设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈[(-3/2)π,π],若向量m·向量n=1,求:(1)sin(Q+π/4)的值(2)cos[Q+(7/12)π]的值不好意思,打错了。正确应为:设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),
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![设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈[(-3/2)π,π],若向量m·向量n=1,求:(1)sin(Q+π/4)的值(2)cos[Q+(7/12)π]的值不好意思,打错了。正确应为:设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),](/uploads/image/z/8550545-41-5.jpg?t=%E8%AE%BE%E5%90%91%E9%87%8Fm%3D%28cosQ%2CsinQ%29%2C%E5%90%91%E9%87%8Fn%3D%282%E2%88%9A2%2BsinQ%2C2%E2%88%9A2-cosQ%29%2CQ%E2%88%88%5B%28-3%2F2%29%CF%80%2C%CF%80%5D%2C%E8%8B%A5%E5%90%91%E9%87%8Fm%C2%B7%E5%90%91%E9%87%8Fn%3D1%2C%E6%B1%82%EF%BC%9A%EF%BC%881%EF%BC%89sin%28Q%2B%CF%80%2F4%29%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89cos%5BQ%2B%287%2F12%29%CF%80%5D%E7%9A%84%E5%80%BC%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%EF%BC%8C%E6%89%93%E9%94%99%E4%BA%86%E3%80%82%E6%AD%A3%E7%A1%AE%E5%BA%94%E4%B8%BA%EF%BC%9A%E8%AE%BE%E5%90%91%E9%87%8Fm%3D%28cosQ%2CsinQ%29%2C%E5%90%91%E9%87%8Fn%3D%282%E2%88%9A2%2BsinQ%2C2%E2%88%9A2-cosQ%29%2C)
设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈[(-3/2)π,π],若向量m·向量n=1,求:(1)sin(Q+π/4)的值(2)cos[Q+(7/12)π]的值不好意思,打错了。正确应为:设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),
设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈[(-3/2)π,π],若向量m·向量n=1,求:
(1)sin(Q+π/4)的值
(2)cos[Q+(7/12)π]的值
不好意思,打错了。
正确应为:设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈((-3/2)π,-π),若向量m·向量n=1,求:
(1)sin(Q+π/4)的值
(2)cos(Q+(7/12)π)的值
设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),Q∈[(-3/2)π,π],若向量m·向量n=1,求:(1)sin(Q+π/4)的值(2)cos[Q+(7/12)π]的值不好意思,打错了。正确应为:设向量m=(cosQ,sinQ),向量n=(2√2+sinQ,2√2-cosQ),
(1)
∵向量m=(cosQ,sinQ),向量n=(2(√2)+sinQ,2(√2)-cosQ)
∴向量m•向量n=2(√2)cosQ+sinQcosQ+2(√2)sinQ-sinQcosQ=4sin(Q+π/4)
∵向量m•向量n=1
∴4sin(Q+π/4)=1
∴sin(Q+π/4)=1/4
(2)
∵Q∈[(-3/2)π,-π]
∴sinQ>0,cosQ
(1) 因为向量m·向量n=1,
所以cosQ(2√2+sinQ)+sinQ(2√2-cosQ)=1,
sinQ+cosQ=√2/4,
√2sin(Q+π/4)=√2/4,
sin(Q+π/4)=1/4.
(2)因为(-3/2)π所以cos(Q+π/4)<0
所以cos(Q+π/4)=-...
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(1) 因为向量m·向量n=1,
所以cosQ(2√2+sinQ)+sinQ(2√2-cosQ)=1,
sinQ+cosQ=√2/4,
√2sin(Q+π/4)=√2/4,
sin(Q+π/4)=1/4.
(2)因为(-3/2)π所以cos(Q+π/4)<0
所以cos(Q+π/4)=-√15/4
cos(Q+(7/12)π)=cos[(Q+π/4)+π/3]=cos(Q+π/4)cosπ/3-sin(Q+π/4)sinπ/3
=-√15/4·1/2-1/4·√3/2
=-(√15+√3)/8
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