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1/ \(3-4\sin^2=4\cos^2x-1\Leftrightarrow4\left(\sin^2x+\cos^2x\right)-4=0\Leftrightarrow4.1-4=0\left(ld\right)\Rightarrow dpcm\)
2/ \(\cos^4x-\sin^4x=\left(\cos^2x+\sin^2x\right)\left(\cos^2x-\sin^2x\right)=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x\)
3/ \(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x=1-2\sin^2x.\cos^2x\)
\(sin\dfrac{3x}{2}\left(cosx+cos4x+cos7x\right)\)
\(=\)\(sin\dfrac{3x}{2}.cosx+sin\dfrac{3x}{2}.cos4x+sin\dfrac{3x}{2}.cos7x=\dfrac{1}{2}\left[sin\dfrac{x}{2}+sin\dfrac{5x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{5x}{2}\right)+sin\dfrac{11x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{11x}{2}\right)+sin\dfrac{17x}{2}\right]\)
\(=\dfrac{1}{2}\left(sin\dfrac{x}{2}+sin\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.2.sin\dfrac{9x}{2}.cos4x=sin\dfrac{9x}{2}.cos4x\)
\(sin\dfrac{3x}{2}\left(sinx+sin4x+sin7x\right)\)
\(=sin\dfrac{3x}{2}.sinx+sin\dfrac{3x}{2}.sin4x+sin\dfrac{3x}{2}.sin7x\)\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{5x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-5x}{2}-cos\dfrac{11x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-11x}{2}-cos\dfrac{17x}{2}\right)\)
\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.-2.sin\dfrac{9x}{2}.sin\left(-4x\right)=sin\dfrac{9x}{2}.sin4x\)
Có \(\dfrac{cos7x+cos4x+cosx}{sin7x+sin4x+sinx}\)
\(=\dfrac{sin\dfrac{3x}{2}\left(cos7x+cos4x+cosx\right)}{sin\dfrac{3x}{2}\left(sin7x+sin4x+sinx\right)}\)\(=\dfrac{sin\dfrac{9x}{2}.cos4x}{sin\dfrac{9x}{2}.sin4x}\)\(=\dfrac{cos4x}{sin4x}\)
\(\Rightarrow\dfrac{cos4x}{sin4x}=\dfrac{1}{2}\)
\(\Leftrightarrow2cos4x=sin4x\)
\(\Leftrightarrow4.cos^24x=sin^24x\)
\(\Leftrightarrow4.cos^24x=1-cos^24x\)\(\Leftrightarrow cos^24x=\dfrac{1}{5}\Leftrightarrow\dfrac{1+cos8x}{2}=\dfrac{1}{5}\)
\(\Leftrightarrow cos8x=-\dfrac{3}{5}\)
Vậy..
\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)
b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)
c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)
d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)
\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)
e/
\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)
Chọn C.
Ta có A = sin4x + cos4x - ¼cos 4x = ( sin2x + cos2x) 2 - 2sin2x.cos2x - ¼cos 4x
= 1 - ½sin22x - ¼cos4x = 1 - ¼(1 – cos4x) - ¼cos4x = 3/4
\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)
\(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=\frac{2sin2x}{sin2x}=2\)
\(\frac{cosx+cos3x+cos2x+cos4x}{sinx+sin3x+sin2x+sin4x}=\frac{2cosx.cos2x+2cosx.cos3x}{2sin2x.cosx+2sin3x.cosx}=\frac{2cosx\left(cos2x+cos3x\right)}{2cosx\left(sin2x+sin3x\right)}\)
\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)
\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)
\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)
\(=\frac{\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)}{\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)}=tan\left(2x+\frac{\pi}{4}\right)\)
\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)
\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)
\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)
\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)
\(=-sin10x-sin6x=-2sin8x.cos2x\)
Đáp án: D
Ta có: sin4x = sin2(2x) = 2sin2xcos2x