Biết \(\sin\alpha+\cos\alpha=\frac{\sqrt{3}+1}{2}\)
Tính \(A=\frac{\cos^2\alpha-\sin^2\alpha}{\cot^2\alpha-\tan^2\alpha}\)
Những câu hỏi liên quan
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.
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)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
1/ Cho \(cot\alpha=\sqrt{5}\) . Tính \(C=sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\)
2/ Cho \(tan\alpha=3\) . Tính \(B=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)
1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)
\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)
\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)
\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)
\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)
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1: \(cota=\sqrt{5}\)
=>\(cosa=\sqrt{5}\cdot sina\)
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+5=6\)
=>\(sin^2a=\dfrac{1}{6}\)
\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)
\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)
2: tan a=3
=>sin a=3*cosa
\(1+tan^2a=\dfrac{1}{cos^2a}\)
=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)
\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)
\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)
\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)
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đơn giản biểu thức:
a, \(\left(\frac{sin\alpha+tan\alpha}{cos\alpha+1}\right)^2+1\)
b, \(tan\alpha\left(\frac{1+cos^2\alpha}{sin\alpha}-sin\alpha\right)\)
c, \(\frac{cot^2\alpha-cos^2\alpha}{cot^2a}+\frac{sin\alpha.cos\alpha}{cot\alpha}\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
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1. Tìm x, biết: a. tan x+cot x2b. sin x.cos xfrac{sqrt{3}}{4}2. a. Biết tanalphafrac{1}{3}Tính Afrac{sinalpha-cosalpha}{sinalpha+cosalpha}b. Biết sinalphafrac{2}{3}Tính B3.sin^2alpha+4.cos^2alphac. Tính Csin^210^o+sin^220^o+sin^270^o+sin^280^od. Tính Dtan20^o.tan35^o.tan55^o.tan70^oe. Tính Esin^6alpha+cos^6alpha+3.sin^2alpha.cos^2alphaf. Tính F3.left(sin^3alpha+cos^3alpharight)-2.left(sin^6alpha+cos^6alpharight)g. Tính Gsqrt{sin^4alpha+4.cos^2alpha}+sqrt{cos^4alpha+4.sin^2alpha}Mọi người giúp mì...
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1. Tìm x, biết:
a. \(\tan x+\cot x=2\)
b. \(\sin x.\cos x=\frac{\sqrt{3}}{4}\)
2.
a. Biết \(\tan\alpha=\frac{1}{3}\)Tính A=\(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)
b. Biết \(\sin\alpha=\frac{2}{3}\)Tính B=\(3.\sin^2\alpha+4.\cos^2\alpha\)
c. Tính C=\(\sin^210^o+\sin^220^o+\sin^270^o+\sin^280^o\)
d. Tính D=\(\tan20^o.\tan35^o.\tan55^o.\tan70^o\)
e. Tính E=\(\sin^6\alpha+\cos^6\alpha+3.\sin^2\alpha.\cos^2\alpha\)
f. Tính F=\(3.\left(\sin^3\alpha+\cos^3\alpha\right)-2.\left(\sin^6\alpha+\cos^6\alpha\right)\)
g. Tính G=\(\sqrt{\sin^4\alpha+4.\cos^2\alpha}+\sqrt{\cos^4\alpha+4.\sin^2\alpha}\)
Mọi người giúp mình với. Mình cảm ơn ạ!
Ta có: sin^2alpha+cos^2alpha1. lại có : sinalphafrac{2}{3} frac{4}{9}+cos^2alpha1 cos^2alphafrac{5}{9}Rightarrowcosalphafrac{sqrt{5}}{3}Mà tanalphafrac{sinalpha}{cosalpha}frac{2}{3}:frac{sqrt{5}}{3}frac{2}{sqrt{5}}mặt khác: tanalpha.cotalpha1Rightarrowcotalphafrac{sqrt{5}}{2}
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Ta có: \(\sin^2\alpha+\cos^2\alpha=1\). lại có : \(\sin\alpha=\frac{2}{3}\)
=> \(\frac{4}{9}+\cos^2\alpha=1\)
=> \(\cos^2\alpha=\frac{5}{9}\Rightarrow\cos\alpha=\frac{\sqrt{5}}{3}\)
Mà \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}:\frac{\sqrt{5}}{3}=\frac{2}{\sqrt{5}}\)
mặt khác: \(\tan\alpha.\cot\alpha=1\Rightarrow\cot\alpha=\frac{\sqrt{5}}{2}\)
Cho \(\tan\alpha-5\cot\alpha+4=0.\). Tính \(A=\frac{4\sin\alpha+2\cos\alpha}{3\sin\alpha-\cos\alpha}\)
\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)
\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)
\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)
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CM các hệ thức sau:
a) \(1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\)
b) \(1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\)
c) \(\cot^2\alpha-\cos^2\alpha=\cot^2\alpha.\cos^2\alpha\)
d) \(\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\)
\(1+tan^2a=1+\frac{sin^2a}{cos^2a}=\frac{cos^2a+sin^2a}{cos^2a}=\frac{1}{cos^2a}\)
\(1+cot^2a=1+\frac{cos^2a}{sin^2a}=\frac{sin^2a+cos^2a}{sin^2a}=\frac{1}{sin^2a}\)
\(cot^2a-cos^2a=\frac{cos^2a}{sin^2a}-cos^2a=cos^2a\left(\frac{1}{sin^2a}-1\right)=cos^2a\left(\frac{1-sin^2a}{sin^2a}\right)\)
\(=cos^2a\left(\frac{cos^2a}{sin^2a}\right)=cos^2a.cot^2a\)
\(\frac{1+cosa}{sina}=\frac{sina\left(1+cosa\right)}{sin^2a}=\frac{sina\left(1+cosa\right)}{1-cos^2a}=\frac{sina\left(1+cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\frac{sina}{1-cosa}\)
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25 tháng 7 2021 lúc 14:59
Cho sin\(\alpha\) + cos\(\alpha\) =\(\sqrt{2}\)
a, Tính cos\(\alpha\), sin\(\alpha\), tan\(\alpha\), cot\(\alpha\).
b, Tính F = \(sin^5\alpha+cos^5\alpha\)