Biết tanB=2 tính
\(A=\frac{2sin\alpha+cos\alpha}{3sin\alpha-4cos\alpha}\)
\(B=sin^2\alpha+2sin\alpha.cos\alpha-3cos^2\alpha\)
\(C=\frac{sin^2\alpha-sin\alpha.cos\alpha-cos^2\alpha}{2sin\alpha.cos\alpha}\)
Giúp mik với, ai làm xong mik sẽ tick cho cảm ơn nhiều
Tính
A= \(\frac{2sin\alpha+cos\alpha}{3sin\alpha-4cos\alpha}\)
B= \(sin^2\alpha+2sin\alpha.cos\alpha-3cos^3\alpha\)
C= \(\frac{sin^2\alpha-sin\alpha.cos\alpha-cos^2\alpha}{2sin\alpha.cos\alpha}\)
Giúp mik với, ai làm được mik sẽ tick cho. Cảm ơn trước nhé
Đơn giản các biểu thức sau:
(1-\(Cos\alpha\)).\(\left(1+Cos\alpha\right)\)
\(1+sin^2\alpha+cos^2\alpha\)
\(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha\)
\(tan^2\alpha-sin^2\alpha.tan^2\alpha\)
\(cos^2\alpha+tan^2\alpha.cos^2\alpha\)
\(tan^2\alpha.\left(2cos^2\alpha+sin^2\alpha-1\right)\)
Gấp!!!:))))
Bài 1: Thu gọn
a) A= sin2\(\alpha\) + sin2\(\alpha\) . tan\(\alpha\)
b) B= \(\dfrac{1+2.sin\alpha.cos\alpha}{sin\alpha+cós\alpha}\)- cos\(\alpha\)
c) C= cos\(\alpha\) + \(\dfrac{sin\alpha}{1+cos\alpha}\)
D = \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
Giúp mk vs please
Cho \(\tan\alpha=\frac{3}{5}\), hãy tính giá trị của:
a) \(M=\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
b) \(N=\frac{\sin\alpha\cos\alpha}{\sin^2\alpha-\cos^2\alpha}\)
c) \(P=\frac{\sin^3\alpha+\cos^3\alpha}{2\sin\alpha\cos^2\alpha+\cos\alpha\sin^2\alpha}\)
Cho góc nhọn α
a) Rút gọn biểu thức S=\(\cos^2\alpha+tg^2.\cos^2\alpha\)
b) Chứng minh:
\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=4\)
Help me plsssssssssss
1. Chứng minh rằng: \(\frac{1-2\sin.\cos\alpha}{sin^2\alpha-\cos^2\alpha}=\frac{sin\alpha-\cos\alpha}{sin\alpha+\cos\alpha}\) (\(\alpha\ne45^o\))
2. Chứng minh: \(\cos^4\alpha+\sin^2\alpha.\cos^2\alpha+\sin^2\alpha\) không phụ thuộc vào x
CMR
a)\(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
b)\(\frac{\tan\alpha+1}{\tan\alpha-1}=\frac{1+\cot\alpha}{1-\cot\alpha}\)
c) \(\tan^2\alpha-\sin^2\alpha=\tan^2\alpha.\sin^2\alpha\)
d)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)