Odpowiedź :
Odpowiedź:
a.
sin120°-cos30° = sin(90°+30°)-cos30°= cos30°-cos30° = 0
b.
cos120° + sin30° =cos(90°+30°) + sin30° = -sin30° + sin30° = 0
c.
sin150° - cos60° = sin(90°+60°) - cos60° = cos60° - cos60° = 0
d.
2tg45° + tg135° = 2tg45° + tg(180°-45°) =2tg45° -tg45° = tg45° =1
Szczegółowe wyjaśnienie:
[tex]a) \ sin120^{o}-cos30^{o} =sin(180^{o}-60^{o})-cos30^{o} = sin60^{o}-cos30^{o} =\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} = 0\\\\b) \ sin150^{o}-cos60^{o} = sin(180^{o}-30^{o}) -cos60^{o} = sin30^{o}-cos60^{o} = \frac{1}{2}-\frac{1}{2} = 0\\\\c) \ cos120^{o}+sin30^{o} = cos(90^{o}+30^{o})+sin30^{o} = -sin30^{o}+sin30^{o} = 0\\\\d) \ 2tg45^{o}+tg135^{o} = 2tg45^{o}+tg(90^{o}+45^{o}) = 2tg45^{o}-ctg45^{o} = 2\cdot1 - 1 =1[/tex]