chung minh voi voi goc nhon \(\alpha\)tuy y ta co
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
\(\cos\alpha=\frac{\cos\alpha}{\sin\alpha}\)
\(\tan\alpha.\cos\alpha=1\)
\(\sin^2\alpha+\cos^2\alpha=1\)
goi y: su dung dinh ly pitago
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Những câu hỏi liên quan
NN
13 tháng 9 2020
a) \(\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos a}\)
\(\Leftrightarrow\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=\sin^2\alpha\)
\(\Leftrightarrow1-\cos^2\alpha=\sin^2\alpha\)
\(\Leftrightarrow\sin^2\alpha+\cos^2\alpha=1\)( luôn đúng )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a)
Ta có:
\({\cos ^4}\alpha {\sin ^4}\alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) \\= {\cos ^2}\alpha - {\sin ^2}\alpha = {\cos ^2}\alpha - (1 - {\cos ^2}\alpha ) \\= {\cos ^2}\alpha - 1 + {\cos ^2}\alpha = 2{\cos ^2}\alpha - 1\)
(đpcm)
b)
Ta có:
\(\frac{{{{\cos }^2}\alpha + {{\tan }^2}\alpha - 1}}{{{{\sin }^2}\alpha }} = \frac{{{{\cos }^2}\alpha \; + {{\tan }^2}\alpha - {{\sin }^2}\alpha - {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{{{{\tan }^2}\alpha - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{\frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} - {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} \\= \frac{1}{{{{\cos }^2}\alpha }} - 1 = {\tan ^2}\alpha \)
(đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 6 2020
\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)
\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)
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/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
QT
Quoc Tran Anh Le
Giáo viên
21 tháng 9 2023
a) Ta có:
\(\begin{array}{l}{\sin ^4}\alpha - {\cos ^4}\alpha = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha - {{\cos }^2}\alpha } \right) = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow {\sin ^2}\alpha - {\cos ^2}\alpha - 1 + 2{\cos ^2}\alpha = 0\\ \Leftrightarrow {\sin ^2}\alpha + {\cos ^2}\alpha - 1 = 0\\ \Leftrightarrow 1 - 1 = 0\\ \Leftrightarrow 0 = 0\end{array}\)
Đẳng thức luôn đúng
b) Ta có:
\(\begin{array}{l}\tan \alpha + \cot \alpha = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{1}{{\sin \alpha .\cos \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\end{array}\)
Đẳng thức luôn đúng
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)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 8 2020
\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)
\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
1) \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
bien doi ve phai ta co
\(\frac{\sin\alpha}{\cos\alpha}=\frac{doi}{huyen}:\frac{ke}{huyen}=\frac{doi}{huyen}.\frac{huyen}{ke}=\frac{doi}{ke}=\tan\alpha\)
2) \(\cot\alpha=\frac{\cos\alpha}{\sin\alpha}\)
bien doi ve phai ta co
\(\frac{\cos\alpha}{\sin\alpha}=\frac{ke}{huyen}:\frac{doi}{huyen}=\frac{ke}{huyen}.\frac{huyen}{doi}=\frac{ke}{doi}=\cot\alpha\)
3) \(\tan\alpha.\cos\alpha=1\)
\(\frac{\cos\alpha}{\sin\alpha}.\frac{\sin\alpha}{\cos\alpha}=1\)
4) \(\sin^2\alpha+\cos^2\alpha=1\)
\(\frac{doi^2}{huyen^2}+\frac{ke^2}{huyen^2}=\frac{huyen^2}{huyen^2}=1\)( su dung dinh ly pitago )