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Chọn C.
Ta có
C = [ ( sin2x + cos2x) – sin2cos2x]2 - [ ( sin4x + cos4x) 2 - 2sin4x.cos4x]
= 2[ 1-sin2x.cos2x]2 - [ ( sin2x + cos2x) 2 - 2sin2x.cos2x]2 + 2sin4x.cos4x
= 2[ 1-sin2x.cos2x]2 - [1-sin2x.cos2x]2 + 2sin4x.cos4x
= 2( 1 - 2sin2x.cos2x + sin4x.cos4x)- ( 1 - 4sin2xcos2x + 4sin4x.cos4x) + 2sin4x.cos4x
= 1.
không phụ thuộc vào x.
\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
M = 3(sin^8x-cos^8x) + 4(cos^6x-2sin^6x)+6sin^4x
Ta có:
sin^8(x) - cos^8(x) = [sin^4(x) ]² - [cos^4(x)]²
= (sin²x + cos²x)(sin²x -cos²x).[ sin^4(x) + cos^4(x) ]
= (sin²x -cos²x)[ sin^4(x) + cos^4(x) ]
= sin^6(x) - cos^6(x) + sin²x.cos^4(x) -cos²x.sin^4(x)
Lúc đó M viết lại là:
M = 3.[sin^6(x) - cos^6(x) + sin²x.cos^4(x) -cos²x.sin^4(x) ] + 4.[ cos^6(x) -2sin^6(x) ] + 6sin^4(x)
M = -5sin^6(x) + cos^6(x) -3sin^4(x).cos²x + 3sin²x.cos^4(x) +6sin^4(x)
M = -3sin^(6)x - 3cos²x.sin^4(x) + cos^4(x).sin²x + cos^6(x) - 2sin^6(x) + 2sin²x.cos^4(x) + 6sin^4(x)
M = -3sin^4(x).(sin²x + cos²x ) + cos^4(x).[sin²x + cos²x ] -2sin²x.[sin^4(x) - cos^4(x) ] + 6sin^4(x)
M = 3sin^4(x) + cos^4(x) -2sin²x.[sin²x - cos²x]
M = 3sin^4(x) + cos^4(x) -2sin^4(x) + 2sin²x.cos²x
M = sin^4(x) + 2sin²x.cos²x + cos^4(x)
M = [sin²x + cos²x ]² = 1
Chọn A.
Ta có:
+ sin4x + cos4x = (sin2x + cos2x)2 - 2sin2x.cos2x = 1 - 2sin2x.cos2x.
+ sin4x + cos4x = 1 - 3sin2x.cos2x.
Do đó
A = 3(1 - 2sin2x.cos2x) - 2(1 - 3sin2x.cos2x) = 1.
\(A=\frac{cos7x-cos8x-cos9x+cos10x}{sin7x-sin8x-sin9x+sin10x}=\frac{(cos10x+cos7x)-\left(cos9x+cos8x\right)}{\left(sin10x+sin7x\right)-\left(sin9x+sin8x\right)}.\)
\(=\frac{2cos\frac{17x}{2}cos\frac{3x}{2}-2cos\frac{17x}{2}cos\frac{x}{2}}{2sin\frac{17x}{2}cos\frac{3x}{2}-2sin\frac{17x}{2}cos\frac{x}{2}}=\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}=cotan\frac{17x}{2}.\)
A=\(\frac{\left(cos7x+cos10x\right)-\left(cos8x+cos9x\right)}{\left(sin7x+sin10x\right)-\left(sin8x+sin9x\right)}\) =\(\frac{2cos\frac{17x}{2}.cos\frac{3x}{2}-2cos\frac{17x}{2}.cos\frac{x}{2}}{2sin\frac{17x}{2}.cos\frac{3x}{2}-2sin\frac{17x}{2}.cos\frac{x}{2}}\)
=\(\frac{2cos\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}{2sin\frac{17x}{2}\left(cos\frac{3x}{2}-cos\frac{x}{2}\right)}\)=\(\frac{cos\frac{17x}{2}}{sin\frac{17x}{2}}\)=cotg\(\frac{17x}{2}\)
\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt
Chọn C.
Ta có: C = 2( sin4x + cos4x + sin2x.cos2x) 2 - ( sin8x + cos8x)
= 2 [ (sin2x + cos2x) 2 - sin2x.cos2x]2 - [ (sin4x + cos4x)2 - 2sin4x.cos4x]
= 2[ 1 - sin2x.cos2x]2 - [ (sin2x+ cos2x) 2 - 2sin2x.cos2x]2 + 2sin4x.cos4x
= 2[ 1- sin2x.cos2x]2 - [ 1 - 2sin2x.cos2x]2 + 2sin4x.cos4x
= 2( 1 - 2sin2xcos2x+ sin4x.cos4x) –( 1- 4sin2xcos2x+ 4sin4xcos4x) + 2sin4x.cos4x
= 1.