§3. Công thức lượng giác

b: \(=2\cdot cos\left(\dfrac{pi}{5}+\dfrac{4}{5}pi\right):2\cdot cos\left(\dfrac{4}{5}pi-\dfrac{pi}{5}\right):2+2\cdot cos\left(\dfrac{2}{5}pi+\dfrac{3}{5}pi\right):2\cdot cos\left(\dfrac{2}{5}pi-\dfrac{3}{5}pi\right):2\)

\(=2\cdot cos\left(\dfrac{pi}{2}\right)\cdot cos\left(\dfrac{3}{10}pi\right)+2\cdot cos\left(\dfrac{pi}{2}\right)\cdot cos\left(-\dfrac{1}{10}pi\right)\)

=0

c: \(=2\cdot cos\left(\dfrac{6}{7}pi+\dfrac{2}{7}pi\right):2\cdot cos\left(\dfrac{6}{7}pi-\dfrac{2}{7}pi\right):2+cos\left(\dfrac{4}{7}pi\right)\)

\(=cos\left(\dfrac{4}{7}pi\right)\left(2\cdot cos\left(\dfrac{2}{7}pi\right)+1\right)\)

Ta có : \(D=cos^2\left(x+y\right)+cos^2\left(x-y\right)-cos2x.cos2y\)

\(=\left[cosx.cosy-sinx.siny\right]^2+\left[cosx.cosy+sinx.siny\right]^2-cos2x.cos2y\)

\(=2\left[cos^2x.cos^2y+sin^2x.sin^2y\right]-\left(cos^2x-sin^2x\right)\left(cos^2y-sin^2y\right)\)

\(=cos^2x.cos^2y+sin^2x.sin^2y+sin^2x.cos^2y+cos^2x.sin^2y\)

\(=\left(cos^2x+sin^2x\right)\left(cos^2y+sin^2y\right)=1\)

1: sin^4x+cos^4x

=(sin^2x+cos^2x)^2-2*sin^2x*cos^2x

=1-1/2*sin^22x

=1/2+1/2(1-sin^2(2x))

=1/2(1+cos^2(2x))

=1/2(1+(1+cos4x)/2)

=3/4+1/4*cos4x

2: cos4x=cos^2(2x)-sin^2(2x)

=(2cos^2x-1)^2-4*sin^2x*cos^2x

=4cos^4x-4cos^2x(1+sin^2x)+1

=4cos^4x-4cos^2x(2-cos^2x)+1

=4cos^4x-8cos^2x+4cos^4x+1

=8cos^4x-8cos^2x+1

b: cot a-tan a

=cosa/sina-sina/cosa

\(=\dfrac{cos^2a-sin^2a}{sina\cdot cosa}=cos2a:\dfrac{sin2a}{2}=2\cdot\dfrac{cos2a}{sin2a}=2\cdot cot2a\)

c: \(VT=tan\left(\dfrac{x}{2}\right)\cdot\left(\dfrac{1}{cosx}+1\right)\)

\(=tan\left(\dfrac{x}{2}\right)\cdot\dfrac{1+cosx}{sinx}\)

\(=\dfrac{sin\left(\dfrac{x}{2}\right)}{cos\left(\dfrac{x}{2}\right)}\cdot\dfrac{2\cdot cos^2\left(\dfrac{x}{2}\right)}{cosx}=\dfrac{2\cdot sin\left(\dfrac{x}{2}\right)\cdot cos\left(\dfrac{x}{2}\right)}{cosx}\)

=sinx/cosx=tanx

8: \(B=-\dfrac{1}{2}\cdot\left[cos72-cos60\right]\cdot\left(-\dfrac{1}{2}\right)\cdot\left[cos120-cos36\right]\)

\(=\dfrac{1}{4}\cdot\left[cos72-\dfrac{1}{2}\right]\left[-\dfrac{1}{2}-cos36\right]\)

\(=\dfrac{1}{4}\cdot\left[-\dfrac{1}{2}\cdot cos72-cos72\cdot cos36+\dfrac{1}{4}+\dfrac{1}{2}\cdot cos36\right]\)

\(=\dfrac{1}{4}\left[\dfrac{1}{4}-\dfrac{1}{2}\cdot\left(2cos^236^0-1\right)-\left(2cos^236^0-1\right)\cdot cos36^0+\dfrac{1}{2}\cdot cos36\right]\)

=1/4*1/4=1/16

10: \(\Leftrightarrow B_{10}\cdot sin20=sin20\cdot cos20\cdot cos40\cdot cos60\cdot cos80\)

\(=\dfrac{1}{2}\cdot sin40\cdot cos40\cdot cos80\cdot\dfrac{1}{2}\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot2\cdot sin40\cdot cos40\cdot cos80\cdot\dfrac{1}{2}\)

\(=\dfrac{1}{8}\cdot sin80\cdot cos80\)

=1/16*sin160

=>B10=1/16

\(P=\left(sin^2x-1\right)tan^2x+\left(cos^2x-1\right)cot^2x\)

\(=-cos^2x.\dfrac{sin^2x}{cos^2x}+\left(-sin^2x\right)\dfrac{cos^2x}{sin^2x}\)

\(=-sin^2x-cos^2x\)

\(=-\left(sin^2x+cos^2x\right)\)

\(=-1\)

\(A=\dfrac{cos^2x-sin^2x}{cosx-sinx}-\dfrac{2cos^2x+2sinx.cosx-2\sqrt{2}\left(sin\left(\dfrac{\pi}{4}\right)cosx+cos\left(\dfrac{\pi}{4}\right)sinx\right)}{2\left(cosx-1\right)}\)

\(=\dfrac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{cosx-sinx}-\dfrac{2cosx\left(sinx+cosx\right)-2\left(sinx+cosx\right)}{2\left(cosx-1\right)}\)

\(=sinx+cosx-\dfrac{2\left(sinx+cosx\right)\left(cosx-1\right)}{2\left(cosx-1\right)}\)

\(=sinx+cosx-\left(sinx+cosx\right)=0\)

\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)

\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)

\(\Rightarrow sin2a=2sina.cosa=2.\left(-\dfrac{4}{5}\right).\left(\dfrac{3}{5}\right)=-\dfrac{24}{25}\)

Câu sau có nhầm đề ko nhỉ?

\(sin\left(\pi-\dfrac{\pi}{3}\right)=sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)