rrút gọn E=cos^2-sin^2/sin^4+cos^4-xin^2 -1
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rút gọn biểu thức sau:
B=\(\dfrac{1-4\sin^2x.\cos^2x}{\left(\sin x+\cos x\right)^2}+2\sin x.\cos x\) , với 0 độ<x<90 độ
\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)
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Rút gọn
M=2(sin6a+cos6a).3(sin4a+cos4a)
N=Sin8a+(sin6a+sin4a+sin2a+1) cos2a
Chứng minh các đẳng thức sau :
a) 1 - cos x/ sin x = sin x/ 1 + cos x
b) ( sin x + cos x - 1 )( sin x + cos x + 1) = 2sin x cos x
c) sin2 x + 2cos x - 1/ 2 + cos x - cos2 x = cos x/ 1 + cos x
d) cos2 x - sin2 x/ cot2 x - tan2x = sin2 x cos2 x
e) 1 - cot4 x = 2/ sin2 x - 1/ sin4x
Lời giải:
a)
\(\frac{1-\cos x}{\sin x}=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}=\frac{1-\cos ^2x}{\sin x(1+\cos x)}=\frac{\sin ^2x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}\)
b)
\((\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x)^2-1^2\)
\(=\sin ^2x+\cos ^2x+2\sin x\cos x-1=1+2\sin x\cos x-1=2\sin x\cos x\)
c)
\(\frac{\sin ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{1-\cos ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{-\cos ^2x+2\cos x}{2+\cos x-\cos ^2x}\)
\(=\frac{\cos x(2-\cos x)}{(2-\cos x)(\cos x+1)}=\frac{\cos x}{\cos x+1}\)
d)
\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)
\(=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}=\frac{\sin ^2x\cos ^2x}{\sin ^2x+\cos ^2x}=\sin ^2x\cos ^2x\)
e)
\(1-\cot ^4x=1-\frac{\cos ^4x}{\sin ^4x}=\frac{\sin ^4x-\cos ^4x}{\sin ^4x}=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin ^4x}\)
\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)
Ta có ddpcm.
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Bài 8: Rút gọn các biểu thức sau:
a/ (1-cos α) . (1+cos α)
b/ 1+sin2 α + cos2 α
c/ sin α - sin α cos2 α
d/ sin4 α + cos4 α + 2sin2 α cos2 α
e/ tan2 α - sin2 α tan2 α
f/ cos2 α + tan2 α cos2 α
giúp mk giải bài này ik mn ơiiiii
tính :
\(E=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cdot\cos^2\alpha\)
\(F=3\sin^3\alpha+\cos^3\alpha-2\sin^6\alpha+\cos^6\alpha\)
\(G=\sqrt{\sin^4\alpha+4\cos^2\alpha}+\sqrt{\cos^4\alpha+4\sin^2\alpha}\)
E = sin^6 + cos^6 + 3sin^2.cos^2
= (sin^2 + cos^2)(sin^4 - sin^2.cos^2 + cos^4) + 3 sin^2.cos^2
= (sin^2 + cos^2)^2 - 3sin^2.cos^2 + 3sin^2.cos^2
= 1
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Rút gọn:
B = sin4 α + cos4 α + 2sin2 α*cos2 α
D = sin2 α*sin2 ß + sin2 α*cos2 ß + cos2 α
hì, hồi nãy câu E cho mk/e sửa đề lại nhé
E = cos6 α + sin6 α + 3 sin2 α* cos2 α
(α: an pha, ß: bê ta)
A = \(\left(sin^2a+cos^2a\right)^2=1^2=1\)
D = \(sin^2\left(sin^2B+cos^2B\right)+cos^2a=sin^2a+cos^2a=1\)
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rút gọn:
1, 1-sin2α
2, (1+cos α)(1-cos α)
3, 1+sin2α+cos2α
4,sin α-sin α.cos2α
5, sin4α+cos4α+2.sin2α.cos2α
6,tan2α-sin2α.tan2α
7, cos2α+tan2α.cos2α
8, tan2α.(2.cos2α+sin2α-1)
\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)
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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 = \(\dfrac{1+\sin\alpha-2\sin^2\left(45^o-\dfrac{\pi}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)
Rút gọn các biểu thức :
a) \(\dfrac{\sin2\alpha+\sin\alpha}{1+\cos2\alpha+\cos\alpha}\)
b) \(\dfrac{4\sin^2\alpha}{1-\cos^2\dfrac{\alpha}{2}}\)
c) \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)
d) \(\dfrac{1+\sin\alpha-2\sin^2\left(45^0-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)
a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
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a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
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b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).
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