Ta có:

\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)

\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)

\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)

\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)

\(=\frac{1}{4}sin3a\)

\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)

\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)

\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)

\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)

\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)

Ta có:

\(\begin{array}{l}\sin \frac{\pi }{{24}}\cos \frac{{5\pi }}{{24}} = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{{24}} + \frac{{5\pi }}{{24}}} \right) + \sin \left( {\frac{\pi }{{24}} - \frac{{5\pi }}{{24}}} \right)} \right]\\ = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{4}} \right) + \sin \left( { - \frac{\pi }{6}} \right)} \right]\\ = \frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} - \frac{1}{2}} \right] = \frac{{\sqrt 2  - 1}}{4}\end{array}\)

Ta có:

\(\begin{array}{l}\sin \frac{{7\pi }}{8}\sin \frac{{5\pi }}{8} = \frac{1}{2}\left[ {\cos \left( {\frac{{7\pi }}{8} - \frac{{5\pi }}{8}} \right) - \cos \left( {\frac{{7\pi }}{8} + \frac{{5\pi }}{8}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{{3\pi }}{2}} \right)} \right]\\ = \frac{1}{2}.\left( {\frac{{\sqrt 2 }}{2} + 0} \right) = \frac{{\sqrt 2 }}{4}\end{array}\)

a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)

b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)

Bạn ơi sin gì đó bạn ?

7.

ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\Leftrightarrow8cosx=\frac{\sqrt{3}cosx+sinx}{sinx.cosx}\)

\(\Leftrightarrow8cosx.sinx.cosx=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow4sin2x.cosx=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)

\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)

5.

\(sin\left(2x+\frac{\pi}{2}+2\pi\right)-2cos\left(x+\frac{\pi}{2}-4\pi\right)=1+2sinx\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)-2cos\left(x+\frac{\pi}{2}\right)=1+2sinx\)

\(\Leftrightarrow cos2x+2sinx=1+2sinx\)

\(\Leftrightarrow cos2x=1\)

\(\Rightarrow x=k\pi\)

6.

\(sin^22x-cos^28x=sin\left(10x+\frac{\pi}{2}+8\pi\right)\)

\(\Leftrightarrow\frac{1-cos4x}{2}-\frac{1+cos16x}{2}=sin\left(10x+\frac{\pi}{2}\right)\)

\(\Leftrightarrow-\left(cos4x+cos16x\right)=2cos10x\)

\(\Leftrightarrow-2cos10x.cos6x=2cos10x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}10x=\frac{\pi}{2}+k\pi\\6x=\pi+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\end{matrix}\right.\)

a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)

Ta có: \({\sin ^2}a + {\cos ^2}a  = 1\)

 \(\Leftrightarrow \frac{1}{9} + {\cos ^2}a  = 1\)

\(\Leftrightarrow {\cos ^2}a =  1 - \frac{1}{9}= \frac{8}{9}\)

\(\Leftrightarrow \cos a  =\pm\sqrt { \frac{8}{9}}  =  \pm \frac{{2\sqrt 2 }}{3}\)

Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)

Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} =  - \frac{{\sqrt 2 }}{4}\)

Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\)

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)

\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} =  - \frac{{4\sqrt 2 }}{7}\)

b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)

\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)

Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 =  - \frac{3}{4}\)

Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)

\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)

\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)

\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 =  - \frac{{\sqrt 7 }}{4}\)

\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)