Rút gọn biểu thức A= \(\frac{sinx+sin3x}{2cosx}\)
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`A=[sin x + sin 2x + sin 3x]/[cos x + cos 2x + cos 3x]`
`A=[2sin2x.cosx+sin2x]/[2cos2x.cosx+cos2x]`
`A=[sin2x(2cosx+1)]/[cos2x(2cosx+1)]`
`A=tan 2x`
\(A=\dfrac{sinx-sin2x+sin3x}{cosx-cos2x+cos3x}\)
\(ĐK\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\) \(A=\dfrac{sinx+sin3x-sin2x}{cosx+cos3x-cos2x}\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}=\dfrac{2sin2x.cosx-sin2x}{2cos2x.cosx-cos2x}\\=\dfrac{sin2x\left(2cosx-1\right)}{cos2x\left(2cosx-1\right)}\end{matrix}\right.\) \(\Rightarrow\) \(A=tan2x\)
\(P=\dfrac{sin2x+sinx}{\dfrac{1}{2}\cdot cosx\cdot sin2x+sin2x}=\dfrac{sinx\left(2cosx+1\right)}{sin2x\left(\dfrac{1}{2}cosx+1\right)}\)
\(=\dfrac{2cosx+1}{2\cdot cosx\cdot\left(\dfrac{1}{2}cosx+1\right)}\)
\(A=\frac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)
\(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=\frac{2sin2x}{sin2x}=2\)
\(\frac{cosx+cos3x+cos2x+cos4x}{sinx+sin3x+sin2x+sin4x}=\frac{2cosx.cos2x+2cosx.cos3x}{2sin2x.cosx+2sin3x.cosx}=\frac{2cosx\left(cos2x+cos3x\right)}{2cosx\left(sin2x+sin3x\right)}\)
\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)
ĐKXĐ:
\(sin3x-sinx\ne0\)
\(\Leftrightarrow sin3x\ne sinx\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne\pi-x+n2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{4}+\frac{n\pi}{2}\end{matrix}\right.\)
TXĐ: \(sin3x-sinx\ne0\)
\(\Leftrightarrow sin3x\ne sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}3x\ne x+k2\pi\\3x\ne\pi-x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
\(A=\frac{sin3x+sinx}{2cosx}=\frac{2sin2x.cosx}{2cosx}=sin2x\)