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[Chii Chii]

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[tex]sin(\frac{\Pi }{4}+\frac{3x}{2})=2sin^{3}(\frac{3\Pi }{4}+\frac{x}{2})[/tex]
@kingsman(lht 2k2) @iceghost

 

[matheverytime]

[tex]sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )=\frac{3sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{3\pi}{4}+\frac{3x}{2} \right )}{2}<=>2sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )-2sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )=sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{3\pi}{4}+\frac{3x}{2} \right )<=>-2cos(\pi+2x)sin\left ( \frac{\pi}{2}-x \right )=cos\left ( \frac{3\pi}{2}+2x \right )sin(-x)<=>2cosx.cos2x=sinx.sin2x<=>cosx=0 \vee cos2x=sin^2x<=>1-2sin^2x=sin^2x=>sin^2x=\frac{1}{3}[/tex]
[tex]cos\left ( \frac{k\pi}{2}+\alpha \right )=sin\alpha[/tex] ( mk có dùng cái này )

 

[Chii Chii]

[tex]sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )=\frac{3sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{3\pi}{4}+\frac{3x}{2} \right )}{2}<=>2sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )-2sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )=sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{3\pi}{4}+\frac{3x}{2} \right )<=>-2cos(\pi+2x)sin\left ( \frac{\pi}{2}-x \right )=cos\left ( \frac{3\pi}{2}+2x \right )sin(-x)<=>2cosx.cos2x=sinx.sin2x<=>cosx=0 \vee cos2x=sin^2x<=>1-2sin^2x=sin^2x=>sin^2x=\frac{1}{3}[/tex]
[tex]cos\left ( \frac{k\pi}{2}+\alpha \right )=sin\alpha[/tex] ( mk có dùng cái này )

sin^3 x = 3sinx - sin3x chứ?

 

[matheverytime]

để mk thửu lại sr bạn

 

[matheverytime]

[tex]sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )=\frac{3sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )}{2}<=>2sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )-2sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )=sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )<=>-4cos(\pi+2x)sin\left ( \frac{\pi}{2}-x \right )=-2cos\left ( \frac{3\pi}{2}+x \right )sin(3\pi+2x)=-2(-sinx).(-sin2x)<=>cos2xcosx=-sin^2x.cosx=>cosx=0 \vee cos2x=-sin^2x[/tex]

 

[Chii Chii]

[tex]sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )=\frac{3sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )}{2}<=>2sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )-2sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )=sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )<=>-4cos(\pi+2x)sin\left ( \frac{\pi}{2}-x \right )=-2cos\left ( \frac{3\pi}{2}+x \right )sin(3\pi+2x)=-2(-sinx).(-sin2x)<=>cos2xcosx=-sin^2x.cosx=>cosx=0 \vee cos2x=-sin^2x[/tex]

Chỗ này mình ko hiểu

 

[matheverytime]

Chỗ này mình ko hiểu
View attachment 58668

nhân chéo chueyenr vế ý

 

[baogiang0304]

Từ đề bài ta có:[tex]sin\frac{\pi }{4}.cos\frac{3x}{2}+cos\frac{\pi }{4}.sin\frac{3x}{2}=2.(sin\frac{3\pi }{4}.cos\frac{x}{2}+cos\frac{3\pi }{4}.sin\frac{x}{2})^{3}[/tex]=>[tex]\frac{1}{\sqrt{2}}.(cos\frac{3x}{2}+sin\frac{3x}{2})=2.(cos\frac{x}{2}-sin\frac{x}{2})^{3}[/tex]=>[tex]4cos^{3}\frac{x}{2}-3cos\frac{x}{2}+3sin\frac{x}{2}-4sin^{3}\frac{x}{2}=2\sqrt{2}.(cos\frac{x}{2}-sin\frac{x}{2})^{3}[/tex]=>[tex](cos\frac{x}{2}-sin\frac{x}{2}).[2\sqrt{2}.(1-2.sin\frac{x}{2}.cos\frac{x}{2})-4.(sin\frac{x}{2}.cos\frac{x}{2}+1)+3]=0[/tex](mình hơi tắt 1 chút)
TH1:[tex]cos\frac{x}{2}=sin\frac{x}{2}=>x=90[/tex](thoả mãn đề bài)
TH2:[tex]2\sqrt{2}-2\sqrt{2}.sinx-2.sinx-1=0[/tex]([tex]2sin\frac{x}{2}.cos\frac{x}{2}=sin x[/tex])
=>[tex]sinx=\frac{2\sqrt{2}-1}{2\sqrt{2}+2}[/tex]

 

[Chii Chii]

[tex]sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )=\frac{3sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )}{2}<=>2sin\left ( \frac{\pi}{4}+\frac{3x}{2} \right )-2sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )=sin\left ( \frac{3\pi}{4}+\frac{x}{2} \right )-sin\left ( \frac{9\pi}{4}+\frac{3x}{2} \right )<=>-4cos(\pi+2x)sin\left ( \frac{\pi}{2}-x \right )=-2cos\left ( \frac{3\pi}{2}+x \right )sin(3\pi+2x)=-2(-sinx).(-sin2x)<=>cos2xcosx=-sin^2x.cosx=>cosx=0 \vee cos2x=-sin^2x[/tex]

làm như bạn thì sinx= sin2x
cosx=cos2x à

 

[dương bình an]

làm như bạn thì sinx= sin2x
cosx=cos2x à

vậy bạn biết làm chưa