CMR :
sinx (1+ 2cos2x + 2cos4x + 2cos6x ) = sin7x
Những câu hỏi liên quan
Chứng minh: \(sin9x=sinx\left(1+2cos2x+2cos4x+2cos6x+2cos8x\right)\)
Rút gọn
A= \(\frac{cosx-cos2x-cos3x+cos4x}{sinx-sin2x-sin3x+sin4x}\)
B= sinx(1+2cos2x+2cos4x+2cos6x)
\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)
\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)
\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)
\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)
\(=sin7x\)
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2sin5x + 1/sinx = 2cos4x + 2cos2x + 3
\(\Leftrightarrow2sin5x.sinx+1=2cos4x.sinx+2cos2x.sinx+3sinx\)
\(\Leftrightarrow2sin5x.sinx+1=sin5x-sin3x+sin3x-sinx+3sinx\)
\(\Leftrightarrow2sin5x.sinx-sin5x-2sinx+1=0\)
\(\Leftrightarrow sin5x\left(2sinx-1\right)-\left(2sinx-1\right)=0\)
(2cos2x+1)(2cos6x+1)(2cos18x+1)=1
Nhận thấy \(sinx=0\) ko phải nghiệm
\(\Leftrightarrow sinx\left(2cos2x+1\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos2x.sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(sin3x-sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos6x.sin3x+sin3x\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos18x.sin9x+sin9x\right)=sinx\)
\(\Leftrightarrow sin27x=sinx\)
Rút gọn các biểu thức sau:
D = \(\frac{1+sin2x+cos2x}{1+sin2x-cos2x}\)E = \(\frac{sin2x+2sin3x+sin4x}{cos3x+2cos4x-cos5x}\)F = \(\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)G = \(\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}\)
\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)
\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)
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\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)
\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)
\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)
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\(G=\frac{cos2x-sin4x-cos6x}{cos2x+sin4x-cos6x}=\frac{-2sin4xsin2x-sin4x}{-2sin4xsin2x+sin4x}\)
\(G=\frac{-sin4x\left(2sin2x+1\right)}{-sin4x\left(2sin2x-1\right)}=\frac{2sin2x+1}{2sin2x-1}\)
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2cos4x + 2cos2x - √2 = 0
1,sin24x+cos26x=sin10x
2,2sin22x+sin7x-1=sinx
Rút gọn:
A=(2sin2x-sin4x)/(2sin2x+sin4x) B=(sin5x-sin3x)/(2cos4x) C=tanx((1+cos²x)/(sinx)-sinx)
\(A=\frac{2sin2x-2sin2x.cos2x}{2sin2x+2sin2x.cos2x}=\frac{1-cos2x}{1+cos2x}=\frac{2sin^2x}{2cos^2x}=tan^2x\)
\(B=\frac{2cos4x.sinx}{2cos4x}=sinx\)
Câu C ko dịch được đề
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Giải pt
\(2sin^22x+sin7x-1=sinx\)
\(\Leftrightarrow1-cos4x+sin7x-1=sinx\)
\(\Leftrightarrow sin7x-sinx-cos4x=0\)
\(\Leftrightarrow2.cos4x.sin3x-cos4x=0\)
\(\Leftrightarrow cos4x\left(2.sin3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{6}+k2\pi\\3x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (\(k\in Z\))
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