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\(cotb=a\Rightarrow\frac{cosb}{sinb}=a\Rightarrow cosb=a.sinb\)
\(sin\left(2b-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(sin2b-cos2b\right)\)
\(=\sqrt{2}sinb.cosb-\frac{\sqrt{2}}{2}\left(1-2sin^2b\right)=a\sqrt{2}sin^2b+\sqrt{2}sin^2b-\frac{\sqrt{2}}{2}\)
\(=\left(a\sqrt{2}+\sqrt{2}\right)sin^2b-\frac{\sqrt{2}}{2}=\left(a\sqrt{2}+\sqrt{2}\right).\frac{1}{1+cot^2b}-\frac{\sqrt{2}}{2}\)
\(=\frac{a\sqrt{2}+\sqrt{2}}{1+a^2}-\frac{\sqrt{2}}{2}\)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Theo định nghĩa ta có :
\(f'\left(x\right)=\lim\limits_{\Delta x\rightarrow0}\frac{f\left(a+\right)-f\left(a\right)}{\Delta x}\)
\(=\lim\limits_{\Delta x\rightarrow0}\frac{\left(a+\Delta x-1\right)\varphi\left(a+\Delta x\right)}{\Delta x}\) do (\(f\left(a\right)=0\))
\(=\lim\limits_{\Delta x\rightarrow0}\varphi\left(a+\Delta x\right)\)
Khi \(\Delta x\rightarrow0\) thì \(a+\Delta x\rightarrow a\) và do \(\varphi\left(x\right)\) là hàm liên tục tại x = a nên có :
\(\lim\limits_{\Delta x\rightarrow0}\varphi\left(a+\Delta x\right)=\varphi\left(a\right)\)
Vậy \(f'\left(a\right)=\varphi\left(a\right)\)
b.
ĐKXĐ: ...
\(\Leftrightarrow\frac{\pi}{3}cot\pi x=\frac{\pi}{6}+k\pi\)
\(\Leftrightarrow cot\pi x=\frac{1}{2}+3k\)
\(\Leftrightarrow\pi x=arccot\left(\frac{1}{2}+3k\right)+n\pi\)
\(\Leftrightarrow x=\frac{1}{\pi}arccot\left(\frac{1}{2}+3k\right)+n\)
c.
\(\Leftrightarrow\left[{}\begin{matrix}\pi tan3x=\frac{\pi}{6}+k2\pi\\\pi tan3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tan3x=\frac{1}{6}+2k\\tan3x=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}arctan\left(\frac{1}{6}+2k\right)+\frac{n2\pi}{3}\\x=\frac{1}{3}arctan\left(\frac{5}{6}+2k\right)+\frac{n2\pi}{3}\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\pi}{2}sin\pi\left(x+1\right)=\frac{\pi}{4}+k\pi\)
\(\Leftrightarrow sin\pi\left(x+1\right)=\frac{1}{2}+2k\)
Do \(-1\le sin\pi\left(x+1\right)\le1\Rightarrow k=0\)
\(\Rightarrow sin\pi\left(x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{6}+k2\pi\\\pi\left(x+1\right)=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{6}+2k\\x+1=\frac{5}{6}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{6}+2k\\x=-\frac{1}{6}+2k\end{matrix}\right.\)
Có: \(sin^2\phi=\frac{1}{1+cot^2\phi}=\frac{1}{a^2+1}\), Từ đây ta được các đẳng thức:
\(sin2\phi=2sin\phi cos\phi=2cot\phi sin^2\phi=\frac{2a}{a^2+1}\)
\(cos2\phi=1-2sin^2\phi=1-\frac{2}{a^2+1}=\frac{a^2-1}{a^2+1}\)
Xét: \(sin\left(2\phi-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(sin2\phi-cos2\phi\right)=\frac{\sqrt{2}}{2}\left(\frac{2a}{a^2+1}-\frac{a^2-1}{a^2+1}\right)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}\left(a+1\right)}{a^2+1}\)