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(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 !
Do \(0< a< \frac{\pi}{2}\Rightarrow sina>0\)
\(sin^2a+cos^2a=1\Rightarrow sina=\sqrt{1-cos^2a}=\sqrt{1-\left(\frac{15}{17}\right)^2}=\frac{8}{17}\)
\(cos2a=2cos^2a-1=2.\left(\frac{15}{17}\right)^2-1=\frac{161}{289}\)
\(A=\frac{cos^2a-sin^2a}{2sin^2a+3sina.cosa}=\frac{\frac{cos^2a}{cos^2a}-\frac{sin^2a}{sin^2a}}{\frac{2sin^2a}{cos^2a}+\frac{3sina.cosa}{cos^2a}}=\frac{1-tan^2a}{2tan^2a+3tana}=\frac{1-2^2}{2.2^2+3.2}=...\)
\(A=cos^2a+cos^2b+2cosa.cosb+sin^2a+sin^2b+2sina.sinb\)
\(=cos^2a+sin^2a+cos^2b+sin^2b+2\left(cosa.cosb+sina.sinb\right)\)
\(=2+2cos\left(a-b\right)=2+2cos\frac{\pi}{3}=3\)
\(\left(cosa+sina\right)^2=\frac{36}{25}\Leftrightarrow1+2sina.cosa=\frac{36}{25}\)
\(\Rightarrow sin2a=\frac{36}{25}-1=\frac{11}{25}\)
\(cos2a=cos^2a-sin^2a=\left(cosa-sina\right)\left(cosa+sina\right)>0\)
\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{6\sqrt{14}}{25}\)
\(o< a< \frac{\pi}{2}\Rightarrow sina>0\)
\(cos2a=-\frac{24}{25}\Leftrightarrow1-2sin^2a=-\frac{24}{25}\)
\(\Rightarrow sin^2a=\frac{49}{50}\Rightarrow sina=\frac{7\sqrt{2}}{10}\)
Lời giải:
$A=2\cos ^3a+\cos ^2a-\sin ^2a+\sin a=2\cos ^3a+2\cos ^2a-1+\sin a$
$=2\cos ^2a(\cos a+1)-(1-\sin a)$
$=2(1-\sin ^2a)(\cos a+1)-(1-\sin a)$
$=2(1-\sin a)(1+\sin a)(\cos a+1)-(1-\sin a)$
$=(1-\sin a)[2(\sin a+1)(\cos a+1)-1]$
$=(1-\sin a)(2\sin a\cos a+2\sin a+2\cos a+1)$
$=(1-\sin a)(\sin 2a+2\sin a+2\cos a+1)$