[Chuyên đề 1 ] Lượng giác

[silvery21]

[tex]\blue {\mathrm{cos^2.(x-a)+sin^2.(x-b)-2cos(x-a).sin(x-b).sin(a-b)=cos^2.(a-b);\forall x[/tex]

[TEX]VT= cos^2.(x-a)+sin^2.(x-b)+ cos (x -a) . ( cos (x-b+a-b)- cos( x- b - a+b))[/TEX]

[TEX]= cos^2.(x-a)+sin^2.(x-b)+cos (x -a) .( cos( x+a-2b) - cos ( x-a))[/TEX]

[TEX]=cos^2.(x-a)+ sin^2.(x-b)+ cos (x -a) . cos( x+a-2b)- cos ( x-a)[/TEX]

[TEX]= sin^2.(x-b)+ cos (x -a) . cos( x+a-2b)[/TEX]

[TEX]= sin^2.(x-b)+ \frac{1}{2} ( cos ( x-a +x+a-2b) +cos( x-a-x-a+2b))[/TEX]

[TEX]=sin^2.(x-b)+ \frac{1}{2} ( cos ( 2x-2b) +cos2(a-b))[/TEX]

[TEX]= sin^2.(x-b)+ \frac{1}{2} (2 cos^2( x-b) -1) + \frac{1}{2} (2 cos^2(a-b) -1)[/TEX]

[TEX]= sin^2.(x-b)+ cos^2( x-b) +cos^2b -1= cos^2 (a-b) =VP........dfcm...[/TEX]

 

[silvery21]

1;tính :

[TEX]A= sin 6^0. sin 42^0. sin 66^0 . sin 78^0[/TEX]

[TEX]B= B= sin 5^0. sin 15^0......sin 85^0[/TEX]

2; tính giá trị của [TEX]tg 37^0 30'[/TEX]

 

[duynhan1]

1;tính :

[TEX]A= sin 6^0. sin 42^0. sin 66^0 . sin 78^0[/TEX]

1.
[TEX]4 A= (cos 72 - cos 60)(cos 120 - cos 36) = \frac14 + \frac12(cos 36 - cos 72) - cos 36. cos 72 \\= \frac14 + \frac12(cos 36 - cos 72) - \frac12 (cos 108 + cos 36) = \frac14 [/TEX]

[TEX]\Leftrightarrow A= \frac{1}{16}[/TEX]

[TEX]B=sin 5. sin 15 . ....sin85[/TEX]

[TEX]B= (sin5.sin85)...(sin55.sin35).sin45[/TEX]

[TEX]16B= cos 80. cos 60. cos40.cos20.sin45[/TEX]

Ta có:

[TEX] cos 20. cos 40. cos 80 = =\frac12 (cos 60 + cos 20).cos80 = \frac14 cos 80 + \frac14(cos 60 + cos 100) = \frac18[/TEX]

[TEX]16B= \frac12. \frac{1}{\sqrt{2}}. \frac18 = \frac{\sqrt{2}}{32}[/TEX]

[TEX]B=\frac{\sqrt{2}}{512}[/TEX]



2.

[TEX]tan75 = tan(45+ 30) = \frac{tan 30 + tan 45}{1- tan30.tan45} = 2+ \sqrt{3}[/TEX]

[TEX]tan 2a = \frac{2 tan a}{1-tan^2a}[/TEX]

Đặt [TEX]x= tan 37^o 30' (DK: 0<x<1)[/TEX]

[TEX](2+\sqrt{3}) x + 2x -(2 +\sqrt{3}) = 0[/TEX]

[TEX]x = \frac{-1 + \sqrt{8+4\sqrt{3}}}{2+\sqrt{3}}[/TEX]

[TEX]tan 37^o 30' = \frac{-1 + \sqrt{8+4\sqrt{3}}}{2+\sqrt{3}} [/TEX]

 

[rua_it]

[TEX]VT= cos^2.(x-a)+sin^2.(x-b)+ cos (x -a) . ( cos (x-b+a-b)- cos( x- b - a+b))[/TEX]

[TEX]= cos^2.(x-a)+sin^2.(x-b)+cos (x -a) .( cos( x+a-2b) - cos ( x-a))[/TEX]

[TEX]=cos^2.(x-a)+ sin^2.(x-b)+ cos (x -a) . cos( x+a-2b)- cos ( x-a)[/TEX]

[TEX]= sin^2.(x-b)+ cos (x -a) . cos( x+a-2b)[/TEX]

[TEX]= sin^2.(x-b)+ \frac{1}{2} ( cos ( x-a +x+a-2b) +cos( x-a-x-a+2b))[/TEX]

[TEX]=sin^2.(x-b)+ \frac{1}{2} ( cos ( 2x-2b) +cos2(a-b))[/TEX]

[TEX]= sin^2.(x-b)+ \frac{1}{2} (2 cos^2( x-b) -1) + \frac{1}{2} (2 cos^2(a-b) -1)[/TEX]

[TEX]= sin^2.(x-b)+ cos^2( x-b) +cos^2b -1= cos^2 (a-b) =VP........dfcm...[/TEX]

[tex]\mathrm{\red{\left{\begin{y^{\prime} =-sin2.(x-a)+sin2.(x-b)-2sin(a-b).cos(2x-a-b)=0}\\{y(b)=cos^2(a-b)[/tex]

:\"&gt;

 

[rua_it]

[tex]\mathrm{\blue{\frac{sin^{10}x+cos^{10}x}{sin^6x+cos^6x}=\frac{4}{sin^22x+4cos^22x}[/tex]

 

[doremon.]

[tex]\mathrm{\blue{\frac{sin^{10}x+cos^{10}x}{sin^6x+cos^6x}=\frac{4}{sin^22x+4cos^22x}[/tex]

\Leftrightarrow[TEX]sin^{10}x+cos^{10}x=\frac{4(1-3sin^2xcos^2x}{4-3sin^22x}[/TEX]

\Leftrightarrow [TEX]sin^{10}x+cos^{10}x=\frac{4(1-3sin^2xcos^2x}{4(1-3sin^2xcos^2x)}=1[/TEX]

\Leftrightarrow[TEX]\left{\begin{sin^{10}x=sin^2x}\\{cos^{10}x=cos^2x} [/TEX]

\Leftrightarrow[TEX]\left{\begin{sinx=0, sinx=-1,sinx=1}\\{cosx=0, cos=1, cos=-1} [/TEX]

\Leftrightarrow[TEX]\left[\begin{x=\frac{\pi}{2}+k.\pi}\\{x=k \pi} [/TEX]

\Rightarrow[TEX]x=\frac{m.\pi}{2}[/TEX]

p/s: đánh CT chồng ngoặc như thế nào nhỉ

 

[doremon.]

1.[TEX] sin^{2007}x+cos^{2008}x=2cos^{2009}x[/TEX]

2.[TEX]3cos2x+4sin2x-9cosx-4sinx+6=0[/TEX]

3.[TEX]2sinx+cotx=2sin2x+1[/TEX]

 

[herrycuong_boy94]

tính a)

b) sin20. sin40.sin80
c.

d.

 

[duoisam117]

b) sin20. sin40.sin80

[TEX]sin20^0.sin40^0.sin80^0[/TEX]

[TEX]= -\frac{1}{2}(cos60^0-cos20^0).sin80^0 [/TEX]

[TEX]= -\frac{1}{2}cos60^0.sin80^0+\frac{1}{2}cos20^0.sin80^0 [/TEX]

[TEX]=-\frac{1}{4}sin80^0+\frac{1}{4}(sin100^0+sin60^0)[/TEX]

[TEX]= \frac{1}{4}sin60^0 [/TEX]

[TEX]= \frac{\sqrt{3}}{8}[/TEX]

c.

[TEX]\frac{1}{sin10^0}-4sin70^0 [/TEX]

[TEX]= \frac{1-4sin70^0.sin10^0}{sin10^0} [/TEX]

[TEX]=[1+4.\frac{1}{2}(cos80^0-cos60^0)].\frac{1}{sin10^0} [/TEX]

[TEX]= [1+2.cos80^0-2.\frac{1}{2}].\frac{1}{cos80^0} \\ \\ =2[/TEX]

 

[duynhan1]

Câu d:

[TEX]A=\frac{tan 80}{ cot 25 + cot 75} - \frac{cot 10}{tan 25 + tan 75} [/TEX]

[TEX]A = tan 80. (\frac{1}{\frac{2.sin 100}{cos 50 - cos 100}} - \frac{1}{\frac{2.sin 100}{cos 50 + cos 100}})[/TEX]


[TEX]A= \frac{sin 80}{cos 80}. \frac{(cos 50 - cos 100)-(cos 100 + cos 50)}{2sin100} [/TEX]

[TEX]A= 1[/TEX]

 

[rua_it]

[tex]\mathrm{\blue{\frac{tanA+tanB+tanC}{3}+\frac{2(sinA+sinB+sinC)}{3} > \pi[/tex]

 

[duynhan1]

tính a)

Với k= 1,2,3...... ta có:

[TEX]2.cos \frac{2k {\pi /} }{7}. sin { {\pi/} }{7} = sin \frac{ (2k+1){\pi/} }{7} - sin \frac{(2k-1){\pi/}}{7} [/TEX]

[TEX]2.A. sin \frac{{\pi/}}{7} = sin {\pi} - sin {\pi/}{7} [/TEX]

[TEX]\Leftrightarrow A = \frac{-1}{2} [/TEX]

 

[rooney_cool]

Một số bất đăng thức lượng giác

Có 10 bài đây. . Làm hết post tiếp. Hình như có mấy bài trùng :|

 

[silvery21]

[tex]\mathrm{\blue{\frac{tanA+tanB+tanC}{3}+\frac{2(sinA+sinB+sinC)}{3} > \pi[/tex]

xét [TEX]x \in ( 0;\pi) f(x) = 2 sin x + tg x - 3x[/TEX] trên nửa khoảng [TEX][0; \pi)[/TEX]

tính đc [TEX] f'(x) >0[/TEX]

[TEX]\forall x \in [ 0;\pi) f(x) >f(0)[/TEX] [TEX]= > 2 sin x + tg x >3x[/TEX]

sdụng vào 3 góc trong 1 tgiác [TEX]=> dfcm[/TEX]

 

[silvery21]

bài của roon.........

5;[TEX] sin {\frac{A}{2} }. sin {\frac{B}{2}}. sin {\frac{C}{2} }= \frac{1}{2} sin {\frac{A}{2}} .( cos {\frac{B-C}{2} } - cos {\frac{B+C}{2}}) \leq \frac{1}{2} sin {\frac{A}{2}} ( 1- sin {\frac{A}{2}} ) \leq \frac{1}{2} ( \frac{sin {\frac{A}{2}} +1-sin {\frac{A}{2}} }{2})^2 =\frac{1}{8}[/TEX]

câu 4 : là max ko fải min .....xem lại đề nhé .....nchung các câu này or đưa về bình phương nhưng sdụng jensen là nhanh nhất

câu 3:

of sdụng

 

[silvery21]

3.[TEX]2sinx+cotx=2sin2x+1[/TEX]

[TEX]dk...........[/TEX]\\

\Leftrightarrow [TEX]2 sin x - 1 + \frac{ sin x}{cos x} - 4 sin x cos x = 0[/TEX]

\Leftrightarrow [TEX]( 2 sin x- 1) ( 1 - \frac{ sin x}{cos x} ( 1 + 2 sin x )) =0[/TEX]

\Leftrightarrow [TEX]2 sin x - 1= 0[/TEX] ...............or... [TEX]sin x - cosx - sin 2x = 0.........done[/TEX]

 

[herrycuong_boy94]

Tính

2. cho tanx, tany là nghiệm của phương trình :

chứng minh

3. cho sin x +siny = 2.sin( x+ y) với x+ y khác k.pi
chứng minh

 

[duynhan1]

Nhận dạng tam giác

1) [TEX]sinC.cos(A-B) = 2.sinA.sinB[/TEX]

2)[TEX] sin A = \frac{cos B + cos C}{cos C + cos A}[/TEX]

 

[duynhan1]

Bài 3 của henry_cuong...

[TEX]sin x +siny = 2.sin( x+ y)[/TEX] [TEX](x+y \not= k. pi)[/TEX]
[TEX]\Leftrightarrow cos {\frac{x-y}{2}} = 2.cos {\frac{x+y}{2}}[/TEX]

[TEX] tan {\frac{x}{2}} . tan {\frac{y}{2}}[/TEX]
[TEX]=\frac{cos {\frac{x-y}{2}} - cos {\frac{x+y}{2}}} {cos {\frac{x+y}{2}} + cos {\frac{x-y}{2}}}[/TEX]
[TEX]= \frac13 [/TEX]

 

[hienzu]

Chứng minh:
cotgx-tgx-2tg2x-4tg4x=8cotg8x
Bài 2
Hãy tính theo m giá trị biểu thức:
B=tana-cota với m=tana+cota
(DH y HN)