求sin(—1320°)cos1110°+cos(—1020°)sin750°+tan495°的值。
要步骤,谢谢合作!!
求sin(—1320°)cos1110°+cos(—1020°)sin750°+tan495°的值。
要步骤,谢谢合作!!
原式=sin(-8π+120°)cos(6π+30°)+cos(-6π+60°)sin(4π+30°)+tan(2π+135°)
=(sin120°)(cos30°)+cos(60°)sin30°+tan135°
=sin60°cos30°+cos60°sin30°-tan45°
=3/4+1/4-1
=0
sin(-7π-60°)COS(6π+150°)+COS(-5π-120°)SIN(4π+30°)+TAN(2π+135°)
=sin(-60°)(-cos150°)+cos(-120°)sin30°+TAN135°
=-sin60°cos30°-sin30°cos60°-cot45°
=-1-1
=-2