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NH
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QL
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HM
21 tháng 9 2023 lúc 15:20

\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)

\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) =  - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B =  - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) =  - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) =  - \frac{3}{8}\end{array}\)

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YL
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LN
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NL
17 tháng 4 2019 lúc 18:09

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

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NN
30 tháng 4 2019 lúc 11:15

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

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TD
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NL
22 tháng 5 2020 lúc 13:24

\(=sin\left(x+\frac{\pi}{2}+42\pi\right)+cos\left(206\pi+\pi+x\right)+sin^2\left(32\pi+\pi+x\right)+sin^2\left(x+\frac{\pi}{2}-2\pi\right)\)

\(=sin\left(x+\frac{\pi}{2}\right)+cos\left(\pi+x\right)+sin^2\left(\pi+x\right)+sin^2\left(x+\frac{\pi}{2}\right)\)

\(=cosx-cosx+sin^2x+cos^2x\)

\(=sin^2x+cos^2x=1\)

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DH
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NL
2 tháng 9 2020 lúc 17:32

\(f\left(-x\right)=\left|-sinx-cosx\right|-\left|-sinx+cosx\right|\)

\(=\left|sinx+cosx\right|-\left|sinx-cosx\right|=-f\left(x\right)\)

\(\Rightarrow f\left(x\right)+f\left(-x\right)=0\)

\(\Rightarrow T=f\left(-\pi\right)+f\left(\pi\right)+f\left(-\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)+...+f\left(-\frac{\pi}{n}\right)+f\left(\frac{\pi}{n}\right)+f\left(0\right)\)

\(=0+0+...+0+f\left(0\right)=f\left(0\right)\)

\(=1-1=0\)

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H24
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NL
19 tháng 8 2020 lúc 12:15

\(\Leftrightarrow1-2sin^2\left(\frac{x}{2}+\frac{\pi}{6}\right)+1-2sin^2\frac{x}{2}=\frac{3}{2}-4sin\left(\frac{x}{2}\right)sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\)

\(\Leftrightarrow\left[sin\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\right]^2=\frac{1}{4}\)

\(\Leftrightarrow4cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right).sin^2\left(\frac{\pi}{12}\right)=\frac{1}{4}\)

\(\Leftrightarrow cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right)=\frac{1}{16sin^2\left(\frac{\pi}{12}\right)}=\frac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow1+cos\left(x+\frac{\pi}{6}\right)=\frac{2+\sqrt{3}}{2}\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow...\)

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JE
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NL
16 tháng 9 2020 lúc 15:00

a.

\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)

\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

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NL
16 tháng 9 2020 lúc 15:04

c.

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)

d.

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)

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TN
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NL
22 tháng 6 2020 lúc 7:25

\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)

\(=2cosx+sinx-cosx-sinx\)

\(=cosx\)

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