Cho \(\tan\alpha=3\)
Tính \(a)M=\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\\ b)B=\frac{\sin15^o+\cos15^o}{\cos15^o}-\cot75^o\)
Cho \(\tan\alpha=3\)
Tính \(a)M=\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\\ b)B=\frac{\sin15^o+\cos15^o}{\cos15^o}-\cot75^o\)
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!
Cho góc \(\alpha \;\;({0^o} < \alpha < {180^o})\) thỏa mãn \(\tan \alpha = 3\)
Tính giá trị biểu thức: \(P = \frac{{2\sin \alpha - 3\cos \alpha }}{{3\sin \alpha + 2\cos \alpha }}\)
\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)
Ta có: \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)
\( \Leftrightarrow {\cos ^2}\alpha = \frac{1}{{10}} \Leftrightarrow \cos \alpha = \pm \frac{{\sqrt {10} }}{{10}}\)
Vì \({0^o} < \alpha < {180^o}\) nên \(\sin \alpha > 0\).
Mà \(\tan \alpha = 3 > 0 \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \frac{{\sqrt {10} }}{{10}}\)
Lại có: \(\sin \alpha = \cos \alpha .\tan \alpha = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)
\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)
Tính:
\(C=\frac{\tan^2\alpha\left(1+\cos^3\alpha\right)+\cot^2\alpha\left(1+\sin^3\alpha\right)}{\left(\sin^3\alpha+\cos^3\alpha\right)\left(1+\sin^3\alpha+\cos\alpha\right)}\)
Biết \(\tan\alpha=\tan35^o.\tan36^o.\tan37^o.....\tan57^o\)
Chứng minh các hệ thức sau:
a) \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\).
b) \(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha \ne {90^o})\)
c) \(1 + {\cot ^2}\alpha = \frac{1}{{{{\sin }^2}\alpha }}\quad ({0^o} < \alpha < {180^o})\)
Tham khảo:
a)
Gọi M(x;y) là điểm trên đường tròn đơn vị sao cho \(\widehat {xOM} = \alpha \). Gọi N, P tương ứng là hình chiếu vuông góc của M lên các trục Ox, Oy.
Ta có: \(\left\{ \begin{array}{l}x = \cos \alpha \\y = \sin \alpha \end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}\alpha = {x^2}\\{\sin ^2}\alpha = {y^2}\end{array} \right.\)(1)
Mà \(\left\{ \begin{array}{l}\left| x \right| = ON\\\left| y \right| = OP = MN\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{x^2} = {\left| x \right|^2} = O{N^2}\\{y^2} = {\left| y \right|^2} = M{N^2}\end{array} \right.\)(2)
Từ (1) và (2) suy ra \({\sin ^2}\alpha + {\cos ^2}\alpha = O{N^2} + M{N^2} = O{M^2}\) (do \(\Delta OMN\) vuông tại N)
\( \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = 1\) (vì OM =1). (đpcm)
b)
Ta có: \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}\;\;(\alpha \ne {90^o})\)
\( \Rightarrow 1 + {\tan ^2}\alpha = 1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}\)
Mà theo ý a) ta có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) với mọi góc \(\alpha \)
\( \Rightarrow 1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }}\) (đpcm)
c)
Ta có: \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }}\;\;\;({0^o} < \alpha < {180^o})\)
\( \Rightarrow 1 + {\cot ^2}\alpha = 1 + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}\)
Mà theo ý a) ta có \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) với mọi góc \(\alpha \)
\( \Rightarrow 1 + {\cot ^2}\alpha = \frac{1}{{{{\sin }^2}\alpha }}\) (đpcm)
Tính :
\(B=\frac{\sin^2\alpha.\cos\left(\frac{\alpha}{2}\right)-\cot\left(\frac{\alpha}{3}\right)}{\frac{1}{\sqrt{2}}\sin\alpha+\sqrt{2}\tan\left(\frac{\alpha}{2}\right)}\) với \(\tan\alpha=\frac{\sin^267^o23'.\cos25^o41'}{\sin45^o16'+\cos^267^o29'}\text{ và }0^o
1. Chứng minh rằng: \(\frac{1-2\sin.\cos\alpha}{sin^2\alpha-\cos^2\alpha}=\frac{sin\alpha-\cos\alpha}{sin\alpha+\cos\alpha}\) (\(\alpha\ne45^o\))
2. Chứng minh: \(\cos^4\alpha+\sin^2\alpha.\cos^2\alpha+\sin^2\alpha\) không phụ thuộc vào x
1.
\(\frac{1-2sin\alpha cos\alpha}{sin^2\alpha-cos^2\alpha}=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)
\(\Leftrightarrow\frac{1-2sin\alpha cos\alpha}{\left(sin\alpha-cos\alpha\right)\left(sin\alpha+cos\alpha\right)}=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)
\(\Leftrightarrow1-2sin\alpha cos\alpha=\left(sin\alpha-cos\alpha\right)^2\)
\(\Leftrightarrow1-2sin\alpha cos\alpha=sin^2\alpha+cos^2\alpha-2sin\alpha cos\alpha\)
\(\Leftrightarrow1-2sin\alpha cos\alpha=1-2sin\alpha cos\alpha\left(đpcm\right)\)
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)
\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)
Đơn giản các biểu thức sau:
a) \(\sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o};\)
b) \(2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) với \({0^o} < \alpha < {90^o}\).
a) Ta có: \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) = - \cos {16^o}\end{array} \right.\)
\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)
b)
Ta có:
\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) = - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) = - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) = - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha - \cos \alpha .\tan \alpha .\cot \alpha \)
\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha - \cos \alpha .1 = \cos \alpha .\)