Cho biểu thức \(A=\dfrac{cos70^o-sin\alpha}{tan60^o-cos70^o}\)( 200 <\(\alpha\) < 900). Chứng minh A < 0
Rút gọn các biểu thức :
a/ \(\frac{2cos^2\alpha-1}{sin\alpha+cos\alpha}\)
b/ \(\frac{sin25+cos70}{sin20+cos65}\)
\(a,=\frac{2cos^2\alpha-cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =cos\alpha-sin\alpha\)
\(b,sin25=cos65;cos70=sin20;Khiđó:B=1\)
1. Tính giá trị biểu thức
S= cos70 +cos50 -cos10
2. Cho a+b=π/4. Cm
(1+tanα).(1+tanβ) =2
3. Tính giá trị biểu thức
P= sin^2 10¤ +sin^2 50¤ +sin^2 70¤
1.
\(cos70+cos50=2cos\dfrac{70+50}{2}.cos\dfrac{70-50}{2}=2.cos60.cos10=2.\dfrac{1}{2}cos10\)
\(cos70+cos50-cos10=0\)
2.\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=1\Rightarrow tana+tanb+tana.tanb+1=2\Leftrightarrow\left(1+tana\right)\left(1+tanb\right)=2\)
Ting giá trị biểu thức: \(\frac{\sin25^o+\cos70^o}{\sin20^o+\cos65^o}\)
GIÚP MÌNH VỚI M.M!!!
Ta có sin25°=cos65°
cos70°=20sin°
=> sịn25°+cos70°/sin20°+cos65°=cos65°+sin20°/sin20°+cos65°=1
Chứng Minh
\(\frac{\sin20^o.\cos70^o+\cos^220^o+\sin160^o-1}{\cos200^o.\cos70^o}=\tan20^o\)
Bt: Tính \(2008\cdot\sin^220^o+\sin20^o+2008\cdot\cos^220^o-\cos70^o+\tan20^o\cdot\tan70^o\)
\(=2008\left(\sin^220^o+\cos^220^o\right)+\cos70^o-\cos70^o+\frac{\sin20^o}{\cos20}.\frac{sin70}{c\text{os}70}\)
\(=2008+1=2009\)
1)Rút gọn biểu thức: sin20-tan40+cotan50-cos70
2)Cho sin a=2/3.Tính giá trị của biểu thức A=2sin^2a+3cos^2a
Bài 1:
\(=\left(\sin20^0-\cos70^0\right)+\left(-\tan40^0+\cot50^0\right)=0+0=0\)
Bài 2:
\(\cos a=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)
\(A=2\cdot\sin^2a+3\cdot\cos^2a=2\cdot\dfrac{4}{9}+3\cdot\dfrac{5}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)
Rút gọn biểu thức:
a, A = \(\dfrac{4\sin^2\alpha}{1-\cos\dfrac{\alpha}{2}}\)
b, B = \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)
c, C = \(\dfrac{1+\sin\alpha-2\sin^2\left(45^o-\dfrac{\pi}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)
tính giá trị biểu thức sau:
\(G=\dfrac{tan30^o+tan40^o+tan50^o+tan60^o}{1-2sin^210^o}\)
Biểu thức này chỉ rút gọn được khi mẫu là \(1-2sin^210^0\)
\(tan40+tan50=\dfrac{sin40}{cos40}+\dfrac{sin50}{cos50}=\dfrac{sin40.cos50+cos50.sin40}{cos40.cos50}\)
\(=\dfrac{sin\left(40+50\right)}{\dfrac{1}{2}\left(cos90+cos10\right)}=\dfrac{2}{cos10}\)
\(\Rightarrow tan30+tan60+tan40+tan50=\dfrac{\sqrt{3}}{3}+\sqrt{3}+\dfrac{2}{cos10}\)
\(=\dfrac{4\sqrt{3}}{3}+\dfrac{2}{cos10}=\dfrac{4\sqrt{3}cos10+6}{3.cos10}=\dfrac{4\sqrt{3}\left(cos10+\dfrac{\sqrt{3}}{2}\right)}{3.cos10}\)
\(=\dfrac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}=\dfrac{8\sqrt{3}cos20.cos10}{3cos10}=\dfrac{8\sqrt{3}}{3}cos20\)
\(\Rightarrow G=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{1-2sin^210}=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{cos20}=\dfrac{8\sqrt{3}}{3}\)
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:
\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) = - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) = - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) = - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha - \cos \alpha .\tan \alpha .\cot \alpha \)
\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha - \cos \alpha .1 = \cos \alpha .\)