7.
a.
\(8sin^4x=2\left(2sin^2x\right)^2=2\left(1-cos2x\right)^2\)
\(=2\left(1-2cos2x+cos^22x\right)=2-4cos2x+2cos^22x\)
\(=2-4cos2x+1+cos4x\)
\(=3-4cos2x+cos4x\)
b.
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x\)
7c.
\(sin^4x+cos^4x-6sin^2x.cos^2x\)
\(=\left(sin^4x+cos^4x-2sin^2x.cos^2x\right)-\left(2sinx.cosx\right)^2\)
\(=\left(cos^2x-sin^2x\right)^2-sin^22x\)
\(=cos^22x-sin^22x=cos4x\)
d.
\(sin^6x+cos^6x=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=1-3sin^2x.cos^2x=1-\dfrac{3}{4}\left(2sinx.cosx\right)^2\)
\(=1-\dfrac{3}{4}sin^22x=1-\dfrac{3}{4}\left(\dfrac{1-cos4x}{2}\right)\)
\(=\dfrac{5}{8}+\dfrac{3}{8}cos4x\)
8a.
\(sin^6x-cos^6x=\left(sin^2x\right)^3-\left(cos^2x\right)^3\)
\(=\left(sin^2x-cos^2x\right)^3+3sin^2x.cos^2x\left(sin^2x-cos^2x\right)\)
\(=\left(-cos2x\right)^3+\dfrac{3}{4}sin^22x.\left(-cos2x\right)\)
\(=-\dfrac{3cos2x+cos6x}{4}-\dfrac{3}{4}\left(\dfrac{1-cos4x}{2}\right)cos2x\)
\(=-\dfrac{3}{4}cos2x-\dfrac{1}{4}cos6x-\dfrac{3}{8}cos2x+\dfrac{3}{8}cos4x.cos2x\)
\(=-\dfrac{9}{8}cos2x-\dfrac{1}{4}cos6x+\dfrac{3}{16}cos6x+\dfrac{3}{16}cos2x\)
\(=-\dfrac{15}{16}cos2x-\dfrac{1}{16}cos6x\) (đề bài bị ngược dấu)
8b.
\(sin^6\dfrac{x}{2}-cos^6\dfrac{x}{2}=\left(sin^2\dfrac{x}{2}\right)^3-\left(cos^2\dfrac{x}{2}\right)^3\)
\(=\left(sin^2\dfrac{x}{2}-cos^2\dfrac{x}{2}\right)\left[\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-sin^2\dfrac{x}{2}.cos^2\dfrac{x}{2}\right]\)
\(=-cosx.\left[1-\dfrac{1}{4}sin^2x\right]\)
\(=\dfrac{1}{4}cosx\left(sin^2x-4\right)\)
8c.
\(cos^3x.cos3x-sin^3x.sin3x=cos^3x\left(4cos^3x-3cosx\right)-sin^3x\left(3sinx-4sin^3x\right)\)
\(=4cos^6x-3cos^4x-3sin^4x+4sin^6x\)
\(=4\left(cos^6x+sin^6x\right)-3cos^4x-3sin^4x\)
\(=4\left(cos^2x+sin^2x\right)\left(cos^4x+sin^4x-sin^2x.cos^2x\right)-3\left(cos^2x+sin^2x\right)^2+6sin^2x.cos^2x\)
\(=4cos^4x+4sin^4x-3+2sin^2x.cos^2x\)
\(=3\left(cos^4x+sin^4x\right)-3+\left(sin^2x+cos^2x\right)^2\)
\(=3\left(sin^4x+cos^4x\right)-2\)
8d.
\(cos^8x-sin^8x=\left(cos^4x-sin^4x\right)\left(cos^4x+sin^4x\right)\)
\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)\left[\left(cos^2x+sin^2x\right)^2-2sin^2x.cos^2x\right]\)
\(=cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)
\(=cos2x\left[1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)\right]\)
\(=cos2x\left(\dfrac{3}{4}+\dfrac{1}{4}cos4x\right)\)
\(=\dfrac{3}{4}cos2x+\dfrac{1}{4}cos2x.cos4x\)
\(=\dfrac{3}{4}cos2x+\dfrac{1}{8}cos2x+\dfrac{1}{8}cos6x\)
\(=\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\)
9a.
\(\dfrac{1-cos2x}{sin2x}=\dfrac{1-\left(1-2sin^2x\right)}{2sinx.cosx}=\dfrac{2sin^2x}{2sinx.cosx}=\dfrac{sinx}{cosx}=tanx\)
b.
\(tanx+cotx=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{\dfrac{1}{2}sin2x}=\dfrac{2}{sin2x}\)
c.
\(cotx-tanx=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=\dfrac{cos^2x-sin^2x}{sinx.cosx}=\dfrac{cos2x}{\dfrac{1}{2}sin2x}=2cot2x\)
d.
\(\dfrac{cosx}{1-sinx}=\dfrac{cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}{sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}.cos\dfrac{x}{2}}=\dfrac{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)}{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)^2}\)
\(=\dfrac{cos\dfrac{x}{2}+sin\dfrac{x}{2}}{cos\dfrac{x}{2}-sin\dfrac{x}{2}}=\dfrac{\sqrt{2}cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}=\dfrac{cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}\)
\(=cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\)
10.a
\(\dfrac{1+cosx}{1-cosx}.tan^2\dfrac{x}{2}=\dfrac{1+2cos^2\dfrac{x}{2}-1}{1-\left(1-2sin^2\dfrac{x}{2}\right)}.tan^2\dfrac{x}{2}\)
\(=\dfrac{cos^2\dfrac{x}{2}}{sin^2\dfrac{x}{2}}.\dfrac{sin^2\dfrac{x}{2}}{cos^2\dfrac{x}{2}}=1\)
b.
\(\dfrac{1}{sin2x}+cot2x=\dfrac{1}{sin2x}+\dfrac{cos2x}{sin2x}=\dfrac{1+cos2x}{sin2x}\)
\(=\dfrac{1+2cos^2x-1}{2sinx.cosx}=\dfrac{cosx}{sinx}=cotx\)
10c. Đề sai
10d.
\(\left(tan2x-tanx\right)cos2x=\left(\dfrac{sin2x}{cos2x}-\dfrac{sinx}{cosx}\right)cos2x\)
\(=\dfrac{\left(sin2x.cosx-cos2x.sinx\right)cos2x}{cos2x.cosx}\)
\(=\dfrac{sin\left(2x-x\right)}{cosx}=\dfrac{sinx}{cosx}=tanx\)
11a.
\(tan\left(\dfrac{\pi}{4}+x\right)-tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{tan\dfrac{\pi}{4}+tanx}{1-tan\dfrac{\pi}{4}.tanx}-\dfrac{tan\dfrac{\pi}{4}-tanx}{1+tan\dfrac{\pi}{4}.tanx}\)
\(=\dfrac{1+tanx}{1-tanx}-\dfrac{1-tanx}{1+tanx}=\dfrac{\left(1+tanx\right)^2-\left(1-tanx\right)^2}{\left(1-tanx\right)\left(1+tanx\right)}\)
\(=\dfrac{4tanx}{1-tan^2x}=2tan2x\)
11b.
\(cos^3x.sinx-sin^3x.cosx=sinx.cosx\left(cos^2x-sin^2x\right)\)
\(=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)
c.
\(cos3x.sin^3x+sin3x.cos^3x=\dfrac{cos3x\left(3sinx-sin3x\right)}{4}+\dfrac{sin3x\left(3cosx+cos3x\right)}{4}\)
\(=\dfrac{3\left(sinx.cos3x+cosx.sin3x\right)-cos3x.sin3x+sin3x.cos3x}{4}\)
\(=\dfrac{3sin\left(x+3x\right)}{4}=\dfrac{3}{4}sin4x\)
11d.
\(tanx.tan\left(\dfrac{\pi}{3}-x\right)tan\left(\dfrac{\pi}{3}+x\right)\)
\(=tanx.\dfrac{sin\left(\dfrac{\pi}{3}-x\right)sin\left(\dfrac{\pi}{3}+x\right)}{cos\left(\dfrac{\pi}{3}-x\right)cos\left(\dfrac{\pi}{3}+x\right)}\)
\(=tanx.\dfrac{cos2x-cos\dfrac{2\pi}{3}}{cos2x-cos\dfrac{2\pi}{3}}=\dfrac{sinx}{cosx}.\dfrac{2cos2x+1}{2cos2x-1}=\dfrac{2sinx.cos2x+sinx}{2cosx.cos2x-cosx}\)
\(=\dfrac{sin3x-sinx+sinx}{cos3x+cosx-cosx}=\dfrac{sin3x}{cos3x}=tan3x\)
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