Cho \(sinx+siny=\sqrt{3}\); \(cosx-cosy=1\). Tính \(cos\left(x+y\right)=?\)
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24 tháng 5 2021
\(2sin\left(x+y\right)=sinx+siny\)
\(\Leftrightarrow2.2.sin\dfrac{x+y}{2}.cos\dfrac{x+y}{2}=2.sin\dfrac{x+y}{2}.cos\dfrac{x-y}{2}\)
\(\Leftrightarrow2cos\dfrac{x+y}{2}=cos\dfrac{x-y}{2}\)
\(\Leftrightarrow2\left(cos\dfrac{x}{2}.cos\dfrac{y}{2}-sin\dfrac{x}{2}.sin\dfrac{y}{2}\right)=cos\dfrac{x}{2}.cos\dfrac{y}{2}+sin\dfrac{x}{2}.sin\dfrac{y}{2}\)
\(\Leftrightarrow cos\dfrac{x}{2}.cos\dfrac{y}{2}=3.sin\dfrac{x}{2}.sin\dfrac{y}{2}\)
\(\Leftrightarrow\left(sin\dfrac{x}{2}:cos\dfrac{x}{2}\right).\left(sin\dfrac{y}{2}:cos\dfrac{y}{2}\right)=\dfrac{1}{3}\)
\(\Leftrightarrow tan\dfrac{x}{2}.tan\dfrac{y}{2}=\dfrac{1}{3}\)
24 tháng 5 2021
2sin(x+y)=sinx+siny2sin(x+y)=sinx+siny
⇔2.2.sinx+y2.cosx+y2=2.sinx+y2.cosx−y2⇔2.2.sinx+y2.cosx+y2=2.sinx+y2.cosx−y2
⇔2cosx+y2=cosx−y2⇔2cosx+y2=cosx−y2
⇔2(cosx2.cosy2−sinx2.siny2)=cosx2.cosy2+sinx2.siny2⇔2(cosx2.cosy2−sinx2.siny2)=cosx2.cosy2+sinx2.siny2
⇔cosx2.cosy2=3.sinx2.siny2⇔cosx2.cosy2=3.sinx2.siny2
⇔(sinx2:cosx2).(siny2:cosy2)=13⇔(sinx2:cosx2).(siny2:cosy2)=13
⇔tanx2.tany2=13⇔tanx2.tany2=13
HP
1 tháng 6 2021
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 \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
HP
1 tháng 6 2021
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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
\(\Leftrightarrow1+2sinx.cosx+1-cosx=2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2\sqrt{3}sin^2x+\left(4-\sqrt{3}\right)sinx-2\right)=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)-\left(2sinx-1\right)\left(\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(cosx+\sqrt{3}sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2021
\(=\dfrac{1+sinx+1-sinx}{\sqrt{\left(1-sinx\right)\left(1+sinx\right)}}=\dfrac{2}{\sqrt{1-sin^2x}}=\dfrac{2}{\sqrt{cos^2x}}=\dfrac{2}{\left|cosx\right|}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\left(sinx+siny\right)^2=3\Leftrightarrow sin^2x+sin^2y+2sinxsiny=3\) (1)
\(\left(cosx-cosy\right)^2=1\Leftrightarrow cos^2x+cos^2y-2cosx.cosy=1\) (2)
Cộng vế với vế (1) và (2):
\(sin^2x+cos^2x+sin^2y+cos^2y-2\left(cosx.cosy-sinx.siny\right)=4\)
\(\Leftrightarrow2-2cos\left(x+y\right)=4\)
\(\Rightarrow cos\left(x+y\right)=-1\)