chứng minh rằng : sin\(^4\)x+cos\(^4\)x=1 - 2cos\(^2\)x.sin\(^2\)x
Những câu hỏi liên quan
chứng minh rằng
a) tanx(cot\(^2\)x - 1) = cotx(1 - tan\(^2\)x)
b) tan\(^2\)x - sin\(^2\)x = tan\(^2\)x.sin\(^2\)x
c) \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}\) - cos\(^2\)x = - cos\(^4\)x
a: tan x(cot^2x-1)
\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)
=cotx-tanx/cotx=cotx(1-tan^2x)
b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)
c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)
\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)
=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)
=-cos^2x*cos^2x=-cos^4x
=>ĐPCM
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Cho 00 x 900. Chứng minh các đẳng thức sau:
1. sin6 x +cos6 x 1 - 3sin2 x cos2 x.
2. sin4 x - cos4 x 1 - 2cos2 x.
3. tan2 x - sin2 x tan2 x.sin2x.
4. cot2 x - cos2 x cot2 x.cos2 x.
5.left(sqrt{dfrac{1+sinx}{1-sinx}}-sqrt{dfrac{1-sinx}{1+sinx}}right)^2 4 tan2 x.
6.left(sqrt{dfrac{1+cosx}{1-cosx}}-sqrt{dfrac{1-cosx}{1+cosx}}right)^2 4 cot2 x.
Đọc tiếp
Cho 00 < x < 900. Chứng minh các đẳng thức sau:
1. sin6 x +cos6 x = 1 - 3sin2 x cos2 x.
2. sin4 x - cos4 x = 1 - 2cos2 x.
3. tan2 x - sin2 x = tan2 x.sin2x.
4. cot2 x - cos2 x = cot2 x.cos2 x.
5.\(\left(\sqrt{\dfrac{1+sinx}{1-sinx}}-\sqrt{\dfrac{1-sinx}{1+sinx}}\right)^2\) = 4 tan2 x.
6.\(\left(\sqrt{\dfrac{1+cosx}{1-cosx}}-\sqrt{\dfrac{1-cosx}{1+cosx}}\right)^2\) = 4 cot2 x.
1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)
=1
2: \(sin^4x-cos^4x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)
\(=1-2\cdot cos^2x\)
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chứng minh đẳng thức
sin4x+cos4x=1-2cos2x nhân sin2x
\(\left(\sin^2x+\cos^2x\right)^2=1\)
\(\sin^4x+\cos^4x+2\sin^2x.\cos^2x=1\)
=> dpcm
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Chứng minh các đẳng thức sau :
a) (sin x + cos x)2 = 1 + 2sin x.cos x
b) sin4 x + cos4 x = 1 - 2sin2 x.cos2 x
c) tan2 x - sin2 x = tan2 x.sin2 x
d) sin6 x + cos6 x = 1 - 3sin2 x.cos2 x
e) sin x.cos x (1 + tan x)(1 + cot x) = 1 + 2sin x .cos x
a)
\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)
b)
\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-2\sin ^2x\cos ^2x\)
c)
\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)
\(=\sin ^2x\left(\frac{1}{\cos ^2x}-1\right)=\sin ^2x. \frac{1-\cos ^2x}{\cos ^2x}=\sin ^2x.\frac{\sin ^2x}{\cos ^2x}\)
\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)
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d)
\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)
\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)
\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)
e)
\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)
\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)
\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)
\(=1+2\sin x\cos x\)
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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)
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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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Giải pt :
cos2x.sin4x + cos2x = 2cosx (sin x + cos x) -1
chứng minh rằng
2) \(cos^4x-sin^4x=2cos^{2^{ }}x-1\)
\(cos^4x-sin^4x=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)\)
\(=cos^2x-sin^2x=cos^2x-\left(1-cos^2x\right)\)
\(=2cos^2x-1\)
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Timg GTNN, GTLN của hàm số:
a) y= 4sin2 x + \(\sqrt{2}\) sin (\(2x+\frac{\pi}{4}\))
b) y= cos x ( 1+cos 2x)
c) y= sin2 x. cos x +cos2 x.sin x
11 tháng 3 2022 lúc 9:59
\(\dfrac{sinx+cosx}{sinx}=\dfrac{sinx+cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}{2cos\dfrac{x}{2}sin\dfrac{x}{2}}\)
\(0< x< 90\), chứng minh
\(cosx=cos2.\left(\dfrac{x}{2}\right)=cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\)
\(sinx=sin2\left(\dfrac{x}{2}\right)=2sin\dfrac{x}{2}cos\dfrac{x}{2}\)
\(\Rightarrow\dfrac{sinx+cosx}{sinx}=\dfrac{sinx+cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}{2sin\dfrac{x}{2}cos\dfrac{x}{2}}\)
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