1.已知tanX=3,求下列各式的值:⑴sinXcosX; ⑵(cos^2) X— 2(sin^2) X2.计算:⑴3sin270°+2cos180°—cos90°+√3 tan0°⑵5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π3.计算:【cos(—45°)cos330°t
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 23:56:02
![1.已知tanX=3,求下列各式的值:⑴sinXcosX; ⑵(cos^2) X— 2(sin^2) X2.计算:⑴3sin270°+2cos180°—cos90°+√3 tan0°⑵5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π3.计算:【cos(—45°)cos330°t](/uploads/image/z/2738947-67-7.jpg?t=1.%E5%B7%B2%E7%9F%A5tanX%3D3%2C%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E7%9A%84%E5%80%BC%EF%BC%9A%E2%91%B4sinXcosX%3B+%E2%91%B5%EF%BC%88cos%5E2%EF%BC%89+X%E2%80%94+2%EF%BC%88sin%5E2%EF%BC%89+X2.%E8%AE%A1%E7%AE%97%EF%BC%9A%E2%91%B43sin270%C2%B0%2B2cos180%C2%B0%E2%80%94cos90%C2%B0%2B%E2%88%9A3+tan0%C2%B0%E2%91%B55sin%EF%BC%88%CF%80%2F2%EF%BC%89%2B2cos0%C2%B0%E2%80%94%EF%BC%884%2F5%EF%BC%89tan%CF%80%E2%80%94%EF%BC%882%2F3%EF%BC%89sin%28%283%CF%80%29%2F2%29%2B4tan2%CF%803.%E8%AE%A1%E7%AE%97%EF%BC%9A%E3%80%90cos%28%E2%80%9445%C2%B0%29cos330%C2%B0t)
1.已知tanX=3,求下列各式的值:⑴sinXcosX; ⑵(cos^2) X— 2(sin^2) X2.计算:⑴3sin270°+2cos180°—cos90°+√3 tan0°⑵5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π3.计算:【cos(—45°)cos330°t
1.已知tanX=3,求下列各式的值:⑴sinXcosX; ⑵(cos^2) X— 2(sin^2) X
2.计算:
⑴3sin270°+2cos180°—cos90°+√3 tan0°
⑵5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π
3.计算:【cos(—45°)cos330°tan585°】/【tan(—120°)】
1.已知tanX=3,求下列各式的值:⑴sinXcosX; ⑵(cos^2) X— 2(sin^2) X2.计算:⑴3sin270°+2cos180°—cos90°+√3 tan0°⑵5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π3.计算:【cos(—45°)cos330°t
1.因为 1/cos²x=(sin²x+cos²x)/cos²x=tan²x+1=4
所以 cos²x=1/4
(1) sinxcosx=tanxcosxcosx=3cos²x=3/4
(2)cos²x-2sin²x=cos²x-2(1-cos²x)=3cos²x-2=3/4-2=-5/4
2
(1)3sin270°+2cos180°—cos90°+√3 tan0°=-3-2-0+0=-5
(2)5sin(π/2)+2cos0°—(4/5)tanπ—(2/3)sin((3π)/2)+4tan2π
=5+2-0+2/3+0=23/3
3.:[cos(—45°)cos330°tan585°]/[tan(—120°)]
=cos45°cos(360°-30°)tan(3×180°+45°)tan(-180°+60°)
=cos45°cos30°tan45°tan60°
=(√2/2)×(√3/2)×1×√3=3√2/4
12