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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2021
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2021
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
15 tháng 6 2016
A=cosa.cos\(\frac{\pi}{3}\)+sina.sin\(\frac{\pi}{3}\)-sina.cos\(\frac{\pi}{6}\)+cosa.sin\(\frac{\pi}{6}\)
A=\(\frac{1}{2}\)cosa+\(\frac{\sqrt{3}}{2}\)sina-\(\frac{\sqrt{3}}{2}\)sina+\(\frac{1}{2}\)cosa
A= cosa
9 tháng 5 2016
a) P = cos(\(\frac{\Pi}{2}\) + x) + cos(2π - x) + cos(3π + x) = -sinx + cosx - cosx = -sinx
5 tháng 7 2021
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)
\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)
\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)
\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)
\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)
\(\Rightarrow A=\dfrac{1}{32}\)
23 tháng 3 2018
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
18 tháng 5 2021
\(x+2y=\dfrac{\pi}{2}\)
\(\Leftrightarrow x+y=\dfrac{\pi}{2}-y\) thay vào A được:
\(A=\dfrac{cos\left(\dfrac{\pi}{2}-y\right)-cosy}{cos\left(\dfrac{\pi}{2}-y\right)+cosy}\)\(=\dfrac{siny-cosy}{siny+cosy}\)\(=\dfrac{\dfrac{\sqrt{2}}{2}.siny-\dfrac{\sqrt{2}}{2}.cosy}{\dfrac{\sqrt{2}}{2}.siny+\dfrac{\sqrt{2}}{2}cosy}\)\(=\dfrac{cos\dfrac{\pi}{4}.siny-sin\dfrac{\pi}{4}.cosy}{sin\dfrac{\pi}{4}.siny+cos\dfrac{\pi}{4}.cosy}\)
\(=\dfrac{sin\left(y-\dfrac{\pi}{4}\right)}{cos\left(y-\dfrac{\pi}{4}\right)}\)\(=tan\left(y-\dfrac{\pi}{4}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 5 2021
\(x+2y=\dfrac{\pi}{2}\Rightarrow x+y=\dfrac{\pi}{2}-y\)
\(\Rightarrow cos\left(x+y\right)=cos\left(\dfrac{\pi}{2}-y\right)\)
\(\Rightarrow cos\left(x+y\right)=siny\)
Do đó: \(A=\dfrac{siny-cosy}{siny+cosy}=\dfrac{\sqrt{2}sin\left(y-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(y-\dfrac{\pi}{4}\right)}=tan\left(y-\dfrac{\pi}{4}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 6 2020
\(A=cos\left(\pi+\frac{\pi}{2}-a\right)-sin\left(\pi+\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}-4\pi\right)-sin\left(a+\frac{\pi}{2}-4\pi\right)\)
\(=-cos\left(\frac{\pi}{2}-a\right)+sin\left(\frac{\pi}{2}-a\right)+cos\left(a+\frac{\pi}{2}\right)-sin\left(a+\frac{\pi}{2}\right)\)
\(=-sina+cosa-sina-cosa=-2sina\)
\(A=cos\left(\dfrac{\pi}{3}+\alpha\right)+cos\left(\dfrac{\pi}{3}-\alpha\right)\)
\(=cos\dfrac{\pi}{3}.cos\alpha-sin\dfrac{\pi}{3}.sin\alpha+cos\dfrac{\pi}{3}.cos\alpha+sin\dfrac{\pi}{3}.sin\alpha\)
\(=2.cos\dfrac{\pi}{3}.cos\alpha=cos\alpha\)
`A=cos(π/3 +α) + cos(π/3-α)`
`=cos π/3 . cos α - sin π/3 . sinα + cos π/3 . cosα + sin π/3 . sinα`
`=2 . 1/2 . cos α`
`= cosα`