cho \(\sin\alpha\frac{1}{2}\) tính cos, tan , cotg
Những câu hỏi liên quan
chứng minh với góc nhọn \(\alpha\) túy ý có;
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)
cotg\(\alpha\)=\(\frac{\cos\alpha}{sin\alpha}\)
\(\tan\alpha\) . cotg \(\alpha\)=1
\(\sin^2\alpha+\cos^2\alpha=1\)
a/ \(\sin\alpha=\frac{C_đ}{C_h}\)
\(\cos\alpha=\frac{C_k}{C_h}\)
\(\Rightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{C_đ}{C_h}}{\frac{C_k}{C_h}}=\frac{C_đ}{C_k}=\tan\alpha\)
b/ \(\frac{\cos\alpha}{\sin\alpha}=\frac{\frac{C_k}{C_h}}{\frac{C_đ}{C_h}}=\frac{C_k}{C_đ}=\cot\alpha\)
c/ \(\tan\alpha.\cot\alpha=\frac{C_đ}{C_k}.\frac{C_k}{C_đ}=1\)
d/ \(\sin^2\alpha=\frac{C_đ^2}{C_h^2}\)
\(\cos^2\alpha=\frac{C_k^2}{C_h^2}\)
\(\Rightarrow\sin^2\alpha+\cos^2\alpha=\frac{C_đ^2+C_k^2}{C_h^2}=\frac{C_h^2}{C_h^2}=1\)
P/s: hok trc lp 9 hay sao mà lm bài bài này?
Đúng 0
Bình luận (1)
7 tháng 9 2019 lúc 17:41
tính \(C=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)+\left(1+cotg^2\alpha\right)\left(1-cos^2\alpha\right)\)
\(C=\left(1+\tan^2\alpha\right).\cos^2\alpha+\left(1+\cot^2\alpha\right).\sin^2\alpha\)
\(=\cos^2\alpha+\cos^2\alpha.\tan^2\alpha+\sin^2\alpha+\sin^2\alpha.\cot^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\cos^2\alpha\right)\)
\(=1+1=2\)
Đúng 0
Bình luận (0)
Em dùng công thức sau 1+tan2x=\(\frac{1}{cos^2}\);\(1+cotg^2x=\frac{1}{sin^2x}\)với sin2x+cos2x=1
Đặt \(\left(sin^2\alpha;cos^2\alpha\right)=\left(a;b\right)\)=>1+a2=\(\frac{1}{b^2}\);\(1+b=\frac{1}{a^2}\);a2+b2=1
Suy ra C=\(\frac{1}{b^2}.\left(1-a^2\right)\)+\(\frac{1}{a^2}.\left(1-b^2\right)\)=\(\frac{1}{b^2}.b^2\)+\(\frac{1}{a^2}.a^2\)=2
Vậy C=2
Đúng 0
Bình luận (0)
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 )
\(\Rightarrow\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\)
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)
Cho \(\tan\alpha=\frac{3}{5}\)
Tính: \(\frac{\sin^3\alpha+\cos^3\alpha}{2\sin\alpha.\cos^2\alpha+\cos\alpha.\sin^2\alpha}\)
Cho \(\tan\alpha-5\cot\alpha+4=0.\). Tính \(A=\frac{4\sin\alpha+2\cos\alpha}{3\sin\alpha-\cos\alpha}\)
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
2 tháng 10 2020 lúc 20:06
cho tan= \(\frac{1}{3}\)tính:
A= \(\frac{\cos^2\alpha+sin\alpha}{cos\alpha-sin\alpha}\)
Cho góc nhọn \(\alpha\)thỏa mãn \(\tan\alpha=\frac{2}{\sqrt{3}}\). Tính: \(B=\frac{\cos^4\alpha+\sin^2\alpha\left(\cos^2\alpha+1\right)}{2\cos^4\alpha+2\sin^2\cos^2-\frac{3}{5}\sin^2\alpha}\)