Các câu hỏi tương tự
cho sinx=3/5 (pi/2<x<pi). Tính cosx ,cos2x, sin(x-pi/4), tan x/2
CMR:
a, \(\frac{\cot^2x-\sin^2x}{\cot^2x-tan^2x}=sin^2x.\cos^2x\)
b, \(\frac{\tan x}{1-\tan^2x}.\frac{\cot^2-1}{\cot x}=1\)
c, \(\frac{1+\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\tan x+1}{\cot x+1}\)
d, \(\frac{\sin x+\cos x-1}{\sin x-cosx+1}=\frac{\cos x}{1+sinx}\)
22 tháng 10 2023 lúc 22:11
C = tan5\(^o\).tan10\(^{^o}\).tan15\(^o\).....tan80\(^o\).tan85\(^o\)
Chứng minh đẳng thức: (tan x /1 - tan^2 x) (cot^2 x - 1/cot x) = 1
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tan x - cot x = 2. Biểu thức tan^2x + cot^2x bằng
Khẳng định nào sau đây đúng:
\(A.\frac{4\tan x\left(1-tan^2x\right)}{\left(1+tan^2x\right)^2}=sin^2x \)
\(B.\frac{4\tan x\left(1-tan^2x\right)}{\left(1+tan^2x\right)^2}=sin2x\)
\(C.\frac{4\tan x\left(1-tan^2x\right)}{\left(1+tan^2x\right)^2}=sin4x\)
\(D.\frac{4\tan x\left(1-tan^2x\right)}{\left(1+tan^2x\right)^2}=sinx\)
1, √x^2-5x+8<=3x-10
2, x^2-4x+3/3-2x<1-x
3, |x^2+x-2|+3x^2-3>0
4, |x+3|>=2(1+x^2)
5, |x^2-1|>x^2+2x-3
6, |2x-1|<x+2
7, 3/2-x<=1
8, 2+3x-x^2<=√4+3x-x^2
9, √x^2-5x+6<5-x
10, x^2-6>5(x+√x^2-5x)
11, |x^2-x|<=x^2-x
1, √x^2-5x+8<=3x-10
2, x^2-4x+3/3-2x<1-x
3, |x^2+x-2|+3x^2-3>0
4, |x+3|>=2(1+x^2)
5, |x^2-1|>x^2+2x-3
6, |2x-1|<x+2
7, 3/2-x<=1
8, 2+3x-x^2<=√4+3x-x^2
9, √x^2-5x+6<5-x
10, x^2-6>5(x+√x^2-5x)
11, |x^2-x|<=x^2-x
Bài 1: Gỉai các phương trình saua) sin3x frac{-sqrt{3}}{2} b) sin4x frac{2}{3} c) cos (x+3) frac{1}{3} d) cos (2x + frac{pi}{3} ) frac{-1}{2}e) tan (2x + 45*) -1f) cot (frac{x}{3}+ frac{pi}{3} ) sqrt[]{3}g) tan ( frac{x}{2}- frac{pi}{4}) tanfrac{pi}{8}
Đọc tiếp
Bài 1: Gỉai các phương trình sau
a) sin3x = \(\frac{-\sqrt{3}}{2}\)
b) sin4x = \(\frac{2}{3}\)
c) cos (x+3) = \(\frac{1}{3}\)
d) cos (2x + \(\frac{\pi}{3}\) ) = \(\frac{-1}{2}\)
e) tan (2x + 45*) = -1
f) cot (\(\frac{x}{3}\)+ \(\frac{\pi}{3}\) ) = \(\sqrt[]{3}\)
g) tan ( \(\frac{x}{2}\)- \(\frac{\pi}{4}\)) = tan\(\frac{\pi}{8}\)