Question

1 a) sin 105 = sin(60° +45)
d) cos 105°=
c) tg15°=tg(45°-30° )=​

Answer 1

Ответ:

$sin 105 = sin(60 +45)=sin60cos45+sin45cos60=\frac{\sqrt{3} }{2}*\frac{\sqrt{2}}{2}+ \frac{1 }{2}*\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2} }{4}\\ cos 105 = cos(60 +45)=cos60cos45-sin45sin60=\frac{1 }{2}*\frac{\sqrt{2}}{2}-\frac{\sqrt{3} }{2}*\frac{\sqrt{2}}{2}=\frac\frac{x}{y} {\sqrt{2}-\sqrt{6} }{4} \\tg15=tg(45-30 )=\frac{tg 45-tg30}{1+tg 45tg30} =\frac{1-\frac{1}{\sqrt{3}} }{1+\frac{1}{\sqrt{3}} } =\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Объяснение: