rut gon bieu thuc (1-2*sin a * cos a)/(sin^2 a - cos^2 a)
Những câu hỏi liên quan
Rut gon bieu thuc:
A=(sin4a+cos4a-1)/(sin6a+cos6a-1)
B=1-sin2a+cos2a
C= 1-sina.cosa.tana
D= cos4a+cos2a.sin2a+sin2a
E=cansin2a.(1+cotx)+cos2a(1+tana)
F= 1+sina/cosa.[1-(1-sina/cosa)2]
Ai giup minh voiii. Minh cam on nhieuu!
B=1-sin2a+cos2a
\(=\sin^2a+\cos^2a-\sin^2a+\cos^2a=2\cos^2a\)
C= 1-sina.cosa.tana
\(=1-\sin a.\cos a.\frac{\sin a}{\cos a}=1-\sin^2a=\cos^2a\)
biết có vậy thôi à
Đúng 0
Bình luận (0)
Chung minh rang voi moi goc luong giac α lam cho bieu thuc xac dinh thi
a) \(\dfrac{1-sin2\alpha}{1+sin2\alpha}\)=cot\(^2\)(\(\dfrac{\pi}{4}\)+α) b) \(\dfrac{sin\alpha+sin\beta cos\left(\alpha+\beta\right)}{cos\alpha-sin\beta sin\left(\alpha+\beta\right)}\)=tan\(\left(\alpha+\beta\right)\).
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
Đúng 3
Bình luận (0)
b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
Đúng 1
Bình luận (2)
cho goc α thoa man π/2<α<π va sin α=1/3.tinh cos α
rut gon bieu thuc F=cosx.tan x/sin2x-cotx .cosx
Bài 1 :
Ta có : a thuộc góc phần tư thứ II .
=> Cos a < 0
- Ta lại có : \(\left\{{}\begin{matrix}sina=\dfrac{1}{3}\\sin^2a+cos^2a=1\end{matrix}\right.\)
\(\Rightarrow cosa=\sqrt{1-\left(\dfrac{1}{3}\right)^2}=-\dfrac{2\sqrt{2}}{3}\)
Bài 2 :
Ta có : \(F=\dfrac{\cos x.\tan x}{\sin^2x-\cot x.\cos x}=\dfrac{\cos x.\dfrac{\sin x}{\cos x}}{\sin^2x-\dfrac{\cos x}{\sin x}.\cos x}\)
\(=\dfrac{\sin x}{\sin^2x-\dfrac{\cos^2x}{\sin x}}=\dfrac{1}{\sin x-\cot^2x}\)
Đúng 1
Bình luận (0)
a. cho góc nhọn anpha, biết sin anpha=2/3 . tinh cos anpha, tan anpha, cot anpha
b.cho tan anpha + cot anpha = 3. tinh gia tri bieu thuc a= sin anpha x cos anpha
a) sin anpha = 2/3 => góc anpha = 42o
cos 42o = 0,743
tan 42o = 0,9
cot 42o = 1/tan 42o = 1/0,9 = 1,111
b) tan anpha + cot anpha = 3
<=> tan anpha + 1/tan anpha = 3
<=> tan2 anpha = 2
<=> tan anpha = \(\sqrt{2}\)
=> góc anpha = 55o
Ta có: a = sin 55o . cos 55o
<=> a = 0,469
Đúng 1
Bình luận (0)
chung minh cac bieu thuc sau khong phu thuoc vao x:
a/ \(3\left(\sin^8x-\cos^8x\right)+4\left(\cos^6x-2\sin^6x\right)+6\sin^4x\)
b/\(\frac{\tan^2x-\cos^2x}{\sin^2x}+\frac{\cot^2x-\sin^2x}{\cos^2x}\)
Cho biet \(\tan a=\frac{2}{3}\)
Tinh gia tri cua bieu thuc M=\(\frac{\sin^3a+3\cos^3a}{27\sin^3a-25\cos^3a}\)
tan a =2/3
=> đặt sin a = 2x thì cos a = 3x
rồi làm tiếp còn cách khác thì k biết làm
Đúng 0
Bình luận (0)
27 tháng 4 2022 lúc 1:49
1. cos 2a + cos 2b - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b 1
3. cos a2 + sin b2 1
4. cos2 a + sin2 a 1
5. cos 2a cos2 a - 2 sin 2a
6. sin 2a - 2 sin a. cos a.
7. sin 2a cos2 a - sin2 a
8. sin 2a - sin 2b 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 1
Câu số mấy đúng?
Đọc tiếp
1. cos 2a + cos 2b = - 2 cos(a+b) cos( a-b)
2. cos2a + sin2b = 1
3. cos a2 + sin b2= 1
4. cos2 a + sin2 a = 1
5. cos 2a = cos2 a - 2 sin 2a
6. sin 2a = - 2 sin a. cos a.
7. sin 2a = cos2 a - sin2 a
8. sin 2a - sin 2b= 2 sin ( a+b) cos ( a - b)
9. sin 2a - sin 2b= 2 cos( a+b) sin ( a - b)
10. cos a2 + sin a2 = 1
Câu số mấy đúng?
Tinh cac gia tri bieu thuc sau:
A= (cota+tana)/(cota-tana) voi sina=3/5
B= (sin^2a-cos^2a)/(sin^2a-3cos^2a) voi cota=-1/3
C1=sin^2a+2cos^2a va C2= sin^4a-cos^4a voi tana=-2
Ai giup minh voii. Minh cam on nhieuu!
\(sina=\frac{3}{5}\Rightarrow sin^2a=\frac{9}{25}\) ; \(cos^2a=1-\frac{9}{25}=\frac{16}{25}\)
\(A=\frac{cota+tana}{cota-tana}=\frac{sina.cosa\left(cota+tana\right)}{sina.cosa\left(cota-tana\right)}=\frac{cos^2a+sin^2a}{cos^2a-sin^2a}=\frac{1}{cos^2a-sin^2a}=\frac{1}{\frac{16}{25}-\frac{9}{25}}=\frac{25}{7}\)
\(B=\frac{sin^2a-cos^2a}{sin^2a-3cos^2a}=\frac{\frac{sin^2a}{sin^2a}-\frac{cos^2a}{sin^2a}}{\frac{sin^2a}{sin^2a}-\frac{3cos^2a}{sin^2a}}=\frac{1-cot^2a}{1-3cot^2a}=\frac{1-\left(-\frac{1}{3}\right)^2}{1-3\left(-\frac{1}{3}\right)^2}=\)
\(C_1=sin^2a+cos^2a+cos^2a=1+cos^2a=1+\frac{1}{1+tan^2a}=1+\frac{1}{1+\left(-2\right)^2}\)
\(C_2=\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)=sin^2a-cos^2a=1-2cos^2a\)
\(=1-\frac{2}{1+tan^2a}=1-\frac{2}{1+\left(-2\right)^2}\)
Đúng 1
Bình luận (0)
Rut gon
A=\(\frac{sin\left(x+y\right)-sinx}{sin\left(x+y\right)+sinx}\) - \(\frac{cos\left(x+y\right)+cosx}{cos\left(x-y\right)-cosx}\)
Lời giải:
$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$
$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$
Đúng 0
Bình luận (0)