Giúp tui bài này vssss
Cho \(\tan\alpha=\frac{8}{15}.\) Tìm \(\sin\alpha,\cos\alpha\)
NM
Những câu hỏi liên quan
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)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
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 )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
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
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) \(\frac{\sin\alpha}{1+\cos\alpha}+\frac{1+\cos\alpha}{\sin\alpha}=\frac{2}{\sin\alpha}\)
\(\frac{1+sin^2a}{1-sin^2a}=\frac{1+sin^2a}{cos^2a}=\frac{1}{cos^2a}+\frac{sin^2a}{cos^2a}=1+tan^2a+tan^2a=1+2tan^2a\)
\(\frac{cosa}{1+sina}+tana=\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina+sin^2a}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sina}{1+cosa}+\frac{1+cosa}{sina}=\frac{sin^2a+cos^2a+2cosa+1}{\left(1+cosa\right)sina}=\frac{2+2cosa}{\left(1+cosa\right)sina}=\frac{2\left(1+cosa\right)}{\left(1+cosa\right)sina}=\frac{2}{sina}\)
Đúng 0
Bình luận (0)
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.
tính
a) \(\tan^2\alpha-\sin^2\alpha-\tan^2\alpha\times\sin^2\alpha\)
b)\(\frac{sin^4\alpha-cos^4\alpha}{sin\alpha+cos\alpha}-sin\alpha+cos\alpha\)
Tính E=\(\frac{8\cos^3\alpha-2\sin^3\alpha+\cos\alpha}{2cos\alpha-sin^3\alpha}\) khi tan α=2, góc α nhọn
Chứng minh các đẳng thức:
a) \({\cos ^4}\alpha - {\sin ^4}\alpha = 2{\cos ^2}\alpha - 1\);
b) \(\frac{{{{\cos }^2}\alpha + {{\tan }^2}\alpha - 1}}{{{{\sin }^2}\alpha }} = {\tan ^2}\alpha \).
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)
Đúng 0
Bình luận (0)
CMR: \(\frac{\sin^4\alpha-\cos^2\alpha+2\cos^4\alpha-\cos^6\alpha}{\cos^4\alpha-\sin^2\alpha+2\sin^4\alpha-\sin^6\alpha}=\tan^6\alpha\)
\(\frac{\sin^4\alpha-\cos^2\alpha+2\cos^4\alpha-\cos^6\alpha}{\cos^4\alpha-\sin^2\alpha+2\sin^4\alpha-\sin^6\alpha}=\frac{\sin^4\alpha-\cos^2\alpha\left(1-\cos^2\alpha\right)^2}{\cos^4\alpha-\sin^2\alpha\left(1-\sin^2\alpha\right)^2}\)
\(=\tan^4\alpha.\frac{1-\cos^2\alpha}{1-\sin^2\alpha}=\tan^6\alpha\)
Đúng 0
Bình luận (0)