pt lượng giác hay
#1
Đã gửi 02-04-2011 - 20:32
BECOME ONE !
#2
Đã gửi 02-04-2011 - 21:01
$ \sqrt{\dfrac{3}{4}+6sin^2(\dfrac{x}{2})sin(\dfrac{x}{2}+\dfrac{ 3\pi }{2})}=\sqrt{3}cos\dfrac{3x}{4}$căn của [ 3/4 + 6sin^2 (x/2).sin(x/2 + 3pi/2) =căn3 . cos (3x/4)
đề như vậy à?
Bài viết đã được chỉnh sửa nội dung bởi Giang1994: 02-04-2011 - 21:05
Don't let people know what you think
#3
Đã gửi 02-04-2011 - 21:17
Post đề sai nha Giang1994$ \sqrt{\dfrac{3}{4}+6sin^2(\dfrac{x}{2})sin(\dfrac{x}{2}+\dfrac{ 3\pi }{2})}=\sqrt{3}cos\dfrac{3x}{4}$
đề như vậy à?
$\begin{array}{l}\sqrt {\dfrac{3}{4} + 6{{\sin }^2}\dfrac{x}{2}.\sin \left( {\dfrac{x}{2} + \dfrac{{3\pi }}{2}} \right)} = \sqrt 3 \cos \left( {\dfrac{{3x}}{4}} \right)\\ \Leftrightarrow \dfrac{3}{4} + 6\left( {1 -{{\cos}^2}\dfrac{x}{2}} \right)\cos \dfrac{x}{2} = \dfrac{3}{2}.2{\cos ^2}\left( {\dfrac{{3x}}{4}} \right)\\
\Leftrightarrow \dfrac{3}{4} + 6\cos \dfrac{x}{2} - 6{\cos ^3}\dfrac{x}{2} = \dfrac{3}{2}\left( {1 + \cos 3\left( {\dfrac{x}{2}} \right)} \right)\\ \Leftrightarrow \dfrac{3}{4} + 6\cos \dfrac{x}{2} - 6{\cos ^3}\dfrac{x}{2} = \dfrac{3}{2}\left( {1 + 4{{\cos }^3}\dfrac{x}{2} - 3\cos \dfrac{x}{2}} \right)\\ \Leftrightarrow 6{\cos ^3}\dfrac{x}{2} - \dfrac{{21}}{2}\cos x + \dfrac{3}{4} = \end{array}=0$
N.HÍCHMÉT
Khó + Lười = Bất lực
#4
Đã gửi 03-04-2011 - 13:55
Đề Có Sai Không???????/Post đề sai nha Giang1994
$\begin{array}{l}\sqrt {\dfrac{3}{4} + 6{{\sin }^2}\dfrac{x}{2}.\sin \left( {\dfrac{x}{2} + \dfrac{{3\pi }}{2}} \right)} = \sqrt 3 \cos \left( {\dfrac{{3x}}{4}} \right)\\ \Leftrightarrow \dfrac{3}{4} + 6\left( {1 -{{\cos}^2}\dfrac{x}{2}} \right)\cos \dfrac{x}{2} = \dfrac{3}{2}.2{\cos ^2}\left( {\dfrac{{3x}}{4}} \right)\\
\Leftrightarrow \dfrac{3}{4} + 6\cos \dfrac{x}{2} - 6{\cos ^3}\dfrac{x}{2} = \dfrac{3}{2}\left( {1 + \cos 3\left( {\dfrac{x}{2}} \right)} \right)\\ \Leftrightarrow \dfrac{3}{4} + 6\cos \dfrac{x}{2} - 6{\cos ^3}\dfrac{x}{2} = \dfrac{3}{2}\left( {1 + 4{{\cos }^3}\dfrac{x}{2} - 3\cos \dfrac{x}{2}} \right)\\ \Leftrightarrow 6{\cos ^3}\dfrac{x}{2} - \dfrac{{21}}{2}\cos x + \dfrac{3}{4} = \end{array}=0$
#5
Đã gửi 01-07-2011 - 10:08
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