(Sina -cosa)^2 =1:25

<=> sin^2a +cos^2a -2sina.cosa =1:25

Ta có sin^2a+cos^2a = 1 

<=> 1-2 sina.cosa =1:25

2sina.cosa =24:25

CT : sin2a= 2sina.cosa=24:25

 Có sin^2 .2a + co^2.2a = 1 

       (24:25)^2 + cos^2.2a =1 

Từ đây rút cos 2a = căn 1-(24:25)^2 =...  bạn  tự làm tiếp nha !

\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)

\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)

\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)

\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)

\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)

\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)

Lời giải:

$\sin ^2a+\cos ^2a=1$

$\cos ^2a=1-\sin ^2a=1-(\frac{-5}{13})^2=\frac{144}{169}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\cos a< 0$

Do đó: $\cos a=-\sqrt{\frac{144}{169}}=\frac{-12}{13}$

$\sin 2a=2\sin a\cos a=2.\frac{-5}{13}.\frac{-12}{13}=\frac{120}{169}$

$\cos 2a=\cos ^2a-\sin ^2a=2\cos ^2a-1=2.\frac{144}{169}-1=\frac{119}{169}$

$\cos a=\cos ^2\frac{a}{2}-\sin ^2\frac{a}{2}$

$=1-2\sin ^2\frac{a}{2}$

$\Leftrightarrow \frac{-12}{13}=1-2\sin ^2\frac{a}{2}$

$\Rightarrow \sin ^2\frac{a}{2}=\frac{25}{26}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\sin \frac{a}{2}>0$

$\Rightarrow \sin \frac{a}{2}=\frac{5}{\sqrt{26}}$

có A=\(\dfrac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)

Do \(0< a< \frac{\pi}{2}\Rightarrow sina>0\)

\(sin^2a+cos^2a=1\Rightarrow sina=\sqrt{1-cos^2a}=\sqrt{1-\left(\frac{15}{17}\right)^2}=\frac{8}{17}\)

\(cos2a=2cos^2a-1=2.\left(\frac{15}{17}\right)^2-1=\frac{161}{289}\)

Chọn C

b)\(P=cos2a-cos(\dfrac{\pi}{3}-a) \\=2cos^2a-1-cos\dfrac{\pi}{3}cosa-sin\dfrac{\pi}{3}sina \\=2.(\dfrac{-2}{5})^2-1-\dfrac{1}{2}.\dfrac{-2}{5}-\dfrac{\sqrt3}{2}.\dfrac{-\sqrt{21}}{5} \\=\dfrac{-24+15\sqrt7}{50}\)

a, Vì : \(\pi< a< \dfrac{3\pi}{2}\)  nên \(cos\alpha< 0\) mà \(cos^2\alpha=1-sin^2\alpha=1-\dfrac{4}{25}=\dfrac{21}{25},\)

do đó : \(cos\alpha=-\dfrac{\sqrt{21}}{5}\)

từ đó suy ra : \(tan\alpha=\dfrac{2}{\sqrt{21}},cot\alpha=\dfrac{\sqrt{21}}{2}\)