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\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
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.
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 \(6cos^2x\)
\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)
\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)
\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)
\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)
\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)
\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)
\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)
\(=-sin10x-sin6x=-2sin8x.cos2x\)
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.
\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)
\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)
\(cos5x.cos3x+sin7x.sinx=\frac{1}{2}cos8x+\frac{1}{2}cos2x-\frac{1}{2}cos8x+\frac{1}{2}cos6x\)
\(=\frac{1}{2}\left(cos6x+cos2x\right)=cos4x.cos2x\)
\(\frac{1-2sin^22x}{1-sin4x}=\frac{cos^22x-sin^22x}{cos^22x+sin^22x-2sin2x.cos2x}\)
\(=\frac{\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)}{\left(cos2x-sin2x\right)^2}=\frac{cos2x+sin2x}{cos2x-sin2x}=\frac{\frac{cos2x}{cos2x}+\frac{sin2x}{cos2x}}{\frac{cos2x}{cos2x}-\frac{sin2x}{cos2x}}=\frac{1+tan2x}{1-tan2x}\)
\(2cosx-3cos\left(\pi-x\right)+5sin\left(4\pi-\frac{\pi}{2}-x\right)+cot\left(\pi+\frac{\pi}{2}-x\right)\)
\(=2cosx+3cosx-5sin\left(\frac{\pi}{2}+x\right)+cot\left(\frac{\pi}{2}-x\right)\)
\(=5cosx-5cosx+tanx=tanx\)
\(A=\frac{1}{2}\left(\frac{sin^2x}{cos^2x}-1\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)
\(A=\frac{-1}{2}\left(\frac{cos^2x-sin^2x}{cos^2x}\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)
\(A=\frac{-cos2x}{2cosx.sinx}+cos4x.cot2x+sin4x\)
\(A=-cot2x+cos4x.cot2x+sin4x\)
\(A=cot2x\left(cos4x-1\right)+sin4x\)
\(A=\frac{cos2x}{sin2x}.\left(1-2sin^22x-1\right)+sin4x\)
\(A=\frac{-2cos2x.sin^22x}{sin2x}+sin4x\)
\(A=-sin4x+sin4x=0\)