9x+sin5x+sin7x}=\frac{2cos6x.cos3x+2cos6x.cosx}{2sin6x.cos3x+2sin6x.cosx}\)
\(=\frac{2cos6x\left(cos3x+cosx\right)}{2sin6x\left(cos3x+cosx\right)}=tan6x\)
\(A=1\Rightarrow tan6x=1\Rightarrow x=\frac{\pi}{24}+\frac{k\pi}{6}\)
1+cosx+cos2x+cos3x=0
sinx+sin3x+sin5x=cosx+cos3x+cos5x
sin^2x + sin^2(3x) = 2sin^2(2x)
mọi người giúp mình giải phương trình này với mình cảm ơn
a/
\(\Leftrightarrow1+cos2x+cos3x+cosx=0\)
\(\Leftrightarrow2cos^2x+2cos2x.cosx=0\)
\(\Leftrightarrow2cosx\left(cosx+cos2x\right)=0\)
\(\Leftrightarrow2cosx\left(2cos^2x+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
b/
\(\Leftrightarrow2sin3x.cosx+sin3x=2cos3x.cosx+cos3x\)
\(\Leftrightarrow sin3x\left(2cosx+1\right)-cos3x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sin3x-cos3x\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
c/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=1-cos4x\)
\(\Leftrightarrow cos6x+cos2x-2cos4x=0\)
\(\Leftrightarrow2cos4x.cos2x-2cos4x=0\)
\(\Leftrightarrow2cos4x\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
Rút gọn biểu thức :
\(A=\dfrac{\sin x+\sin3x+\sin5x}{\cos x+\cos3x+\cos5x}\)
Rút gọn biểu thức :
\(A=\dfrac{\sin x+\sin3x+\sin5x}{\cos x+\cos3x+\cos5x}\)
Giải các phương trình sau:
1+cosx+cos2x+cos3x=0
sinx+sin3x+sin5x=cosx+cos3x+cos5x
MỌI NGƯỜI GIÚP MÌNH VỚI MÌNH CẢM ƠN
9. Rút gọn các biểu thức sau
A= cos7x - cos8x - cos9x + cos10x / sin7x - sin8x - sin9x + sin10x
B = sin2x + 2sin3x + sin4x / sin3x +2sin4x + sin5x
C= 1+cosx + cos2x + cos3x / cosx + 2cos^2 . x -1
D = sin4x + sin5x + sin6x / cos4x + cos5x + cos6x
\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
CM các đẳng thức:
a) \(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=tan\left(2x+\frac{15}{4}\right)\)
b) \(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2=-2sin8x.cos2x\)
\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)
\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)
\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)
\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)
\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)
\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)
\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)
\(=-sin10x-sin6x=-2sin8x.cos2x\)