Biết tan \(\alpha\)= tan 35 độ * tan 36 độ *...* tan 52 độ *tan 53 độ
Tính \(M=\frac{tan^2\alpha\left(1+cos^3\alpha\right)+cot^3\alpha\left(1+sin^3\alpha\right)}{\left(sin^3\alpha+cos^3\alpha\right)\left(1+sin\alpha+cos\alpha\right)}\)
Những câu hỏi liên quan
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 đẳng thức sau:
a, sin^4alpha-cos^4alpha+12sin^2alpha
b,dfrac{sin^2alpha+2cos^2alpha-1}{cot^2alpha}sin^2alpha
c, dfrac{1-sin^2alpha.cos^2alpha}{cos^2alpha}-cos^2alphatan^2alpha
d, dfrac{sin^2alpha-tan^2alpha}{cos^2alpha-cot^2alpha}tan^6alpha
e, left(1+cotalpharight)sin^3alpha+left(1+tanalpharight)cos^3alphasinalpha.cosalpha
f,dfrac{left(sinalpha+cosalpharight)^2-1}{cotalpha-sinalpha.cosalpha}2tan^2alpha
Đọc tiếp
Chứng minh các đẳng thức sau:
a, \(\sin^4\alpha-\cos^4\alpha+1=2\sin^2\alpha\)
b,\(\dfrac{\sin^2\alpha+2\cos^2\alpha-1}{\cot^2\alpha}=\sin^2\alpha\)
c, \(\dfrac{1-\sin^2\alpha.\cos^2\alpha}{\cos^2\alpha}-\cos^2\alpha=\tan^2\alpha\)
d, \(\dfrac{\sin^2\alpha-\tan^2\alpha}{\cos^2\alpha-\cot^2\alpha}=\tan^6\alpha\)
e, \(\left(1+\cot\alpha\right)\sin^3\alpha+\left(1+\tan\alpha\right)\cos^3\alpha=\sin\alpha.\cos\alpha\)
f,\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-1}{\cot\alpha-\sin\alpha.\cos\alpha}=2\tan^2\alpha\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
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d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
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e)
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)
\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)
\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)
\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)
\(=\sin a+\cos a\)
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Chứng minh các đẳng thức :
a) tan3alpha-tan2alpha-tanalphatanalphatan2alphatan3alpha
b) dfrac{4tanalphaleft(1-tan^2alpharight)}{left(1+tan^2alpharight)^2}sin4alpha
c) dfrac{1+tan^4alpha}{tan^2alpha+cot^2alpha}tan^2alpha
d) dfrac{cosalphasinleft(alpha-3right)-sinalphacosleft(alpha-3right)}{cosleft(3-dfrac{pi}{6}right)-dfrac{1}{2}sin3}-dfrac{2tan3}{sqrt{3}}
Đọc tiếp
Chứng minh các đẳng thức :
a) \(\tan3\alpha-\tan2\alpha-\tan\alpha=\tan\alpha\tan2\alpha\tan3\alpha\)
b) \(\dfrac{4\tan\alpha\left(1-\tan^2\alpha\right)}{\left(1+\tan^2\alpha\right)^2}=\sin4\alpha\)
c) \(\dfrac{1+\tan^4\alpha}{\tan^2\alpha+\cot^2\alpha}=\tan^2\alpha\)
d) \(\dfrac{\cos\alpha\sin\left(\alpha-3\right)-\sin\alpha\cos\left(\alpha-3\right)}{\cos\left(3-\dfrac{\pi}{6}\right)-\dfrac{1}{2}\sin3}=-\dfrac{2\tan3}{\sqrt{3}}\)
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
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b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
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c) \(\dfrac{1+tan^4\alpha}{tan^2\alpha+cot\alpha}=\left(1+tan^4\alpha\right):\left(tan^2\alpha+cot^2\alpha\right)\)
\(=\left(1+\dfrac{sin^4\alpha}{cos^4\alpha}\right):\left(\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}\right)\)
\(=\dfrac{sin^4\alpha+cos^4\alpha}{cos^4\alpha}:\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) (Đpcm).
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Xem thêm câu trả lời
\(cos^225-\sin23+\cos67-\frac{3\tan54}{\cot36}+\cos^265\)\(\sin^4\alpha+\cos^4\alpha+2\sin\alpha.\cos\alpha\)\(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)\(\left(\tan\alpha+\cot\alpha\right)^2-\left(\tan\alpha-\cot\alpha\right)^2\)
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Vẽ tam giác vuông ra, đặt 3 cạnh là a,b,c rồi tính
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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)\(\frac{1-4\sin^2\alpha.\cos^2\alpha}{\left(\sin\alpha-\cos\alpha\right)^2}=\left(\sin\alpha+\cos\alpha\right)^2\)
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. 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 : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
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CMR: \(\frac{\sin^2\alpha}{\cos\alpha\left(1+\tan\alpha\right)}-\frac{\cos^2\alpha}{\sin\alpha\left(1+\cot\alpha\right)}=\sin\alpha-\cos\alpha\)
\(\frac{sin^2\alpha}{cos\alpha.\left(1+\frac{sin\alpha}{cos\alpha}\right)}-\frac{cos^2\alpha}{sin\alpha.\left(1+\frac{cos\alpha}{sin\alpha}\right)}=\frac{sin^2\alpha}{cos\alpha+sin\alpha}-\frac{cos^2\alpha}{sin\alpha+cos\alpha}=\frac{\left(sin\alpha+cos\alpha\right).\left(sin\alpha-cos\alpha\right)}{sin\alpha+cos\alpha}=sin\alpha-cos\alpha\)
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tính \(\frac{2}{\tan\alpha-1}+\frac{\cos\alpha+1}{\cot\alpha-1}\)
b) \(2\left(\sin^6\alpha+\cos^6\alpha\right)-3\left(\sin^4\alpha+\cos^4\alpha\right)\)
Câu a chắc bạn ghi nhầm \(\frac{cota+1}{cota-1}\) thành \(\frac{cosa+1}{cota-1}\)
\(\frac{2}{tana-1}+\frac{cota+1}{cota-1}=\frac{2cota}{1-cota}+\frac{cota+1}{cota-1}=\frac{-2cota+cota+1}{cota-1}=\frac{1-cota}{-\left(1-cota\right)}=-1\)
\(2\left(sin^6x+cos^6x\right)-3\left(sin^4x+cos^4x\right)\)
\(=2\left(sin^2x+cos^2x\right)^3-6sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3\left(sin^2x+cos^2x\right)^2+6sin^2x.cos^2x\)
\(=-1-6sin^2x.cos^2x+6sin^2x.cos^2x=-1\)
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