\(=cos\left(\dfrac{4}{3}pi\right)+sin\left(\dfrac{pi}{6}\right)+tan\left(-\dfrac{3}{4}pi\right)\)

\(=-\dfrac{1}{2}+\dfrac{1}{2}+1=1\)

\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)

\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)

\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)

\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)

\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)

\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)

\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)

\(\Rightarrow A=\dfrac{1}{32}\)

\(H.sin\dfrac{\pi}{19}=sin\dfrac{\pi}{19}.cos\dfrac{\pi}{19}+sin\dfrac{\pi}{19}cos\dfrac{3\pi}{19}+...+sin\dfrac{\pi}{19}cos\dfrac{17\pi}{19}\)

\(=\dfrac{1}{2}sin\dfrac{2\pi}{19}+\dfrac{1}{2}sin\dfrac{4\pi}{19}-\dfrac{1}{2}sin\dfrac{2\pi}{19}+...+\dfrac{1}{2}sin\dfrac{18\pi}{19}-\dfrac{1}{2}sin\dfrac{16\pi}{19}\)

\(=\dfrac{1}{2}sin\dfrac{18\pi}{19}=\dfrac{1}{2}sin\left(\pi-\dfrac{\pi}{19}\right)=\dfrac{1}{2}sin\dfrac{\pi}{19}\)

\(\Rightarrow H=\dfrac{1}{2}\)

A\(=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}\)

Đặt tử là Y; mẫu là U

Có \(Y=\)\(cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)\)

\(=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)\)

\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}\)\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}\)

\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}\)\(=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}\)

\(=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}\)

\(\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)\)

\(=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)\)\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\)

\(\Rightarrow Y=-\dfrac{1}{2}\)

Có \(sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}\)

\(=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}\)

\(=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)\)

\(=\dfrac{1}{8}.sin\dfrac{8\pi}{7}\)\(=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}\)

\(\Rightarrow U=-\dfrac{1}{8}\) 

Vậy \(A=\dfrac{Y}{U}=4\)

CHÚC BẠN HỌC TỐT NHÉ

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.

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

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

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

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

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

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