• 1) известно что:
    2) упростите выражения 
    3) докажите тождество

    question img

Ответы 2

  • sin(\frac{\pi}{2}+a)=-\frac{1}{2}\\
sin(30+a)\\
\\
sin(\frac{\pi}{2}+a)=cosa\\
cosa=-\frac{1}{2}\\
a=-\frac{2\pi}{3}+2\pi*n=\frac{4pi}{3}\\\\
sin(30+240)=sin(270)=-1\\\\
2cos^2a-(tga*cosa)^2-(ctga*sina)^2=\\
2cos^2a-(\frac{sina}{cosa}*cosa)^2-(\frac{cosa}{sina}*sina)^2=\\
2cos^2a-sin^2a-cos^2a=\\
cos^2a-sin^2a=cos2a\\
\\\frac{sina+sin3a}{cosa+cos3a}*(1+cos4a)=\\
\frac{2sin2a*cosa}{2cos2a*cosa}*(1+cos4a)=\\
\frac{sin2a}{cos2a}*(1+cos(2*2a))=\\
\frac{sin2a}{cos2a}*(2cos^2(2a))=\\
2sin2a*cos2a=sin4a\\
\\
ctg2a*\frac{2tga}{1+tg^2a}=\\
ctg2a*sin2a=\\
\frac{cos2a}{sin2a}*sin2a=cos2a
  • 1)\sin(\frac{\pi}{2}+\alpha)=-\frac{1}{2};  \\
\pi<\alpha<\frac{3\pi}{2};\ \alpha\in(\pi;\frac{2\pi}{3});\\
\sin(30^0+\alpha)-?;\\
\frac{\pi}{2}+\alpha=(-1)^n\arcsin(-\frac{1}{2})+\pi n;\ n\in Z\\
\alpha=-(-1)^n\arcsin(\frac{1}{2})-\frac{\pi}{2}+\pi n;\\
\alpha=(-1)^{n+1}\cdot\frac{\pi}{6}-\frac{\pi}{2}+\pi n;\\
n=2: \ \alpha=-\frac{\pi}{6}-\frac{\pi}{2}+2\pi=\frac{-\pi-3\pi+12\pi}{6}=\frac{8\pi}{6}=\frac{4\pi}{3}\in(\pi;\frac{2\pi}{3});\\\sin(30^0+\alpha)=\sin(\frac{\pi}{6}+\frac{4\pi}{3})=\sin(\frac{\pi+8\pi}{6})=\\
=\sin\frac{9\pi}{6}=\sin{\frac{3\pi}{2}}-1;\\
or: \sin\frac{\pi}{6}\cos\frac{4\pi}{3}+\sin{\frac{4\pi}{3}}\cos\frac{\pi}{6}=\\
=\frac{1}{2}\cdot(-\frac{1}{2})+(-\frac{\sqrt{3}}{2})\cdot\frac{\sqrt{3}}{2}=-\frac{1}{4}-\frac{3}{4}=-\frac{4}{4}=-1\sin(30^0+\alpha)=-1;\\2)a)2\cos^2\alpha-(tg\alpha\cdot\cos\alpha)^2-(ctg\alpha\cdot\sin\alpha)^2=\\
=2\cos^2\alpha-\left(\left(\frac{\sin\alpha}{\cos\alpha}\cdot\cos\alphaight)^2+\left(\frac{\cos\alpha}{\sin\alpha}\cdot sin\alphaight)^2ight)=\\
=2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alphaight)=2\cos^2\alpha-1=\cos2\alpha;\\
б)\frac{\sin\alpha+\sin3\alpha}{\cos\alpha+\cos3\alpha}\cdot(1+\cos4\alpha)=\\
=\frac{\sin\alpha+\sin(\alpha+2\alpha)}{\cos\alpha+\cos(\alpha+2\alpha)}\cdot(1+2\cos^22\alpha-1)=\\
=\frac{\sin\alpha+\sin\alpha\cos2\alpha+\cos\alpha\sin2\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-\sin\alpha\sin2\alpha}\cdot2\cos^22\alpha=\\
==\frac{\sin\alpha+\sin\alpha\cos2\alpha+2\cos^2\alpha\sin\alpha}{\cos\alpha+\cos\alpha\cos2\alpha-2\sin^2\alpha\cos\alpha}\cdot2\cos^22\alpha=\\=\frac{\sin\alpha(1+\cos2\alpha+2\cos^2\alpha)}{\cos\alpha(1+\cos2\alpha-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\
\frac{\sin\alpha(1+2\cos^2\alpha-1+2\cos^2\alpha)}{\cos\alpha(\cos2\alpha+1-2\sin^2\alpha)}\cdot2\cos^22\alpha=\\
=\frac{\sin\alpha\cdot4\cos^2\alpha}{\cos\alpha\cdot2\cos2\alpha}\cdot2\cos^22\alpha=\\
=4\sin\alpha\cos\alpha\cos2\alpha=2\sin2\alpha2\cos2\alpha=\sin4\alpha;\\3)ctg2\alpha\frac{2tg\alpha}{1+tg^2\alpha}=\cos2\alpha;\\
\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{sin\alpha}{\cos\alpha}}{1+(\frac{\sin\alpha}{\cos\alpha})^2}=\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{2\frac{\sin\alpha}{\cos\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}=2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\frac{\sin\alpha}{\cos\alpha}}{\frac{1}{\cos^2\alpha}}=\\
=2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\frac{\sin\alpha\cdot\cos^2\alpha}{\cos\alpha}==2\cdot\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin\alpha\cdot\cos\alpha=\frac{\cos2\alpha}{\sin2\alpha}\cdot2\sin\alpha\cos\alpha=\frac{\cos2\alpha}{\sin2\alpha}\cdot\sin2\alpha=\cos2\alpha;
    • Автор:

      rykerbscy
    • 6 лет назад
    • 0
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