作业帮计算：sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
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计算：sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
计算：sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
计算：sin(-1200°)×cos1230°+cos(-1020°)sin(-1050°)+tan1305°
sin(-1200+360*4)*cos(1230-360*3)+cos(-1020+3*360)*-sin(-1050+360*3)+tan(1305-180*7)
=sin(240)*cos(150)+cos(60)*sin(30)+tan(45)
=[-sin(60)]*[-cos(30)]+cos(60)*sin(30)+tan(45)
=-((根号)3)/2*-((根号)3)/2+1/2*1/2+1
=3/4+1/4+1
=2
计算 1:sin（-11π/6）+cos12π/5 *tan4π 2：sin11ts40*cos（-690）+tan1845第二题是度数1140度
sin(-11/6π)+cos12/5π×tan4π化简
求值sin42°-cos12°+sin54°
求值:sin(-11π/6)+cos12π/5*tan4π+1/cos7π/3
sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)
cos57°cos12°+sin57°sin12°=
求值cos72°cos12°+sin72°sin12°
sin18°cos12°+cos18°cos78°
化简cos12°cos24°cos48°cos84°=
cos12°cos24°cos48°cos84°=
cos12°-cos24°-cos48°+cos84°
sin6°cos24°cos48°cos12°=
sin6°*cos12°*cos24°*cos48°
sin42°-cos12°+sin54°的值是
证明：Sin42°-cos12°+sin54°=½
化简cos18°cos12°-sin18sin12所得的值为
cos12°cos24°cos48°cos96°详细过程..cos12°cos24°cos48°cos96°=?
（tan12°-根号3）/【（4*cos12°*cos12°）*sin12°希望可以多说说解题思路