rút gọn biểu thức sau
P=sin2x + sin2(\(\dfrac{pi}{3}-x\)) =sinx*sin(\(\dfrac{pi}{3}-x\))
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15 tháng 6 2021
a1)\(\dfrac{sin110}{cos110}+\dfrac{cos20}{sin20}\)
\(=\dfrac{sin\left(180-70\right)}{cos\left(180-70\right)}+\dfrac{cos\left(90-70\right)}{sin\left(90-70\right)}\)
\(=\dfrac{sin70}{-cos70}+\dfrac{sin70}{cos70}=0\)
a2) \(sin^2x+sin^2\left(\dfrac{\pi}{3}-x\right)+sinx.sin\left(\dfrac{\pi}{3}-x\right)\)
\(=\dfrac{1}{2}\left(1-cos2x\right)+\dfrac{1}{2}\left[1-cos\left(\dfrac{2\pi}{3}-2x\right)\right]+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(\dfrac{\pi}{3}\right)\right]\)
\(=\dfrac{1}{2}-\dfrac{1}{2}.cos2x+\dfrac{1}{2}-\dfrac{1}{2}.cos\left(\dfrac{2\pi}{3}-2x\right)+\dfrac{1}{2}.cos\left(2x-\dfrac{\pi}{3}\right)-\dfrac{1}{4}\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x+cos\left(\dfrac{2\pi}{3}-2x\right)-cos\left(2x-\dfrac{\pi}{3}\right)\right]\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left[cos2x-2.sin\dfrac{\pi}{6}.sin\left(\dfrac{\pi-4x}{2}\right)\right]\)
\(=\dfrac{3}{4}-\dfrac{1}{2}\left(cos2x-cos2x\right)\)
\(=\dfrac{3}{4}\)
a3) \(sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(=\dfrac{1-cos2x}{2}+\dfrac{1}{2}\left[cos\left(-2x\right)+cos\left(\dfrac{2\pi}{3}\right)\right]\)
\(=\dfrac{1-cos2x}{2}+\dfrac{cos2x}{2}-\dfrac{1}{4}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}\)
\(=\dfrac{1}{4}\)
NM
1
KQ
29 tháng 7 2023
\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)
\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)
\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)
\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)+cos\left(x+\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)+cos\left(x\right)cos\left(\dfrac{\pi}{2}\right)-sin\left(x\right)sin\left(\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)-sin\left(x\right)\)
\(=0\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2022
\(A=\dfrac{1-cos2x}{2}+\dfrac{1-cos\left(\dfrac{2\pi}{3}-2x\right)}{2}+\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{3}\right)-\dfrac{1}{2}cos\left(\dfrac{\pi}{3}\right)\)
\(=\dfrac{3}{4}-\dfrac{1}{2}cos2x+\dfrac{1}{2}\left(cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(\dfrac{2\pi}{3}-2x\right)\right)\)
\(=\dfrac{3}{4}-cos2x-sin\left(\dfrac{\pi}{6}\right).sin\left(2x-\dfrac{\pi}{2}\right)\)
\(=\dfrac{3}{4}-cos2x+cos2x=\dfrac{3}{4}\)
15 tháng 6 2023
\(P=\dfrac{sin2x+sinx}{\dfrac{1}{2}\cdot cosx\cdot sin2x+sin2x}=\dfrac{sinx\left(2cosx+1\right)}{sin2x\left(\dfrac{1}{2}cosx+1\right)}\)
\(=\dfrac{2cosx+1}{2\cdot cosx\cdot\left(\dfrac{1}{2}cosx+1\right)}\)
19 tháng 8 2023
a: pi<x<3/2pi
=>cosx<0
=>\(cosx=-\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(tanx=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}\)
cot x=1:3/4=4/3
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{-3}{5}\cdot\dfrac{-4}{5}=\dfrac{24}{25}\)
\(cos2x=1-2\cdot sin^2x=1-2\cdot\left(-\dfrac{3}{5}\right)^2=\dfrac{7}{25}\)
\(tan2x=\dfrac{24}{25}:\dfrac{7}{25}=\dfrac{24}{7}\)
cot 2x=1:24/7=7/24
b: \(sin\left(x+\dfrac{pi}{3}\right)=sinx\cdot cos\left(\dfrac{pi}{3}\right)+sin\left(\dfrac{pi}{3}\right)\cdot cosx\)
\(=\dfrac{-3}{5}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{-4}{5}=\dfrac{-3-4\sqrt{3}}{10}\)
2
22 tháng 8 2023
`a)sin x =4/3`
`=>` Ptr vô nghiệm vì `-1 <= sin x <= 1`
`b)sin 2x=-1/2`
`<=>[(2x=-\pi/6+k2\pi),(2x=[7\pi]/6+k2\pi):}`
`<=>[(x=-\pi/12+k\pi),(x=[7\pi]/12+k\pi):}` `(k in ZZ)`
`c)sin(x - \pi/7)=sin` `[2\pi]/7`
`<=>[(x-\pi/7=[2\pi]/7+k2\pi),(x-\pi/7=[5\pi]/7+k2\pi):}`
`<=>[(x=[3\pi]/7+k2\pi),(x=[6\pi]/7+k2\pi):}` `(k in ZZ)`
`d)2sin (x+pi/4)=-\sqrt{3}`
`<=>sin(x+\pi/4)=-\sqrt{3}/2`
`<=>[(x+\pi/4=-\pi/3+k2\pi),(x+\pi/4=[4\pi]/3+k2\pi):}`
`<=>[(x=-[7\pi]/12+k2\pi),(x=[13\pi]/12+k2\pi):}` `(k in ZZ)`
22 tháng 8 2023
a: sin x=4/3
mà -1<=sinx<=1
nên \(x\in\varnothing\)
b: sin 2x=-1/2
=>2x=-pi/6+k2pi hoặc 2x=7/6pi+k2pi
=>x=-1/12pi+kpi và x=7/12pi+kpi
c: \(sin\left(x-\dfrac{pi}{7}\right)=sin\left(\dfrac{2}{7}pi\right)\)
=>x-pi/7=2/7pi+k2pi hoặc x-pi/7=6/7pi+k2pi
=>x=3/7pi+k2pi và x=pi+k2pi
d: 2*sin(x+pi/4)=-căn 3
=>\(sin\left(x+\dfrac{pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
=>x+pi/4=-pi/3+k2pi hoặc x-pi/4=4/3pi+k2pi
=>x=-7/12pi+k2pi hoặc x=19/12pi+k2pi
19 tháng 8 2023
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: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)
\(=\dfrac{tanx-1}{1+tanx}\)
\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)
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