Chứng minh đạo hàm của y không phụ thuộc vào x:
y=\(\dfrac{Sin^6x+C\text{os}^6x-1}{Sin^4x+C\text{os}^4x-1}\)
chứng minh biểu thức sau không phụ thuộc vào x:
\(\dfrac{sin^6x+cos^6x+2}{sin^4x+cos^6x+1}\)
Sửa đề: sin^4x+cos^4x+1
\(A=\dfrac{\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x+2}{\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x+1}\)
\(=\dfrac{3\left(1-sin^2xcos^2x\right)}{2\left(1-sin^2xcos^2x\right)}=\dfrac{3}{2}\)
Chứng minh các biểu thức sau không phụ thuộc vào x:
1, \(A=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
2, \(B=cos^6x+2sin^4x.cos^2x+3sin^2x.cos^4x+sin^4x\)
3, \(C=cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
4, \(D=cos^2x+cos^2\left(x+\dfrac{2\pi}{3}\right)+cos^2\left(\dfrac{2\pi}{3}-x\right)\)
5, \(E=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)-\left(sin^8x+cos^8x\right)\)
6, \(F=cos\left(\pi-x\right)+sin\left(\dfrac{-3\pi}{2}+x\right)-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\dfrac{3\pi}{2}-x\right)\)
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
chứng minh rằng:
a)\(\frac{c\text{os}a.cot\text{a}-sin\text{a}.t\text{ana}}{\frac{1}{sin\text{a}}-\frac{1}{c\text{os}a}}=1+sin\text{a}.c\text{os}a\)
b)\(\frac{c\text{os}a+sin\text{a}-1}{c\text{os}a-sin\text{a}+1}=\frac{sin\text{a}}{1+c\text{os}a}\)
c)\(\frac{sin\text{a}}{1+c\text{os}a}+\frac{1+c\text{os}a}{sin\text{a}}=\frac{2}{sin\text{a}}\)
\(sin^4x+c\text{os}^4x=\frac{3}{4}+\frac{1}{4}c\text{os}4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)
\(=1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{3}{4}+\frac{1}{4}cos4x\)
chứng minh rằng
a)
\(\frac{sin\text{a}}{1+c\text{os}a}+cot\text{a}=\frac{1}{sin\text{a}}\)
b)\(\frac{1}{c\text{os}a}-\frac{c\text{os}a}{1+sin\text{a}}=t\text{ana}\)
c) \(\frac{t\text{ana}-sin\text{a}}{sin^3a}=\frac{1}{c\text{os}a\left(1+c\text{os}a\right)}\)
d) \(\frac{sin\text{a}+c\text{os}a-1}{sin\text{a}-c\text{os}a+1}=\frac{c\text{os}a}{1+sin\text{a}}\)
Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
Mấu chốt trong các bài này là việc sử dụng công thức $\sin ^2a+\cos ^2a=1$
Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
Chứng minh các biểu thức sau không phụ thuộc x:
a) A = \(2\left(sin^6x+cos^6x\right)-3\left(sin^4x+cos^4x\right)\)
b) \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}\)
c) C = \(2cos^4x-sin^4x+sin^2x.cos^2x+3sin^2x\)
Giả sử các biểu thức đều có nghĩa
\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)
\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)
\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)
\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)
b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)
\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)
c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)
\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)
Chứng minh rằng các hàm số sau có đạo hàm không phụ thuộc vào x :
a) \(y=\sin^6x+\cos^6x+3\sin^2x.\cos^2x\)
b) \(y=\cos^2\left(\dfrac{\pi}{3}-x\right)+\cos^2\left(\dfrac{\pi}{3}+x\right)+\cos^2\left(\dfrac{2\pi}{3x}-x\right)+\cos^2\left(\dfrac{2\pi}{3x}+x\right)-2\sin^2x\)
a) Cách 1: Ta có:
y' = 6sin5x.cosx - 6cos5x.sinx + 6sinx.cos3x - 6sin3x.cosx = 6sin3x.cosx(sin2x - 1) + 6sinx.cos3x(1 - cos2x) = - 6sin3x.cos3x + 6sin3x.cos3x = 0.
Vậy y' = 0 với mọi x, tức là y' không phụ thuộc vào x.
Cách 2:
y = sin6x + cos6x + 3sin2x.cos2x(sin2x + cos2x) = sin6x + 3sin4x.cos2x + 3sin2x.cos4x + cos6x = (sin2x + cos2x)3 = 1
Do đó, y' = 0.
b) Cách 1:
Áp dụng công thức tính đạo hàm của hàm số hợp
(cos2u)' = 2cosu(-sinu).u' = -u'.sin2u
Ta được
y' =[sin - sin] + [sin - sin] - 2sin2x = 2cos.sin(-2x) + 2cos.sin(-2x) - 2sin2x = sin2x + sin2x - 2sin2x = 0,
vì cos = cos = .
Vậy y' = 0 với mọi x, do đó y' không phụ thuộc vào x.
Cách 2: vì côsin của hai cung bù nhau thì đối nhau cho nên
cos2 = cos2 '
cos2 = cos2 .
Do đó
y = 2 cos2 + 2cos2 - 2sin2x = 1 +cos + 1 +cos - (1 - cos2x) = 1 +cos + cos + cos2x = 1 + 2cos.cos(-2x) + cos2x = 1 + 2cos2x + cos2x = 1.
Do đó y' = 0.
Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=2\left(cos^6x+sin^6x\right)-3\left(cos^4x+sin^4x\right)\)
b) \(B=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-sin^8x-cos^8x\)
c) \(C=\dfrac{sin^2x}{1+cotgx}+\dfrac{cos^2x}{1+tgx}+sinx.cosx\)
d) \(D=\dfrac{cotg^2a-cos^2x}{cotg^2x}+\dfrac{sinx.cosx}{cotgx}\)
e) \(E=3\left(sin^8x-cos^8x\right)+4\left(cos^6x-2sin^6x\right)+6sin^4x\)
f) \(F=\dfrac{tg^2x}{sin^2x.cos^2x}-\left(1+tg^2x\right)^2\)
chứng minh biểu thức không phụ thuộc vào x:
\(3\left(sin^8x-cos^8x\right)+4\left(cos^6x-2sin^6x\right)+6sin^4x\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-8sin^6x+6sin^4x\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6sin^4x\left(1-sin^2x\right)\)
\(=sin^6x+3sin^4x.cos^2x+3cos^2x.sin^4x+cos^6x\)
\(=\left(sin^2x+cos^2x\right)^3=1\)