\(\int_0^{\frac{\pi}{2}}\left(x+\sin\left(x\right)\right)\cos\left(x\right)\)
cho hàm số y=f(x) liên tục trên [0;π/2] thỏa \(\int_0^{\frac{\pi}{2}}f^2\left(x\right)dx=3\pi\) , \(\int_0^{\pi}\left(\sin x-x\right)f'\left(\frac{x}{2}\right)dx=6\pi\) ; \(f\left(\frac{\pi}{2}\right)=0\) Tính \(\int_0^{\frac{\pi}{2}}\left(f''\left(x\right)\right)^3dx\)
giúp em với ạ.
Cho \(\cos 2x = \frac{1}{4}\).
Tính: \(A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)\); \(B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)\)
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
1.\(\int_0^1x\left(e^{2x^{ }}-\frac{x^2}{\sqrt{4-x^2}}\right)dx\)
2.Giải pt \(\sin\left(2x+\frac{\pi}{2}\right)+\cos x-\sin x=0\)
rút gọn biểu thức:
A= cosa.sin( b-c )+ cosb. sin(c-a) + cosc.sin( a-b)
B= \(sin^2x+cos\left(\frac{\pi}{3}-x\right).cos\left(\frac{\pi}{3}+x\right)\)
C=\(sin^2x+sin^2\left(\frac{2\pi}{3}+x\right)+sin^2\left(\frac{2\pi}{3}-x\right)\)
D=\(sin^2\left(\frac{\pi}{4}+x\right)-sin^2x-2sinx.sin\frac{\pi}{4}.cos\left(\frac{\pi}{4}+x\right)\)
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
sin\(\left(x+\frac{85\Pi}{2}\right)+cos\left(207\Pi+x\right)+sin^2\left(33\Pi+x\right)+sin^2\left(x-\frac{3\Pi}{2}\right)\)
\(=sin\left(x+\frac{\pi}{2}+42\pi\right)+cos\left(206\pi+\pi+x\right)+sin^2\left(32\pi+\pi+x\right)+sin^2\left(x+\frac{\pi}{2}-2\pi\right)\)
\(=sin\left(x+\frac{\pi}{2}\right)+cos\left(\pi+x\right)+sin^2\left(\pi+x\right)+sin^2\left(x+\frac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x\)
\(=sin^2x+cos^2x=1\)
Cho hàm số \(y=f\left(x\right)=\left|\sin x-\cos x\right|-\left|\sin x+\cos x\right|\) .Với mọi số nguyên dương n tính \(T=f\left(-\pi\right)+f\left(-\frac{\pi}{2}\right)+...+f\left(-\frac{\pi}{n}\right)+f\left(0\right)+f\left(\frac{\pi}{n}\right)+...+f\left(\frac{\pi}{2}\right)+f\left(\pi\right)\)
\(f\left(-x\right)=\left|-sinx-cosx\right|-\left|-sinx+cosx\right|\)
\(=\left|sinx+cosx\right|-\left|sinx-cosx\right|=-f\left(x\right)\)
\(\Rightarrow f\left(x\right)+f\left(-x\right)=0\)
\(\Rightarrow T=f\left(-\pi\right)+f\left(\pi\right)+f\left(-\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)+...+f\left(-\frac{\pi}{n}\right)+f\left(\frac{\pi}{n}\right)+f\left(0\right)\)
\(=0+0+...+0+f\left(0\right)=f\left(0\right)\)
\(=1-1=0\)
\(\cos\left(x+\frac{\pi}{3}\right)+\cos x=\frac{3}{2}-4\sin\left(\frac{x}{2}\right)\cdot\sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(\frac{x}{2}+\frac{\pi}{6}\right)+1-2sin^2\frac{x}{2}=\frac{3}{2}-4sin\left(\frac{x}{2}\right)sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[sin\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\right]^2=\frac{1}{4}\)
\(\Leftrightarrow4cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right).sin^2\left(\frac{\pi}{12}\right)=\frac{1}{4}\)
\(\Leftrightarrow cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right)=\frac{1}{16sin^2\left(\frac{\pi}{12}\right)}=\frac{2+\sqrt{3}}{4}\)
\(\Leftrightarrow1+cos\left(x+\frac{\pi}{6}\right)=\frac{2+\sqrt{3}}{2}\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow...\)
GPT
a) \(sin\left(2x+1\right)+cos\left(3x-1\right)=0\)
b) \(sin\left(2x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{4}\right)\)
c) \(sin\left(3x+\frac{2\pi}{3}\right)+sin\left(x-\frac{7\pi}{5}\right)=0\)
d) \(cos\left(4x+\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{4}\right)=0\)
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.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
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.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
Chứng minh đẳng thức
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)=cosx\)
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
\(=2cosx+sinx-cosx-sinx\)
\(=cosx\)