\(\left(\sin a+\cos a\right)^2=\sin^2a+\cos^2a+2\cdot\sin a\cdot\cos a\)
\(=1+2\cdot\dfrac{1}{2}=2\)
Do đó: \(\sin a+\cos a=\pm\sqrt{2}\)
=>Chọn D
\(\left(\sin a+\cos a\right)^2=\sin^2a+\cos^2a+2\cdot\sin a\cdot\cos a\)
\(=1+2\cdot\dfrac{1}{2}=2\)
Do đó: \(\sin a+\cos a=\pm\sqrt{2}\)
=>Chọn D
Chứng minh:
a)\(cot^2\alpha-cos^2\alpha\cdot cot^2\alpha=cos^2\alpha\)
b)\(tan^2\alpha-sin^2\alpha\cdot tan^2\alpha=sin^2\alpha\)
c) \(\dfrac{1-cos^2}{sin\alpha}\) = \(\dfrac{sin\alpha}{1+cos\alpha}\)
d)\(tan^2\alpha-sin^2\alpha=tan^2\cdot sin^2\alpha\)
e) \(\sin^6\alpha+cos^6\alpha+3sin^2\cdot cos^2\alpha=1\)
TÍNH
a) A= tan 1 độ* tan 2 độ * tan 3 độ.....tan 89 độ
b) Cho góc nhọn α,tan α=\(\dfrac{1}{2}\) tính:
B=\(\dfrac{\sin\alpha+2\cos\alpha}{3\sin\alpha-4\cos\alpha}\)
D=\(\dfrac{2\sin^2\alpha-3\cos^2\alpha}{4\cos^2\alpha-5\sin^2\alpha}\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
rút gọn biểu thức :
A = 1 + \(\dfrac{2\sin\alpha.\cos\alpha}{\cos^2\alpha-\sin^2\alpha}\)
B = \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
Rút gọn .
\(A=\dfrac{1+2\sin\alpha\cos\alpha}{\sin\alpha+\cos\alpha}\)
\(B=\left(\sin\alpha+\cos\alpha\right)^2-\left(\cos\alpha-\sin\alpha\right)^2\)
\(C=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)}{\sin\alpha\cos\alpha}\)
Mấy bạn giúp đỡ được phần nào thì giúp , giúp hết thì tốt quá .
Hãy đơn giản các biểu thức:
a) 1-sin2α
b) (1-cosα)(1+cosα)
c) 1+cos2α+sin2α
d) sinα-sinα cos2α
e) sin4α+cos4α+2sin2α cos2α
f) tan2α-sin2α tan2α
g) cos2α+tan2α cos2α
h) tan2α (2cos2α+sin2α-1)
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)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
CM các hệ thức sau:
a) \(1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\)
b) \(1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\)
c) \(\cot^2\alpha-\cos^2\alpha=\cot^2\alpha.\cos^2\alpha\)
d) \(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
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