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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2cosx-3cosx-sin\left(3\pi+\frac{\pi}{2}-x\right)+tan\left(\pi+\frac{\pi}{2}-x\right)\)
\(A=-cosx+sin\left(\frac{\pi}{2}-x\right)+tan\left(\frac{\pi}{2}-x\right)\)
\(A=-cosx+cosx+cotx=cotx\)
\(B=2cosx+sin\left(4\pi+\pi-x\right)+sin\left(2\pi-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sin\left(\pi-x\right)+sin\left(-\frac{\pi}{2}+x\right)-sinx\)
\(B=2cosx+sinx-sin\left(\frac{\pi}{2}-x\right)-sinx\)
\(B=2cosx-cosx=cosx\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\pi}{2}< a< \frac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{3}}{2}\)
\(A=cosa.cos\frac{4\pi}{3}+sina.sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}.\left(-\frac{1}{2}\right)+\frac{1}{2}.\left(-\frac{\sqrt{3}}{2}\right)=0\)
\(B=cos\left(2a+2019.2\pi\right)=cos2a=1-2sin^2a=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F\left(x\right)=sin\left(2\pi-\frac{\pi}{2}+x\right)+cos\left(14\pi-\frac{\pi}{2}-x\right)+sin\left(2x+\pi+x\right)-cos\left(6\pi+\pi-x\right)\)
\(=-sin\left(\frac{\pi}{2}-x\right)+cos\left(\frac{\pi}{2}+x\right)+sin\left(\pi+x\right)-cos\left(\pi-x\right)\)
\(=-cosx-sinx-sinx+cosx=-2sinx\)
b/ \(F\left(x\right)=-1\Leftrightarrow-2sinx=-1\)
\(\Rightarrow sinx=\frac{1}{2}\Rightarrow x=30^0\)
Tìm điểm biểu diễn trên đường tròn lượng giác của góc α = \(\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)
\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)
\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)
\(=\frac{1}{4}sin3a\)
\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)
\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)
\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)
\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=cos\left(6\pi+\pi-x\right)+sin\left(2\pi+\frac{\pi}{2}-x\right)+tan^2\left(\pi+\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(7\pi+\pi+x\right)}\)
\(=cos\left(\pi-x\right)+sin\left(\frac{\pi}{2}-x\right)+tan^2\left(\frac{\pi}{2}-x\right)-\frac{1}{sin^2\left(\pi+x\right)}\)
\(=-cosx+cosx+cot^2x-\frac{1}{sin^2x}\)
\(=cot^2x-\left(1+cot^2x\right)=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)\(P=cos2a-cos(\dfrac{\pi}{3}-a) \\=2cos^2a-1-cos\dfrac{\pi}{3}cosa-sin\dfrac{\pi}{3}sina \\=2.(\dfrac{-2}{5})^2-1-\dfrac{1}{2}.\dfrac{-2}{5}-\dfrac{\sqrt3}{2}.\dfrac{-\sqrt{21}}{5} \\=\dfrac{-24+15\sqrt7}{50}\)
a, Vì : \(\pi< a< \dfrac{3\pi}{2}\) nên \(cos\alpha< 0\) mà \(cos^2\alpha=1-sin^2\alpha=1-\dfrac{4}{25}=\dfrac{21}{25},\)
do đó : \(cos\alpha=-\dfrac{\sqrt{21}}{5}\)
từ đó suy ra : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)
Bài 1:
\(\Leftrightarrow-cosa-cosa+sina+cosa=0\Leftrightarrow sina=cosa\)
\(\Rightarrow a=\frac{\pi}{4}+k\pi\Rightarrow a\) thuộc cung thứ nhất và thứ 3
Bài 2:
Ta có \(\frac{5\pi}{3}-\left(-\frac{\pi}{3}\right)=\frac{6\pi}{3}=2\pi\Rightarrow\) góc \(\frac{5\pi}{3}\) và \(-\frac{\pi}{3}\) cùng cung biểu diễn