lượng giác 11

K

khongphaibang

Ta có :

$\frac{1}{8}\tan \left( {x - \frac{\pi }{6}} \right)\tan \left( {x + \frac{\pi }{3}} \right)$

\Leftrightarrow $\frac{1}{8}\tan \left( {x - \frac{\pi }{6}} \right)\tan \left[ {\left( {x - \frac{\pi }{6}} \right) + \frac{\pi }{2}} \right]$

\Leftrightarrow $\frac{1}{8}\tan \left( {x - \frac{\pi }{6}} \right)\cot \left( {x - \frac{\pi }{6}} \right) = \frac{1}{8}$

Khi đó phương trình trở thành

${\sin ^3}x.\sin 3x + {\cos ^3}x.\cos 3x = \frac{1}{8}$

Sau đó bạn dùng công thức nhân ba để gải tiếp
 
N

nguyenbahiep1

1. Sin^3(x)sin3x+cos^3(x)cos3x= 1/8 tan(x- π/6)tan(x+ π/3)

[laTEX]sin^3x.sin3x + cos^3x.cos3x = \frac{1}{8}.tan(x - \frac{\pi}{6})cot(\frac{\pi}{6}-x) \\ \\ \frac{(3sinx-sin3x)sin3x}{4}+\frac{(3cosx+cos3x)cos3x}{4} = - \frac{1}{8} \\ \\ 3(sinx.sin3x+cosx.cos3x) + cos^23x -sin^23x = - \frac{1}{2} \\ \\ 6cos2x + 2cos6x +1 = 0 \\ \\ 2(4cos^32x - 3cos2x) + 6cos2x +1 = 0 \\ \\ cos2x = - \frac{1}{2}[/laTEX]
 
C

conga222222

$\eqalign{
& {\sin ^3}x\sin 3x + {\cos ^3}x\cos 3x = \frac{{\tan \left( {x - \frac{\pi }{6}} \right)\tan \left( {x + \frac{\pi }{3}} \right)}}{8} \cr
& dk... \cr
& co \cr
& \tan \left( {x - \frac{\pi }{6}} \right) = - \cot \left( {x + \frac{\pi }{3}} \right) \cr
& \to {\sin ^3}x\sin 3x + {\cos ^3}x\cos 3x = \frac{{\tan \left( {x - \frac{\pi }{6}} \right)\tan \left( {x + \frac{\pi }{3}} \right)}}{8} = - \frac{1}{8} \cr
& \leftrightarrow 8{\sin ^3}x\sin 3x + 8{\cos ^3}x\cos 3x = - 1 \cr
& \leftrightarrow - 2\left( {1 - \cos 2x} \right)\left( {\cos 4x - \cos 2x} \right) + 2\left( {1 + \cos 2x} \right)\left( {\cos 4x + \cos 2x} \right) = - 1 \cr
& \leftrightarrow - 2\left( {1 - \cos 2x} \right)\left( {2{{\cos }^2}2x - 1 - \cos 2x} \right) + 2\left( {1 + \cos 2x} \right)\left( {2{{\cos }^2}2x - 1 + \cos 2x} \right) = - 1 \cr
& dat\;cos2x = t\;\left( {\left| t \right| \leqslant 1} \right) \cr
& \to - 2\left( {1 - t} \right)\left( {2{t^2} - t - 1} \right) + 2\left( {1 + t} \right)\left( {2{t^2} + t - 1} \right) = - 1 \cr
& giai\;phuong\;trinh\;bac\;3\;cua\;t \cr
& \to .... \cr} $
 
K

khongphaibang

Mà lần sau nhớ học viết Mytex đi nha

1.

$\begin{array}{l}
{\sin ^3}x.\sin 3x + {\cos ^3}x.\cos 3x = \frac{1}{8}tan\left( {x - \frac{\pi }{6}} \right)\tan \left( {x + \frac{\pi }{3}} \right)\\
8{\cos ^3}\left( {x + \frac{{5\pi }}{{12}}} \right) = \cos \left( {3x + \frac{\pi }{4}} \right)
\end{array}$
 
Top Bottom