Question

Ребята помогите докажите тождество 470 задача

Answer 1

1) 4sinα*cos³α-2sin2α*sin²α=sin4α;
4sinα*cos³α-2sin2α*sin²α=4sinα*cos³α-4sinα*cosα*sin²α=
=4sinα*cosα(cos²α-sin²α)=4sinα*cosα*cos2α=2sin2α*cos2α=sin4α;
$2) frac{tgx}{1+ tg^{2}x}+ frac{ctgx}{1+ctg^2x}=sin2x; \ tgx*cos^2x+ctgx*sin^2x=sinx*cosx+sinx*cosx= \ =2sinx*cosx=sin2x;$
3) tg^4α(8cos²(π-α)-cos(π+4α)-1)=8sin^4α;
tg^4α(8cos²α+cos4α-1)=tg^4α(8cos²α+cos²2α-sin²2α-sin²2α-cos²2α)=
=tg^4α(8cos²α-2sin²2α)=tg^4α(8cos²α-8sin²α*cos²α)=8sin^4α/cos²α(1-sin²α)=
=8sin^4α/cos²α*cos²α=8sin^4α;
$4) 2tg frac{ alpha }{2}(tg alpha +ctg alpha )(1-tg^2 frac{ alpha }{2})= frac{1}{ cos^{4} frac{ alpha }{2}}; \ 2tg frac{ alpha }{2}( frac{2tg frac{ alpha }{2}}{1-tg^2 frac{ alpha }{2}}+ frac{1-tg^2 frac{ alpha }{2} }{2tg^2 frac{ alpha }{2}})(1-tg^2 frac{ alpha }{2})= ( frac{4tg^2 frac{ alpha }{2} }{1-tg^2 frac{ alpha }{2}}+(1-tg^2 frac{ alpha }{2}))(1-tg^2 frac{ alpha }{2})$
$=4tg^2 frac{ alpha }{2}+ (1-tg^2 frac{ alpha }{2} )^{2}=4tg^2 frac{ alpha }{2}+1-2tg^2 frac{ alpha }{2}+tg^4 frac{ alpha }{2}= \ =1+2tg^2 frac{ alpha }{2}+tg^4 frac{ alpha }{2}= (1+tg^2 frac{ alpha }{2} )^{2}= frac{1}{cos^4 frac{ alpha }{2}}.$