2sin(750°)-√3cos(930°)+tg²(60°)/2

Question

2sin(750°)-√3cos(930°)+tg²(60°)/2

Answer 1

2sin(750°)-√3cos(930°)+tg²(60°)/2=2sin(360°*2+30°)-√3cos(360°*2+210°)+tg²(60°)/2=sin30°-√3cos210°+tg²(60°)/2=1/2-√3cos(180°+30°)+(√3)²/2=1/2+√3cos30°+3/2=2+√3*√3/2=2+3/2=3 1/2