Question

Решите номер 93 и 94 пожалуйста))

Answer 1

93
$sin2 alpha =2sin alpha cos alpha \ cos2 alpha = cos^2 alpha -sin^2 alpha \ sin alpha = sqrt{ frac{1-cos2 alpha }{2} } \ cos alpha = sqrt{ frac{1+cos2 alpha }{2} } \ alpha in[-1;1]: \ sin(arcsin alpha )= alpha \ cos(arccos alpha )= alpha \ sin(arccos alpha )= sqrt{1- alpha ^2} \ cos(arcsin alpha )= sqrt{1- alpha ^2}$
а)
$cos(2arcsin(- frac{1}{2} ))=cos^2(arcsin(- frac{1}{2} ))-sin^2(arcsin(- frac{1}{2} ))= \ =(cos(arcsin(- frac{1}{2} )))^2-(sin(arcsin(- frac{1}{2} )))^2=( sqrt{1-( frac{1}{2} )^2} )^2-(- frac{1}{2} )^2= \ =1- frac{1}{4} - frac{1}{4}= frac{1}{2}$
б)
$mathrm{tg}(2arcsin frac{2}{3} )= cfrac{sin(2arcsin frac{2}{3} )}{cos(2arcsin frac{2}{3} )} = cfrac{2sin(arcsin frac{2}{3} )cos(arcsin frac{2}{3} )}{cos^2(arcsin frac{2}{3} )-sin^2(arcsin frac{2}{3})} = \ = cfrac{2cdot frac{2}{3}cdot sqrt{1-( frac{2}{3} )^2} }{(sqrt{1-( frac{2}{3} )^2)}^2-( frac{2}{3} )^2} =cfrac{frac{4}{3}cdot frac{ sqrt{5} }{3} }{ frac{ 5}{9}-frac{4}{9} } = cfrac{frac{4sqrt{5} }{9} }{ frac{ 1}{9} } =4 sqrt{5}$
в)
$sin(2arccos frac{1}{4} )=2sin(arccos frac{1}{4} )cos(arccos frac{1}{4} )= \ =2cdot sqrt{1-( frac{1}{4} )^2} cdot frac{1}{4} = frac{1}{2} cdot frac{ sqrt{15} }{4} = frac{ sqrt{15} }{8}$
г)
$mathrm{ctg}( frac{1}{2} arccos(- frac{1}{3} ))= cfrac{cos( frac{1}{2} arccos(- frac{1}{3} ))}{sin( frac{1}{2} arccos(- frac{1}{3} ))} = cfrac{ sqrt{ cfrac{1+ cos(arccos(- frac{1}{3} )}{2} }}{sqrt{ cfrac{1- cos(arccos(- frac{1}{3} )}{2} }} = \ = sqrt{ cfrac{1+cos(arccos(- frac{1}{3}) }{1-cos(arccos(- frac{1}{3})} } = sqrt{ cfrac{1- frac{1}{3} }{1-(- frac{1}{3} )} } = sqrt{ cfrac{ frac{2}{3} }{ frac{4}{3} } } = sqrt{ frac{1}{2} } =frac{ sqrt{2} }{2}$
94
Под знаком арксинуса/арккосинуса может стоять число, по модулю не больше 1:
а))
$arccos(2+b) \ -1 leq 2+b leq 1 \ -1-2 leq b leq 1-2 \ -3 leq b leq -1$
б)
$arcsin(b-4) \ -1 leq b-4 leq 1 \ -1+4 leq b leq 1+4 \ 3 leq b leq 5$
в)
$arcsin(1-3b) \ -1 leq 1-3b leq 1 \ -2 leq -3b leq 0 \ 0 leq 3b leq 2 \ 0 leq b leq frac{2}{3}$
г)
$arccos(6-5b) \ -1 leq 6-5b leq 1 \ -7 leq -5b leq -5 \ 5 leq 5b leq 7 \ 1 leq b leq frac{7}{5}$