Biết rằng \(sin\left(x-\frac{\pi}{3}\right)+sin\frac{13\pi}{2}=sin\left(x+\frac{\pi}{3}\right)\). Tính gtri của \(cosx\)
Chứng minh đẳng thức
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)=cosx\)
\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
\(=2cosx+sinx-cosx-sinx\)
\(=cosx\)
Tính \(\sin \left( {\alpha + \frac{\pi }{6}} \right),\cos \left( {\frac{\pi }{4} - \alpha } \right)\) biết \(\sin \alpha = - \frac{5}{{13}},\pi < \alpha < \frac{{3\pi }}{2}\)
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
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
Cho \(\cos 2x = \frac{1}{4}\).
Tính: \(A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)\); \(B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)\)
\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)
\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)
Giải các pt lượng giác sau
1) \(cos^2\left(x-\frac{\pi}{6}\right)-sin^2\left(x-\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\)
2) \(sin^4-sin^4\left(x+\frac{\pi}{2}\right)=sin\left(x+\frac{\pi}{3}\right)\)
3) \(8cos^3\left(x-\frac{\pi}{3}\right)-1=0\)
\(\text{1) }cos^2\left(x-\frac{\pi}{6}\right)-sin^2\left(x-\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+m2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{m2\pi}{3}\\x=\frac{\pi}{6}+n2\pi\end{matrix}\right.\\\Leftrightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3} \)
\(2\text{) }sin^4x-sin^4\left(x+\frac{\pi}{2}\right)=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow sin^2x-cos^2x=sin\left(x+\frac{\pi}{3}\right)\\ \Leftrightarrow cos\left(\pi-2x\right)=cos\left(\frac{\pi}{6}-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}\pi-2x=\frac{\pi}{6}-x+m2\pi\\\pi-2x=x-\frac{\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}-m2\pi\\x=\frac{7\pi}{18}-\frac{n2\pi}{3}\end{matrix}\right.\)
\(3\text{) }pt\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+m2\pi\\x=n2\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)
b/
\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
c/
\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)
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.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
sin\(\left(x+\frac{85\Pi}{2}\right)+cos\left(207\Pi+x\right)+sin^2\left(33\Pi+x\right)+sin^2\left(x-\frac{3\Pi}{2}\right)\)
\(=sin\left(x+\frac{\pi}{2}+42\pi\right)+cos\left(206\pi+\pi+x\right)+sin^2\left(32\pi+\pi+x\right)+sin^2\left(x+\frac{\pi}{2}-2\pi\right)\)
\(=sin\left(x+\frac{\pi}{2}\right)+cos\left(\pi+x\right)+sin^2\left(\pi+x\right)+sin^2\left(x+\frac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x\)
\(=sin^2x+cos^2x=1\)
Rút gọn các biểu thức sau:
1) \(A=2cosx+3cosx\left(\pi-x\right)-sin\left(\frac{7\pi}{2}-x\right)+tan\left(\frac{3\pi}{2}-x\right)\)
2) \(B=2sin\left(\frac{\pi}{2}+x\right)+sin\left(5\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)
\(A=2cosx-3cosx-sin\left(3\pi+\frac{\pi}{2}-x\right)+tan\left(\pi+\frac{\pi}{2}-x\right)\)
\(A=-cosx+sin\left(\frac{\pi}{2}-x\right)+tan\left(\frac{\pi}{2}-x\right)\)
\(A=-cosx+cosx+cotx=cotx\)
\(B=2cosx+sin\left(4\pi+\pi-x\right)+sin\left(2\pi-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sin\left(\pi-x\right)+sin\left(-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sinx-sin\left(\frac{\pi}{2}-x\right)-sinx\)
\(B=2cosx-cosx=cosx\)
\(\sin^2\left(x+\frac{\pi}{3}\right)+\sin^2\left(x+\frac{2\pi}{3}\right)=\frac{3-\sin x}{2}\)
\(\Leftrightarrow1-cos\left(2x+\frac{2\pi}{3}\right)+1-cos\left(2x+\frac{4\pi}{3}\right)=3-sinx\)
\(\Leftrightarrow cos\left(2x+\frac{2\pi}{3}\right)+cos\left(2x+\frac{4\pi}{3}\right)=sinx-1\)
\(\Leftrightarrow2cos\left(2x+\pi\right)cos\left(\frac{\pi}{3}\right)=sinx-1\)
\(\Leftrightarrow-cos2x=sinx-1\)
\(\Leftrightarrow-\left(1-2sin^2x\right)=sinx-1\)
\(\Leftrightarrow2sin^2x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)