\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)

\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)

\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)

\(=0\)

\(C=2sin3x.cos3x.\frac{cos3x}{sin3x}-\left(cos^23x-sin^23x\right)\)

\(=2cos^23x-cos^23x+sin^23x=cos^23x+sin^23x=1\)

\(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)\)

\(=\sqrt{2}\left(sinx.sin\frac{\pi}{4}-cosx.cos\frac{\pi}{4}\right)=-\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

Câu này bạn ghi nhầm đề (lưu ý rằng \(sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\))

Câu 2b bạn cũng xem lại đề, chắc chắn ko đúng

\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}sin\frac{C}{2}\)

Câu 3 bạn cũng ghi sai đề luôn

Trong 1 ngày đẹp trời thì câu 4 cũng sai luôn cho đỡ lạc lõng đồng đội:

\(sin\left(a+b-b\right)=sin\left(a+b\right)cosb-cos\left(a+b\right)sinb=2sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(a+b\right)\left[cosb-2\right]=cos\left(a+b\right).sinb\)

\(\Leftrightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{sinb}{cosb-2}\Leftrightarrow tan\left(a+b\right)=\frac{sinb}{cosb-2}\)

4 câu bạn ghi đúng đề bài duy nhất câu 1, kinh thiệt :(

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

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

\(A=0\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

Ta có :
\(\frac{sinx+sin\left(\frac{x}{2}\right)}{1+cosx+cos\left(\frac{x}{2}\right)}=\frac{2sin\left(\frac{x}{2}\right).cos\left(\frac{x}{2}\right)+sin\left(\frac{x}{2}\right)}{2cos^2\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)}\)

\(=\frac{sin\left(\frac{x}{2}\right)\left(2cos\left(\frac{x}{2}\right)+1\right)}{cos\left(\frac{x}{2}\right)\left(2cos\left(\frac{x}{2}\right)+1\right)}=\frac{sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)}\)

\(=tan\left(\frac{x}{2}\right)\left(đpcm\right)\)

Lời giải:
\(\sin (a+b)=\sin (a+b+c-c)=\sin (a+b+c).\cos c-\cos (a+b+c)\sin c\)

\(\sin (a+c)=\sin (a+c+b-b)=\sin (a+b+c)\cos b-\cos (a+b+c)\sin b\)

Do đó:

\(\text{VT}=\sin (a+b+c)\cos b\cos c-\cos (a+b+c)\sin c\cos b-\sin (a+b+c)\cos b\cos c+\cos (a+b+c)\sin b\cos c\)

\(=\sin (a+b+c)(\cos b\cos c-\cos b\cos c)+\cos (a+b+c)(\sin b\cos c-\sin c\cos b)\)

\(=\cos (a+b+c)(\sin b\cos c-\cos b\sin c)=\cos (a+b+c)\sin (b-c)\)

\(=\text{VP}\)

Ta có đpcm.

\(cosa.sina=\frac{1}{5}\Rightarrow\frac{cosa.sina}{sin^2a}=\frac{1}{5sin^2a}=\frac{sin^2a+cos^2a}{5sin^2a}\)

\(\Rightarrow\frac{cosa}{sina}=\frac{1}{5}+\frac{1}{5}.\frac{cos^2a}{sin^2a}\)

\(\Rightarrow cota=\frac{1}{5}+\frac{1}{5}cot^2a\)

\(\Rightarrow cot^2a-5cota+1=0\)

\(\Rightarrow cota=\frac{5\pm\sqrt{21}}{2}\)

Câu 2:

\(\frac{cosa}{1-sina}=\frac{cosa\left(1+sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\frac{cosa\left(1+sina\right)}{1-sin^2a}=\frac{cosa\left(1+sina\right)}{cos^2a}=\frac{1+sina}{cosa}\)

b/

\(\frac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}\)

\(=\frac{sin^2a+cos^2a+2sina.cosa-\left(sin^2a+cos^2a-2sina.cosa\right)}{sina.cosa}\)

\(=\frac{4sina.cosa}{sina.cosa}\)

\(=4\)

\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)

Bạn ghi đề ko đúng

\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)

\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)

\(=sin^2a-sin^2b\)

\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)

Câu này bạn cũng ghi đề ko đúng

\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)

\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)

\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)