Hãy đơn giản các biểu thức sau:
a) \(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
b) \(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)\)
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:
\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) = - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) = - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) = - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha - \cos \alpha .\tan \alpha .\cot \alpha \)
\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha - \cos \alpha .1 = \cos \alpha .\)
Hãy đơn giản các biểu thức:
a) \(\sin\alpha-\sin\alpha.\cos^2\alpha\)
b) \(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
c) \(\cos^2\alpha+\tan^2\alpha.\cos^2\alpha\)
d) \(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
alibaba nguyễn bạn có thể giải cụ thể đc ko v ?
Rút gọn các biểu thức sau:
a, \(\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right) - cos\alpha \),
b, \({\left( {cos\alpha + \sin \alpha } \right)^2} - \sin 2\alpha \)
\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)
\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)
1. Đơn giản biểu thức
a. \(\sin\alpha\cdot\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
b. \(\left(\sin^2\alpha+\cos^2\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
c. \(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\)
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
Cho góc bất kì \(\alpha \). Chứng minh các đẳng thức sau:
a) \({\left( {\sin \alpha + \cos \alpha } \right)^2} = 1 + \sin 2\alpha ;\;\)
b) \({\cos ^4}\alpha - {\sin ^4}\alpha = \cos 2\alpha .\)
a) Ta có: \({\left( {\sin \alpha + \cos \alpha } \right)^2} = {\sin ^2}\alpha + 2\sin \alpha \cos \alpha + {\cos ^2}\alpha = 1 + \sin 2\alpha \;\)
b) \({\cos ^4}\alpha - {\sin ^4}\alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = \cos 2\alpha \;\)
Đơn giản các biểu thức sau:
\(a,\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
\(b,\sin\alpha\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
a, = \(\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha\)+ \(\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\)
= \(2\sin^2\alpha+2\cos^2\alpha\)= 4
b,=\(\sin\alpha\cos\alpha\)(\(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\))
= \(\sin\alpha\cos\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}\)
=1
#mã mã#
b, =1
mk viết thiếu
#mã mã#
Hãy đơn giản biểu thức :
a) \(1-\sin^2\alpha\)
b) \(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cos^2\alpha\)
c) \(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)
d) \(tg^2\alpha-\sin^2\alpha.tg^2\alpha\)
e) \(1+\sin^2\alpha+\cos^2\alpha\)
g) \(\cos^2\alpha+tg^2\alpha.\cos^2\alpha\)
h) \(\sin\alpha-\sin\alpha.\cos^2\alpha\)
i) \(tg^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
đáp án :
a) \(cos^2\alpha\)
b) 1
c) \(sin^2\alpha\)
d) \(sin^2\alpha\)
e) 2
g) 1
h) \(sin^3\alpha\)
i) \(sin^2\alpha\)
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, \(\frac{cot^2\alpha-cos^2\alpha}{cot^2a}+\frac{sin\alpha.cos\alpha}{cot\alpha}\)
\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)
\(=tan^2a+1=\frac{1}{cos^2a}\)
\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)
\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)
\(=1-sin^2a+sin^2a=1\)
Rút gọn các biểu thức sau:
A= \(\dfrac{cos^2\alpha-sin^2\alpha}{cot^2\alpha-tan^2\alpha}-cos^2\alpha\)
B= \(\sqrt{sin^4\alpha+6cos^2\alpha+3cos^4\alpha}+\sqrt{cos^4\alpha+6sin^2\alpha+3sin^4\alpha}\)
\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)
\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)
\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)
\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)
\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)
\(=2\left(sin^2a+cos^2a\right)+2=4\)