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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)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)
\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)
\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)
\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)
\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)
\(=sin2x.cosx-cos2x.sinx=sinx\)
![](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)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(cos\left(2x+\frac{\pi}{6}\right)cos\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos4x+cos\frac{\pi}{3}\right)=\frac{1}{2}\left(cos4x+\frac{1}{2}\right)\)
\(sin\left(x+\frac{\pi}{6}\right)sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\left(cos\frac{\pi}{3}-cos2x\right)=\frac{1}{2}\left(\frac{1}{2}-cos2x\right)\)
\(\Rightarrow C=\frac{1}{2}sinx.cos4x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{2}sin3x.cos2x\)
\(=\frac{1}{4}sin5x-\frac{1}{4}sin3x+\frac{1}{4}sinx+\frac{1}{4}sin3x-\frac{1}{4}sin5x+\frac{1}{4}sinx\)
\(=\frac{1}{2}sinx\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(cos^2\left(\frac{\pi}{4}-x\right)=sin^2\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=sin^2\left(x+\frac{\pi}{4}\right)\)
\(\Rightarrow\frac{1-sin^2x}{2cot\left(\frac{\pi}{4}+x\right).cos^2\left(\frac{\pi}{4}-x\right)}=\frac{cos^2x}{2cot\left(\frac{\pi}{4}+x\right).sin^2\left(\frac{\pi}{4}+x\right)}=\frac{cos^2x}{2.cos\left(\frac{\pi}{4}+x\right).sin\left(\frac{\pi}{4}+x\right)}\)
\(=\frac{cos^2x}{sin\left(\frac{\pi}{2}+2x\right)}=\frac{cos^2x}{cos2x}\)???
Đến đây thì đoán là bạn ghi sai đề, tử số phải là \(cos^2x-sin^2x\) chứ ko phải \(1-sin^2x\) vì \(cos^2x-sin^2x=cos2x\) mới rút gọn hết với mẫu
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{2}-x-\frac{\pi}{6}\right)sin\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{3}-x\right)sin\left(-x-\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(x-\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}-x-\frac{\pi}{4}\right)=cos\left(-\frac{7\pi}{12}\right)=cos\frac{7\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}\)
Ủa thì \(cos\frac{\pi}{2}=0\) là 1 giá trị lượng giác quen thuộc mà bạn, thay trực tiếp vô thôi :D
\(cos\left(\frac{17\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)+sin^2x\)
\(=cos\left(4\pi+\frac{\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)+sin^2x\)
\(=cos\left(\frac{\pi}{4}+x\right)cos\left(\frac{\pi}{4}-x\right)+sin^2x\)
\(=\frac{1}{2}\left(cos\frac{\pi}{2}+cos2x\right)+sin^2x\)
\(=\frac{1}{2}cos2x+sin^2x=\frac{1}{2}\left(1-2sin^2x\right)+sin^2x\)
\(=\frac{1}{2}-sin^2x+sin^2x=\frac{1}{2}\)