Nhờ các bạn giải hộ mình. Khó quá!!! .Tks
1/ Chứng minh biểu thức sau không phụ thuộc vào biến số x.
A = sin2x + sin2\(\left(\frac{2\pi}{3}+x\right)+sin^2\left(\frac{2\pi}{3}-x\right)\)
Chứng minh biểu thức sau không phụ thuộc vào x:
Q = \(sinx-sin\left(x+\dfrac{\pi}{5}\right)+sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)+sin\left(x+\dfrac{4\pi}{5}\right)\)
1. Chứng minh rằng: \(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
2. Chứng minh biểu thức sau không phụ thuộc \(x\): \(A=sin\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\), nếu \(cosx=\frac{1}{2}\) với \(\frac{3\pi}{2}< x< 2\pi\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
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 các biểu thức sau không phụ thuộc x :
a) \(A=\sin\left(\dfrac{\pi}{4}+x\right)-\cos\left(\dfrac{\pi}{4}-x\right)\)
b) \(B=\cos\left(\dfrac{\pi}{6}-x\right)-\sin\left(\dfrac{\pi}{3}+x\right)\)
c) \(C=\sin^2x+\cos\left(\dfrac{\pi}{3}-x\right).\cos\left(\dfrac{\pi}{3}+x\right)\)
d) \(D=\dfrac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x}.\cot x\)
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
CMR: biểu thức sau không phụ thuộc vào x
P=sin4x+sin4\(\left(x+\frac{\pi}{4}\right)+sin^4\left(x+\frac{\pi}{2}\right)+sin^4\left(x+\frac{3\pi}{4}\right)\)
\(P=sin^4x+\left(sin^2\left(x+\frac{\pi}{4}\right)\right)^2+cos^4x+\left(cos^2\left(x+\frac{\pi}{4}\right)\right)^2\)
\(=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{4}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}+\frac{1}{2}sin2x+\frac{1}{4}sin^22x+\frac{1}{4}+\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=1+\frac{1}{2}\left(sin^22x+cos^22x\right)=\frac{3}{2}\)
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 biểu thức sau không phụ thuộc vào x:
P = \(sinxcos\left(2x+\dfrac{\pi}{6}\right)cos\left(2x-\dfrac{\pi}{6}\right)+sin3xsin\left(x+\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)\)
Chứng minh đẳng thức sau :
1) \(sin^2\left(\frac{\pi}{8}+x\right)-sin^2\left(\frac{\pi}{8}-x\right)=\frac{\sqrt{2}}{2}sin2x\)
2) \(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=tanx\)
Chứng minh biểu thức sau không phụ thuộc vào x \(cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+cos\left(x+\frac{\pi}{6}\right)cos\left(x+\frac{3\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{2}-x-\frac{\pi}{6}\right)sin\left(\frac{\pi}{2}-x-\frac{3\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(\frac{\pi}{3}-x\right)sin\left(-x-\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{4}\right)+sin\left(x-\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{4}\right)\)
\(=cos\left(x-\frac{\pi}{3}-x-\frac{\pi}{4}\right)=cos\left(-\frac{7\pi}{12}\right)=cos\frac{7\pi}{12}=\frac{\sqrt{2}-\sqrt{6}}{4}\)