Rút gọn biểu thức
A= cos 2x . Cos 4x . Cos 8x . Cos 16x
1) Rút gọn biểu thức :
\(M=2\left(sin^4x+cos^4x+cos^2.sin^2x\right)^2-\left(sin^8x+cos^8x\right)\)
\(\left(sin^4x+cos^4x+cos^2x.sin^2x\right)^2-sin^8x\)
\(=\left(sin^4x+cos^2x\left(cos^2x+sin^2x\right)\right)^2-sin^8x\)
\(=\left(sin^4x+cos^2x\right)^2-sin^8x=\left(sin^4x+cos^2x-sin^4x\right)\left(sin^4x+cos^2x+sin^4x\right)\)
\(=cos^2x\left(2sin^4x+cos^2x\right)=2sin^4x.cos^2x+cos^4x\)
Tương tự: \(\left(sin^4x+cos^4x+sin^2xcos^2x\right)^2-cos^8x\)
\(=\left(cos^4x+sin^2x\left(sin^2x+cos^2x\right)\right)^2-cos^8x\)
\(=\left(cos^4x+sin^2x\right)^2-cos^8x\)
\(=\left(cos^4x+sin^2x-cos^4x\right)\left(cos^4x+sin^2x+cos^4x\right)\)
\(=sin^2x\left(2cos^4x+sin^2x\right)=2sin^2x.cos^4x+sin^4x\)
\(\Rightarrow M=2sin^2x.cos^4x+2sin^2x.cos^2x+sin^2x+cos^4x\)
\(M=2sin^2x.cos^2x\left(cos^2x+sin^2x\right)+sin^4x+cos^4x\)
\(M=2sin^2x.cos^2x+sin^4x+cos^4x\)
\(M=\left(sin^2x+cos^2x\right)^2=1\)
Rút gọn các biểu thức sau
1, \(\dfrac{1+\cot x}{1-\cot x}-\dfrac{2+2\cot^2x}{\left(\tan x-1\right)\left(\tan^2x+1\right)}\)
2, \(\sqrt{\sin^4x+6\cos^2x+3\cos^4x}+\sqrt{\cos^4x+6\sin^2x+3\sin^4x}\)
Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý
Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)
Câu 1 đề vẫn có vấn đề:
\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)
\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)
\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)
Có thể coi như ko thể rút gọn tiếp
2.
\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)
\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)
\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)
\(=2\left(cos^2x+sin^2x\right)+2=4\)
rút gọn
\(\dfrac{\sin^2x-\cos^2x+\cos^4x}{\cos^2x-\sin^2x+\sin^4x}\)
\(A=\dfrac{sin^2x-cos^2x.\left(1-cos^2x\right)}{cos^2x-sin^2x.\left(1-sin^2x\right)}=\dfrac{sin^2x-cos^2x.sin^2x}{cos^2x-sin^2x.cos^2x}\\ =\dfrac{sin^2x.\left(1-cos^2x\right)}{cos^2x.\left(1-sin^2x\right)}=\dfrac{sin^2x.sin^2x}{cos^2x.cos^2x}=\dfrac{sin^4x}{cos^4x}.\)
rút gọn a= sin^4x+cos^4x+2sin^2x+cos^2x
`A=sin^4x+cos^4x+2sin^2x+cos^2x`
`=(sin^2x+cos^2x)^2-2sin^2xcos^2x+sin^2x+(sin^2x+cos^2x)`
`=1-1/2 sin^2 2x + sin^2 x+1`
`=2-1/2 sin^2 2x + sin^2x`
Rút gọn biểu thức A= sin x + sin 2 x + sin 3 x cos x + cos 2 x + cos 3 x
A. tan4x
B. tan 3x
C. tan 2x
D. tan x + tan 2x
Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=\cos^4x-\sin^4x+2\sin^2x+\tan2x.\cot2x\)
b) \(B=\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
c) \(C=3\left(\sin^8x-\cos^8x\right)+4\left(\cos^6x-2\sin^6x\right)+6\sin^4x\)
d) \(D=2\left(\sin^4x+\cos^4x+\sin^2x.\cos^2x\right)-\left(\sin^8x+\cos^8x\right)\)
chứng minh biểu thức ko phụ thuộc vào x
A= \(\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
B= \(3\left(\sin^8x-\cos^8x\right)+4\left(\cos^6x-2\sin^6x\right)+6\sin^4x\)
\(A=\sqrt{\left(1-cos^2x\right)^2+4cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4sin^2x}\)
\(=\sqrt{cos^4x+2cos^2x+1}+\sqrt{sin^4x+2sin^2x+1}\)
\(=\sqrt{\left(cos^2x+1\right)^2}+\sqrt{\left(sin^2x+1\right)^2}\)
\(=sin^2x+cos^2x+2=3\)
b/
\(3\left(sin^8x-cos^8x\right)=3\left(sin^4x+cos^4x\right)\left(sin^4x-cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)\)
\(=3sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x-3cos^6x\)
\(\Rightarrow B=-5sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x+cos^6x+6sin^4x\)
\(=-5sin^6x-3sin^4x\left(1-sin^2x\right)+3cos^4x\left(1-cos^2x\right)+cos^6x+6sin^4x\)
\(=-2sin^6x-2cos^6x+3sin^4x+3cos^4x\)
\(=-2\left(1-3sin^2x.cos^2x\right)+3\left(1-2sin^2x.cos^2x\right)\)
\(=-2+3=1\)
Rút gọn :
a) \(\cos\dfrac{x}{5}\cos\dfrac{2x}{5}\cos\dfrac{4x}{5}\cos\dfrac{8x}{5}\)
b) \(\sin\dfrac{x}{7}+2\sin\dfrac{3x}{7}+\sin\dfrac{5x}{7}\)
a)\(\eqalign{ & A\sin {x \over 5} = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} = {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over {16}}\sin {{16x} \over 5} \cr} \)
Suy ra biểu thức rút gọn \(A =\sin{{16x} \over 5}:16\sin {x \over 5}\)
b)\(\eqalign{ & B = \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7} = 2\sin {{3x} \over 7} + (\sin {x \over 7} + \sin {{5x} \over 7}) \cr & = 2\sin {{3x} \over 7} + 2\sin {1 \over 2}({{5x} \over 7} + {x \over 7})cos{1 \over 2}({{5x} \over 7} - {x \over 7}) \cr & = 2\sin {{3x} \over 7}(1 + \cos {{2x} \over 7}) = 4\sin {{3x} \over 7}{\cos ^2}{x \over 7} \cr}\)
Rút gọn \(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{\left(1-cos^2x\right)^2+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{1-cos^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{sin^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=2\sqrt{sin^2x+cos^4x}\)