Cho góc $\alpha$ tm : $\frac{\pi }{2}<\alpha  < \pi $ và $sin\pi =\frac{3}{5}$
 Tính : 
                          $ A= \frac{tan\alpha }{1+tan^{2}\alpha }$

Sin a = 3/5 => Cos a = -4/5 ( \pi / 2 < a < \pi )

A = 1/2.( Tan2a ) = 1/2.( Sin2a / Cos2a ) = 1/2.(2.Sina.Cosa / (2Cos^2a - 1) ) = -12/7

$A=tan.cos^2=(sin/cos).cos^2=sin.cos=sin\sqrt{1-sin^2}$
đến đây lm típ đc òi!
ok ok thanks! –  Confusion 16-03-16 08:26 PM
bh cũng đk :D nhưng đừng lâu quá –  ☼SunShine❤️ 16-03-16 07:57 PM
để m nghĩ cách khác thử xem! bao h cần z? –  Confusion 16-03-16 07:51 PM
chỉ sợ nó ko liên tục vì mình lm cái này còn non tay lắm! –  Confusion 16-03-16 07:50 PM
dài lắm! m gợi ý nha: bn đặt căn y=a rồi đạo hàm f(a)=a^3-a –  Confusion 16-03-16 07:49 PM
zâng :3 mà bn lm bài kia đuy –  ☼SunShine❤️ 16-03-16 07:47 PM
cách của mình hay đúng ko? heheh –  Confusion 16-03-16 07:46 PM
:)) chắc dễ quá –  ☼SunShine❤️ 16-03-16 07:45 PM
uầy, sao mn lm nhanh ghê! –  Confusion 16-03-16 07:43 PM
Do  $\pi/2<\alpha < \pi \Rightarrow cos\alpha <0 $. Ta có: $sin\alpha^{2}+cos\alpha^{2}=1 \Rightarrow cos\alpha= \frac{-4}{5}$.
$tan\alpha=\frac{sin\alpha }{cos\alpha}=\frac{-3}{4}$.....
Thay vào đề.....
sin^2 + cos^2 =1 => cos ( loại được 1 nghiệm theo dk giả thiết )
=> tính dc tan => tính A

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