Question

Срочно!!!!!! Помогите пожалуйста

Answer 1

sin(pi/3-pi/4)=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)=√3/2*√2/2-1/2*√2/2=

=√2/2(√3/2-1/2)=(√6-√2)/4

Answer 2

$1); ; a=frac{pi}{3}-frac{pi}{4}=frac{pi}{12}=15^circ \\sina=sin(frac{pi}{3}-frac{pi}{4})=sinfrac{pi}{3}cdot cosfrac{pi}{4}-sinfrac{pi}{4}cdot cosfrac{pi}{3}=\\=frac{sqrt3}{2}cdot frac{sqrt2}{2}-frac{sqrt2}{2}cdot frac{1}{2}=frac{sqrt3cdot sqrt2-sqrt2}{4}=frac{sqrt6-sqrt2}{4}=frac{sqrt3-1}{2sqrt2}\\cosa=cos(frac{pi }{3}-frac{pi}{4})=cosfrac{pi}{3}cdot cosfrac{pi}{4}+sinfrac{pi}{3}cdot sinfrac{pi}{4}=\\=frac{1}{2}cdot frac{sqrt2}{2}+frac{sqrt3}{2}cdot frac{sqrt2}{2}=frac{sqrt2+sqrt6}{4}=frac{sqrt3+1}{2sqrt2}$

$tga=frac{sina}{cosa}=frac{sqrt3-1}{sqrt3+1}=frac{(sqrt3-1)^2}{(sqrt3-1)(sqrt3+1)}=frac{4-2sqrt3}{3+1}=frac{2(2-sqrt3)}{4}=2-sqrt3\\ctga=frac{1}{tga}=frac{1}{2-sqrt3}=frac{2+sqrt3}{(2-sqrt3)(2+sqrt3)}=frac{2+sqrt3}{4-3}=2+sqrt3\\\sinfrac{pi }{12}=sin15^circ =frac{sqrt3-1}{2sqrt2}=frac{sqrt6-sqrt2}{4}; ,; ; cosfrac{pi }{12}=cos15^circ =frac{sqrt3+1}{2sqrt2}=frac{sqrt6+sqrt2}{4}; ,\\tg{frac{pi}{12}=tg15^circ =2-sqrt3; ; ,; ; ctg frac{pi }{12}=2+sqrt3$

$2); ; cosa=-frac{5}{13}\\pi <a<frac{3pi}{2}; ; to ; ; sina<0; ,; ; tga>0; ,; ; ctga>0\\sina=-sqrt{1-cos^2a}=-sqrt{1-frac{25}{169}}=-frac{12}{13}\\sin2a=2, sina, cosa=2cdot (- frac{12}{13})cdot (-frac{5}{13})=frac{120}{169}\\cos2a=2cos^2a-1=2cdot frac{25}{169}-1=-frac{119}{169}\\tg2a=frac{sin2a}{cos2a}=-frac{120}{119}; ; ,; ; ctga=-frac{119}{120}$

$3); ; 2sin(frac{pi}{4}+a)cdot sin(frac{pi}{4}-a)+sin^2a=\\=2cdot frac{1}{2}cdot Big (cos(frac{pi}{4}+a-frac{pi}{4}+a)-cos(frac{pi }{4}-a+frac{pi }{4}-a)Big )+sin^2a=\\=cos2a-cosfrac{pi }{2}+sin^2a=(cos^2a-sin^2a)-0+sin^2a=cos^2a; ;\\cos^2a=cos^2a; .\\star ; ; sinacdot sinbeta =frac{1}{2}cdot (cos(a-beta )-cos(a+beta )); ; star$