1.求三角函数值：⑴sin(—（43π）/6)⑵cos(—(83π)/6)⑶tan(—（41π）/3)⑷cos((11π)/4)⑸sin930°2.化简：⑴Cos(x—π)tan(x—2π)tan(2π—x)/sin(π+x)⑵((sin^2)(—x))—tan(360°—x)*tan(—x)—sin(180°

Question

1.求三角函数值：⑴sin(—（43π）/6) ⑵cos(—(83π)/6)
⑶tan(—（41π）/3)
⑷cos((11π)/4)
⑸sin930°
2.化简：
⑴【Cos(x—π) tan(x—2π) tan(2π—x)】/【sin(π+x)】
⑵((sin^2 )(—x))—tan(360°—x) *tan(—x)—sin(180°—x) cos(360°—x) tan(180°+x)

Answer 1

你这题真够长的!看图,放大了.