pi/2<a,b<pi

=>cos a<0; cos b<0; sin a>0; sin b>0

\(cosa=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5};sina=\sqrt{1-\left(-\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=-3/5:4/5=-3/4; tan b=12/13:(-5/13)=-12/5

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana\cdot tanb}\)

\(=\dfrac{-\dfrac{3}{4}+\dfrac{-12}{5}}{1-\dfrac{-3}{4}\cdot\dfrac{-12}{5}}=\dfrac{63}{16}\)

sin(a-b)=sina*cosb-sinb*cosa

\(=\dfrac{3}{5}\cdot\dfrac{-5}{13}-\dfrac{-4}{5}\cdot\dfrac{12}{13}=\dfrac{-15+48}{65}=\dfrac{33}{65}\)

a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k\pi\\2x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+x+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=cos\left(x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+x+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)

c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)

=>\(cos\left(x-\dfrac{pi}{6}\right)=sin\left(-2x-\dfrac{pi}{3}\right)\)

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)

=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

1: cos(2x+pi/6)=cos(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi

=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi

=>x=pi/30+k2pi/5 hoặc x=pi-k2pi

2: sin(2x+pi/6)=sin(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi

=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi

=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi

1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

3.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)

b.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)

\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)

a: 3/2pi<x<2pi

=>sin x<0

=>\(sinx=-\sqrt{1-\left(\dfrac{1}{6}\right)^2}=-\dfrac{\sqrt{35}}{6}\)

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{6}\cdot\dfrac{-\sqrt{35}}{6}=\dfrac{-\sqrt{35}}{18}\)

\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{1}{36}-1=\dfrac{1}{18}-1=\dfrac{-17}{18}\)

\(tan2x=\dfrac{-\sqrt{35}}{18}:\dfrac{-17}{18}=\dfrac{\sqrt{35}}{17}\)

\(cot2x=1:\dfrac{\sqrt{35}}{17}=\dfrac{17}{\sqrt{35}}\)

b: \(sin\left(\dfrac{pi}{3}-x\right)\)

\(=sin\left(\dfrac{pi}{3}\right)\cdot cosx-cos\left(\dfrac{pi}{3}\right)\cdot sinx\)

\(=\dfrac{1}{2}\cdot\dfrac{-\sqrt{35}}{6}-\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{-\sqrt{35}-1}{12}\)

c: \(cos\left(x-\dfrac{3}{4}pi\right)\)

\(=cosx\cdot cos\left(\dfrac{3}{4}pi\right)+sinx\cdot sin\left(\dfrac{3}{4}pi\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{-\sqrt{2}}{2}+\dfrac{-\sqrt{35}}{6}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{-\sqrt{2}-\sqrt{70}}{12}\)

d: tan(pi/6-x)

\(=\dfrac{tan\left(\dfrac{pi}{6}\right)-tanx}{1+tan\left(\dfrac{pi}{6}\right)\cdot tanx}\)

\(=\dfrac{\dfrac{\sqrt{3}}{3}-\sqrt{35}}{1+\dfrac{\sqrt{3}}{3}\cdot\left(-\sqrt{35}\right)}\)