Tìm \(\alpha\) biết:
a) \(\sin\alpha=0\)
b) \(\sin\alpha=1\)
c) \(\sin\alpha=-1\)
d) \(\cos\alpha=0\)
e) \(\cos\alpha=1\)
f) \(\cos\alpha=-1\)
K
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AH
Akai Haruma
Giáo viên
1 tháng 10 2018
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
AH
Akai Haruma
Giáo viên
1 tháng 10 2018
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
29 tháng 10 2018
a) 1- \(sin^2\alpha\)= \(cos^2\alpha\)
b) (\(1-cos\alpha\))(\(1+cos\alpha\)) = 1 - cos2\(\alpha\) = sin2\(\alpha\)
c) 1 + cos2\(\alpha\) + sin2\(\alpha\) = \(1+1=2\)
d) sin\(\alpha\) - sin\(\alpha.cos^2\alpha\)
= \(sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
e) \(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha\)
= \(\left(sin^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha+\left(cos^2\alpha\right)^2\)
= \(\left(sin^2\alpha+cos^2\alpha\right)^2=1^2=1\)
f) \(tan^2\alpha-sin^2\alpha.tan^2\alpha\)
= \(tan^2\alpha\left(1-sin^2\alpha\right)=tan^2\alpha.cos^2\alpha=sin^2\alpha\)
g) \(cos^2\alpha+tan^2\alpha.cos^2\alpha\)
= \(cos^2\alpha\left(1+tan^2\alpha\right)=cos^2\alpha.\dfrac{1}{cos^2\alpha}=1\)
h) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)
= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)
= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)
= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)
31 tháng 10 2017
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)
17 tháng 4 2017
a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
BT
9 tháng 5 2017
a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 10 2020
\(\frac{sin^2a-cos^2a+cos^4a}{cos^2a-sin^2a+sin^4a}=\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^2a-cos^2a.sin^2a}{cos^2a-sin^2a.cos^2a}\)
\(=\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^2a.sin^2a}{cos^2a.cos^2a}=tan^4a\)
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-sin^2a.cos^2a=1-2sin^2a.cos^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2020
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
a/ \(\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{3-1}{3+1}\)
b/ \(\frac{2sina+3cosa}{3sina-5cosa}=\frac{3tana+3}{3tana-5}=\frac{3.3+3}{3.3-5}\)
c/ \(\frac{1+2cos^2a}{1-cos^2a-cos^2a}=\frac{1+2cos^2a}{1-2cos^2a}=\frac{1+2.\frac{1}{10}}{1-2.\frac{1}{10}}\)
d/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \({\cos ^2}\alpha + {\sin ^2}\alpha = 1\)
b) \(\tan \alpha .\cot \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\)
c) \(\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha + 1\)
d) \(\frac{1}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha \)
c, \(sin\alpha=-1\Leftrightarrow\alpha=-\dfrac{\pi}{2}+k2\pi\)
a, \(sin\alpha=0\Leftrightarrow\alpha=k\pi\)