Cho \(\tan\alpha=\frac{1}{3}\)
Tính \(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
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)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
Cho tan \(\alpha\)=\(\frac{3}{5}\). Tính
A= \(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
B=\(\frac{\sin\alpha\cdot\cos\alpha}{\sin^2\alpha-\cos^2\alpha}\)
C=\(\frac{\sin^3\alpha\cdot\cos^3\alpha}{2\sin\alpha\cdot\cos^2\alpha+\cos\alpha\cdot\sin^2\alpha}\)
Giúp mình với . MÌnh cảm ơn
Cho \(\tan\alpha=\frac{3}{5}\)
Tính: \(\frac{\sin^3\alpha+\cos^3\alpha}{2\sin\alpha.\cos^2\alpha+\cos\alpha.\sin^2\alpha}\)
Biết tan α=3. Tính giá trị các biểu thức sau:
a)\(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
b)\(\frac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-5\cos\alpha}\)
c)\(\frac{1+2\cos^2\alpha}{\sin^2\alpha-\cos^2\alpha}\)
d)\(\frac{\sin^4\alpha+\cos^4\alpha}{1+\sin^2\alpha}\)
\(\frac{1}{cos^2a}=1+tan^2a\Rightarrow cos^2a=\frac{1}{1+tan^2a}=\frac{1}{10}\)
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/ \(\frac{\left(1-cos^2a\right)^2+\left(cos^2a\right)^2}{1+1-cos^2a}=\frac{\left(1-\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2}{2-\frac{1}{10}}\)
Cho \(\tan\alpha-5\cot\alpha+4=0.\). Tính \(A=\frac{4\sin\alpha+2\cos\alpha}{3\sin\alpha-\cos\alpha}\)
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
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) \(P=\frac{\sin^3\alpha+\cos^3\alpha}{2\sin\alpha\cos^2\alpha+\cos\alpha\sin^2\alpha}\)
\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)
\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
Chứng minh các hệ thức sau:
a) \(\frac{1-cos\alpha}{sin\alpha}=\frac{sin\alpha}{1+cos\alpha}\)
b) \(tan^2\alpha-sin^2\alpha=tan^2\alpha.sin^2\alpha\)
c) \(\frac{1-tan\alpha}{1+tan\alpha}=\frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}\)
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 )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
cm các đẳng thức:
a) \(\frac{1+\sin^2\alpha}{1-\sin^2\alpha}=1+2\tan^2\alpha\)
b) \(\frac{\cos\alpha}{1+\sin\alpha}+\tan\alpha=\frac{1}{\cos\alpha}\)
c) \(\frac{\sin\alpha}{1+\cos\alpha}+\frac{1+\cos\alpha}{\sin\alpha}=\frac{2}{\sin\alpha}\)
\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)
\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)
a) Biết sinα= \(\frac{1}{2}\). Tính cosα, tanα, cotα.
b) Biết cosα= \(\frac{2}{5}\). Tính sinα, tanα, cotα.
c) Biết tanα= 3. Tính cosα, sinα, cotα.
d) Biết cotα=\(\sqrt{3}\). Tính cosα, tanα, sinα.
e) Biết sinα= \(\frac{1}{\sqrt{3}}\). Tính cosα, tanα, cotα.
cho \(\tan\alpha=3.Tính\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)
\(\frac{cosa+sina}{cosa-sina}=\frac{1+tana}{1-tana}=\frac{1+3}{1-3}=-2\)
=>1+sina/1-sina=1+tana/1-tana=1+3/1-3=4/-2=-2
bạn chỉ cần tính góc an-pha ra rồi thay vào thôi