Chứng minh đẳng thức sau: \(\frac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}=tan3x\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=1-cos4x\)
\(\Leftrightarrow cos6x+cos2x-2cos4x=0\)
\(\Leftrightarrow2cos4x.cos2x-2cos4x=0\)
\(\Leftrightarrow2cos4x\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
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/
\(\Leftrightarrow2sin3x.cosx+sin3x=2cos3x.cosx+cos3x\)
\(\Leftrightarrow sin3x\left(2cosx+1\right)-cos3x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sin3x-cos3x\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{cos^3x-cos3x}{cosx}+\frac{sin^3x+sin3x}{sinx}=cos^2x-\frac{cos3x}{cosx}+sin^2x+\frac{sin3x}{sinx}\)
\(=1+\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=1+\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=1+\frac{2sin2x}{sin2x}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c/
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=cosx\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)
\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)
\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)
b/
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
2.
\(sinx-\sqrt{3}cosx=2sin5\text{}x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
\(\frac{sinx+sin5x+sin3x}{cosx+cos5x+cos3x}=\frac{2sin3x.cos2x+sin3x}{2cos3x.cos2x+cos3x}=\frac{sin3x\left(2cos2x+1\right)}{cos3x\left(2cos2x+1\right)}=\frac{sin3x}{cos3x}=tan3x\)