Question

va rog muuuult 50 puncte ​

Answer 1

Răspuns:

$a$∈$(\pi ,\frac{3\pi }{2})=> sin<0, cos<0$

b∈$(\frac{3\pi }{2},2\pi )=> sin<0, cos>0$

$a) sin2a=2sina*cosa$

$sin^2a+cos^2a=1=>sin^2a=1-cos^2a<=>1-\frac{4}{9}=> sin^2a=\frac{5}{9}$

$sina=-\frac{\sqrt{5} }{3}$

$=>sin2a=2*(-\frac{\sqrt{5} }{3} )*(-\frac{2}{3} )=2*\frac{2\sqrt{5} }{9} =\frac{4\sqrt{5} }{9}$

$cos2a=cos^2a-sin^2a=\frac{4}{9}-\frac{5}{9} =-\frac{1}{9}$

$b)sin(a+b)=sina*cosb+sinb*cosa$

$sin^2b+cos^2b=1=>sin^2b=1-cos^2b<=>1-\frac{1}{25}=>sin^2b=\frac{24}{25}$

$sinb=-\frac{2\sqrt{6} }{5}$

$sin(a+b)=sina*cosb+sinb*cosa<=>(-\frac{\sqrt{5} }{3} )*\frac{1}{5}+(-\frac{2\sqrt{6} }{5})*(-\frac{2}{3} )=-\frac{\sqrt{5} }{15} +\frac{4\sqrt{6} }{15}=\frac{4\sqrt{6}-\sqrt{5} }{15}$