sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)
sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)
化简cos^2(-α)-[tan(360°+α)/sin(-α)]
化简,cos²(-α)-tan(360°+α)/sin(-α)
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3
求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)
tan(90°-α)等于多少 化简就行了 用sin/cos =tan 的关系
①sin(α+180°)cos(-α)sin(-α-180°)和②sin³(-α)cos(2π+α)tan(-
设cos(180°+α)=1/3 且 sinα>α 求sin(180°+α)-tanα/cos(180°-α)+tan(
已知α=1050°,化简sin(540°-α)/tan(α-180°)×sin(-α-360°)/cos(720°-α)
化简:tanα*(cosα-sinα)+[sinα(sinα+tanα)/1+cosα]
化简:tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα.
已知tanα=2,计算:[sin(90°+α)-cos(180°-α)]/[sin(90°-α)-sin(180°-α)