Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: 2*sin x+1<>0
=>sin x<>-1/2
=>x<>-pi/6+k2pi và x<>7/6pi+k2pi
b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)
mà 1+cosx>=0
nên 2-cosx>=0
=>cosx<=2(luôn đúng)
c ĐKXĐ: tan x>0
=>kpi<x<pi/2+kpi
d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)
=>cos(x-pi/4)<>1/2
=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi
=>x<>7/12pi+k2pi và x<>-pi/12+k2pi
e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi
=>x<>5/6pi+kpi và x<>kpi-pi/4
f: ĐKXĐ: cos^2x-sin^2x<>0
=>cos2x<>0
=>2x<>pi/2+kpi
=>x<>pi/4+kpi/2
\(a,y'=\left(tanx\right)'=\left(\dfrac{sinx}{cosx}\right)'\\ =\dfrac{\left(sinx\right)'cosx-sinx\left(cosx\right)'}{cos^2x}\\ =\dfrac{cos^2x+sin^2x}{cos^2x}\\ =\dfrac{1}{cos^2x}\\ b,\left(cotx\right)'=\left[tan\left(\dfrac{\pi}{2}-x\right)\right]'\\ =-\dfrac{1}{cos^2\left(\dfrac{\pi}{2}-x\right)}\\ =-\dfrac{1}{sin^2\left(x\right)}\)
a: pi/2<x<pi
=>cosx<0
=>\(cosx=-\sqrt{1-\left(\dfrac{1}{5}\right)^2}=-\dfrac{2\sqrt{6}}{5}\)
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{5}\cdot\dfrac{-2\sqrt{6}}{5}=\dfrac{-4\sqrt{6}}{25}\)
\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{24}{25}-1=\dfrac{48}{25}-1=\dfrac{23}{25}\)
\(tan2x=-\dfrac{4\sqrt{6}}{25}:\dfrac{23}{25}=-\dfrac{4\sqrt{6}}{23}\)
\(cot2x=1:\dfrac{-4\sqrt{6}}{23}=\dfrac{-23}{4\sqrt{6}}\)
b: \(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=sinx\cdot\dfrac{\sqrt{3}}{2}-cosx\cdot\dfrac{1}{2}\)
\(=\dfrac{1}{5}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{-2\sqrt{6}}{5}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}+2\sqrt{6}}{10}\)
c: \(cos\left(x-\dfrac{pi}{3}\right)=cosx\cdot cos\left(\dfrac{pi}{3}\right)+sinx\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=-\dfrac{2\sqrt{6}}{5}\cdot\dfrac{1}{2}+\dfrac{1}{5}\cdot\dfrac{1}{2}=\dfrac{-2\sqrt{6}+1}{10}\)
d: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)
\(=\dfrac{tanx-1}{1+tanx}\)
\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)
a)
$cos\left(x+\frac{\pi }{6}\right)=\frac{4}{5}cos\left(\frac{\pi }{6}\right)-\left(-\frac{3}{5}\right)sin\left(\frac{\pi }{6}\right)=\frac{4}{5}.\frac{\sqrt{3}}{2}+\frac{3}{5}.\frac{1}{2}=\frac{3+4\sqrt{3}}{10}$
b) $tan(x + \frac{\pi}{4}) = \frac{-3/5 + 1}{1 + (-3/5)(1)} = \frac{-2/5}{2/5} = -1$
1. \(D=R\)
2. \(sinx\ne0\Leftrightarrow x\ne k\pi\Rightarrow D=R\backslash\left\{k\pi|k\in R\right\}\)
3. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+\dfrac{k\pi}{2}|k\in R\right\}\)
4. \(cos\left(x+\dfrac{\pi}{4}\right)\ne0\Leftrightarrow x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\Rightarrow D=R\backslash\left\{\dfrac{\pi}{4}+k\pi|k\in R\right\}\)
