1.
\(cosa+cos\left(a+\dfrac{2\pi}{3}\right)+cos\left(a-\dfrac{2\pi}{3}\right)\)
\(=cosa+2cosa.cos\left(\dfrac{2\pi}{3}\right)\)
\(=cosa+2.cosa.\left(-\dfrac{1}{2}\right)=cosa-cosa=0\)
b.
\(sin^2a+sin^2\left(a-\dfrac{\pi}{3}\right)-sina.sin\left(a-\dfrac{\pi}{3}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}cos2a+\dfrac{1}{2}-\dfrac{1}{2}cos\left(2a-\dfrac{2\pi}{3}\right)-sina.sin\left(a-\dfrac{\pi}{3}\right)\)
\(=1-\dfrac{1}{2}\left[cos2a+cos\left(2a-\dfrac{2\pi}{3}\right)\right]-sina.sin\left(a-\dfrac{\pi}{3}\right)\)
\(=1-cos\left(2a-\dfrac{\pi}{3}\right).cos\left(\dfrac{\pi}{3}\right)-\dfrac{1}{2}cos\left(\dfrac{\pi}{3}\right)+\dfrac{1}{2}cos\left(2a-\dfrac{\pi}{3}\right)\)
\(=1-\dfrac{1}{2}cos\left(2a-\dfrac{\pi}{3}\right)-\dfrac{1}{4}+\dfrac{1}{2}cos\left(2a-\dfrac{\pi}{3}\right)\)
\(=\dfrac{3}{4}\)
2.
\(sin3a=sin\left(2a+a\right)=sin2a.cosa+cos2a.sina\)
\(=2sina.cosa.cosa+\left(1-2sin^2a\right).sina\)
\(=2sina.cos^2a+sina-2sin^3a\)
\(=2sina\left(1-sin^2a\right)+sina-2sin^3a\)
\(=3sina-4sin^3a\)
b.
\(cos3a=cos\left(2a+a\right)=cos2a.cosa-sin2a.sina\)
\(=\left(2cos^2a-1\right)cosa-2sin^2a.cosa\)
\(=2cos^3a-cosa-2\left(1-cos^2a\right)cosa\)
\(=2cos^3a-cosa-2cosa+2cos^3a\)
\(=4cos^3a-3cosa\)
2c.
\(tanx+cotx=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}\)
\(=\dfrac{1}{sinx.cosx}=\dfrac{2}{2sinx.cosx}=\dfrac{2}{sin2x}\)
2d.
\(sin^4x+cos^4x=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-\dfrac{1}{2}\left(2sinx.cosx\right)^2\)
\(=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1-cos4x}{2}\right)\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}cos4x\)
\(=\dfrac{3}{4}+\dfrac{1}{4}cos4x\)
3.
a.
\(sin2a=2sina.cosa=\dfrac{2sina}{cosa}.cos^2a=2tana\left(\dfrac{cos^2a}{1}\right)\)
\(=2tana\left(\dfrac{cos^2a}{sin^2a+cos^2a}\right)=2tana\left(\dfrac{1}{\dfrac{sin^2a}{cos^2a}+\dfrac{cos^2a}{cos^2a}}\right)=\dfrac{2tana}{tan^2a+1}\)
b.
\(cos2a=cos^2a-sin^2a=\dfrac{cos^2a-sin^2a}{1}=\dfrac{cos^2a-sin^2a}{cos^2a+sin^2a}\)
\(=\dfrac{\dfrac{cos^2a}{cos^2a}-\dfrac{sin^2a}{cos^2}a}{\dfrac{cos^2a}{cos^2a}+\dfrac{sin^2a}{cos^2a}}=\dfrac{1-tan^2a}{1+tan^2a}\)
c.
\(\dfrac{tan2a}{tan4a-tan2a}=\dfrac{tan2a}{\dfrac{sin4a}{cos4a}-\dfrac{sin2a}{cos2a}}=\dfrac{tan2a.cos2a.cos4a}{sin4a.cos2a-cos4a.sin2a}\)
\(=\dfrac{\dfrac{sin2a}{cos2a}.cos2a.cos4a}{sin\left(4a-2a\right)}=\dfrac{sin2a.cos4a}{sin2a}=cos4a\)
4.
\(cos2x.cosx=\dfrac{1}{2}cos3x+\dfrac{1}{2}cosx\)
\(cos3x.sin2x=\dfrac{1}{2}sin5x-\dfrac{1}{2}sinx\)
\(sin4x.cosx=\dfrac{1}{2}sin5x+\dfrac{1}{2}sin3x\)
\(sin3x.sin5x=\dfrac{1}{2}cos2x-\dfrac{1}{2}cos8x\)
5.
\(cosx+cos3x=2cos2x.cosx\)
\(cos4x-cos3x=2cos\dfrac{7x}{2}cos\dfrac{x}{2}\)
\(sin2x+sinx=2sin\dfrac{3x}{2}cos\dfrac{x}{2}\) (hoặc bằng \(sinx\left(2cosx+1\right)\))
\(sin5x-sin3x=2cos4x.sinx\)
5.
\(A=\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}\)
\(=\dfrac{sin3a}{cos3a}=tan3a\)
\(B=\dfrac{sin3a+sina+sin2a}{cos3a+cosa+cos2a}=\dfrac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\dfrac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=tan2a\)
\(C=\dfrac{sin4x+sin2x+2sin3x}{sin5x+sin3x+2sin4x}=\dfrac{2sin3x.cosx+2sin3x}{2sin4x.cosx+2sin4x}\)
\(=\dfrac{2sin3x\left(cosx+1\right)}{2sin4x\left(cosx+1\right)}=\dfrac{sin3x}{sin4x}\)
6.
\(sina.sin\left(\dfrac{\pi}{3}-a\right)sin\left(\dfrac{\pi}{3}+a\right)=\dfrac{1}{2}sina\left[cos2a-cos\left(\dfrac{2\pi}{3}\right)\right]\)
\(=\dfrac{1}{2}sina\left(cos2a+\dfrac{1}{2}\right)=\dfrac{1}{2}sina\left(\dfrac{3}{2}-2sin^2a\right)\)
\(=\dfrac{1}{4}sina\left(3-4sin^2a\right)=\dfrac{1}{4}\left(3sina-4sin^3a\right)=\dfrac{1}{4}sin3a\)
b.
\(sin5a-2sina\left(cos4a+cos2a\right)=sin\left(4a+a\right)-2sina.cos4a-2sina.cos2a\)
\(=sin4a.cosa+cos4a.sina-2sina.cos4a-2sina.cos2a\)
\(=sin4a.cosa-cos4a.sina-2sina.cos2a\)
\(=sin\left(4a-a\right)-\left(sin3a-sina\right)\)
\(=sin3a-sin3a+sina=sina\)
7.
a.
\(sinA+sinB+sinC=2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}+2sin\dfrac{C}{2}cos\dfrac{C}{2}\)
\(=2cos\dfrac{C}{2}cos\dfrac{A-B}{2}+2sin\dfrac{C}{2}cos\dfrac{C}{2}\)
\(=2cos\dfrac{C}{2}\left(cos\dfrac{A-B}{2}+sin\dfrac{C}{2}\right)=2cos\dfrac{C}{2}\left(cos\dfrac{A-B}{2}+cos\dfrac{A+B}{2}\right)\)
\(=4cos\dfrac{C}{2}cos\dfrac{A}{2}cos\dfrac{B}{2}\)
b.
\(cosA+cosB+cosC=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}+1-2sin^2\dfrac{C}{2}\)
\(=2sin\dfrac{C}{2}cos\dfrac{A-B}{2}-2sin^2\dfrac{C}{2}+1=2sin\dfrac{C}{2}\left(cos\dfrac{A-B}{2}-sin\dfrac{C}{2}\right)+1\)
\(=2sin\dfrac{C}{2}\left(cos\dfrac{A-B}{2}-cos\dfrac{A+B}{2}\right)+1\)
\(=4sin\dfrac{C}{2}sin\dfrac{A}{2}sin\dfrac{B}{2}+1\)
