--.--  \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ

\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)

\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)

\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)

\(\cos2a=2\cos^2a-1=\frac{7}{25}\)

\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)

\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)

\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)

\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)

\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)

Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)

\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)

\(\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)

\(cos^2\left(a-\frac{\pi}{4}\right)-sin^2\left(a-\frac{\pi}{4}\right)=cos\left(2a-\frac{\pi}{2}\right)\)

\(=cos\left(\frac{\pi}{2}-2a\right)=sin2a\)

\(\cos \alpha  =  - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}}  =  - \frac{{12}}{{13}}\) (vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\))

\(\sin \left( {\alpha  + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)

\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha  + \sin \frac{\pi }{4}sin\alpha  = \frac{{ - 17\sqrt 2 }}{{26}}\)

\(\frac{a}{2}\in\left(\frac{\pi}{2};\frac{3\pi}{4}\right)\Rightarrow tan\frac{a}{2}< 0\) ; \(sin\frac{a}{2}>0;cos\frac{a}{2}< 0\)

Đặt \(tan\frac{a}{2}=x< 0\)

\(\frac{2x}{1-x^2}=3\Leftrightarrow3x^2+2x-3=0\Rightarrow tan\frac{a}{2}=x=\frac{-1-\sqrt{10}}{3}\)

\(tan2a=\frac{2tana}{1-tan^2a}=\frac{6}{1-9}=-\frac{3}{4}\)

\(tan4a=\frac{2tan2a}{1-tan^22a}=-\frac{24}{7}\)

\(cos\frac{a}{2}=-\frac{1}{\sqrt{1+tan^2\frac{a}{2}}}=\) số thật kinh khủng

\(sin\frac{a}{2}=\sqrt{1-cos^2\frac{a}{2}}=...\)

\(sin\left(\frac{a}{2}+\frac{\pi}{2}\right)=\sqrt{2}\left(sin\frac{a}{2}+cos\frac{a}{2}\right)=...\)

Ta có:

\({\cos ^2}a + {\sin ^2}a = 1 \Rightarrow \sin a =  \pm \frac{4}{5}\)

Do \(0 < a < \frac{\pi }{2} \Leftrightarrow \sin a = \frac{4}{5}\)

\(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{4}{3}\)

Ta có;

\(\begin{array}{l}\sin \left( {a + \frac{\pi }{6}} \right) = \sin a.\cos \frac{\pi }{6} + \cos a.\sin \frac{\pi }{6} = \frac{4}{5}.\frac{{\sqrt 3 }}{2} + \frac{3}{5}.\frac{1}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\cos \left( {a - \frac{\pi }{3}} \right) = \cos a.\cos \frac{\pi }{3} + \sin a.\sin \frac{\pi }{3} = \frac{3}{5}.\frac{1}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3 + 4\sqrt 3 }}{{10}}\\\tan \left( {a + \frac{\pi }{4}} \right) = \frac{{\tan a + \tan \frac{\pi }{4}}}{{1 - \tan a.tan\frac{\pi }{4}}} = \frac{{\frac{4}{3} + 1}}{{1 - \frac{4}{3}}} =  - 7\end{array}\)

a/ \(\frac{\pi}{2}\le y\le\pi\Rightarrow cosy< 0\)

\(\Rightarrow cosy=-\sqrt{1-sin^2y}=-\frac{2\sqrt{2}}{3}\)

\(sin2y=2siny.cosy=2.\left(\frac{1}{3}\right).\left(-\frac{2\sqrt{2}}{3}\right)=-\frac{4\sqrt{2}}{9}\)

\(cos\left(\frac{\pi}{3}-y\right)=cos\frac{\pi}{3}cosy+sin\frac{\pi}{3}siny=\frac{\sqrt{3}-2\sqrt{2}}{6}\)

\(tany+5=\frac{siny}{cosy}+5=5-\frac{\sqrt{2}}{4}\)

b/ \(-\frac{\pi}{2}\le a\le9\Rightarrow sina\le0\)

\(\Rightarrow sina=\sqrt{1-cos^2a}=-\frac{4}{5}\)

\(sin2a=2sina.cosa=-\frac{24}{25}\)

\(cos2a=cos^2a-sin^2a=-\frac{7}{25}\)

\(tan2a=\frac{sin2a}{cos2a}=\frac{24}{7}\)

c/ \(\pi\le a\le\frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina\le0\\cosa\le0\end{matrix}\right.\)

\(\Rightarrow cosa=-\frac{1}{\sqrt{1+tan^2a}}=-\frac{1}{2}\Rightarrow sina=-\frac{\sqrt{3}}{2}\)

\(\Rightarrow sin2a=2sina.cosa=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left(\sqrt{3}-sin2a\right)sin\frac{2\pi}{3}=\frac{3}{4}\)

Câu 1:

\(tan\left(a+\frac{\pi}{4}\right)=1\Rightarrow a+\frac{\pi}{4}=\frac{\pi}{4}+k\pi\Rightarrow a=k\pi\) (\(k\in Z\) )

Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow\frac{\pi}{2}< k\pi< 2\pi\Rightarrow\frac{1}{2}< k< 2\Rightarrow k=1\Rightarrow a=\pi\)

\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+sin\pi=-\frac{\sqrt{3}}{2}\)

Câu 2:

\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}=cot\left(-\frac{\pi}{6}\right)\)

\(\Rightarrow a+\frac{\pi}{3}=-\frac{\pi}{6}+k\pi\Rightarrow a=-\frac{\pi}{2}+k\pi\) (\(k\in Z\))

\(\Rightarrow\frac{\pi}{2}< -\frac{\pi}{2}+k\pi< 2\pi\Rightarrow-\pi< k\pi< \frac{5\pi}{2}\)

\(\Rightarrow-1< k< \frac{5}{2}\Rightarrow k=\left\{0;1;2\right\}\Rightarrow a=\left\{-\frac{\pi}{2};\frac{\pi}{2};\frac{3\pi}{2}\right\}\) \(\Rightarrow cosa=0\)

\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)

Câu 1:

\(tan\left(a+\frac{\pi}{4}\right)=1\Leftrightarrow\frac{sin\left(a+\frac{\pi}{4}\right)}{cos\left(a+\frac{\pi}{4}\right)}=1\Leftrightarrow sin\left(a+\frac{\pi}{4}\right)=cos\left(a+\frac{\pi}{4}\right)\)

\(\Leftrightarrow sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}=cosa.cos\frac{\pi}{4}-sina.sin\frac{\pi}{4}\)

\(\Leftrightarrow\frac{\sqrt{2}}{2}sina+\frac{\sqrt{2}}{2}cosa=\frac{\sqrt{2}}{2}cosa-\frac{\sqrt{2}}{2}sina\)

\(\Rightarrow\sqrt{2}sina=0\Rightarrow sina=0\)

\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+0=-cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}\)

Câu 2:

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

\(\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)+\sqrt{3}sin\left(a+\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}+\sqrt{3}sina.cos\frac{\pi}{3}+\sqrt{3}cosa.sin\frac{\pi}{3}=0\)

\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)

\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\)

\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)