1.求函数y=sinx+cosx+2sinxcosx+2的最大值和最小值,又若x∈[0,π]呢?2.若4cosAcosB=√6,4sinAsinB=√2,求(1-cos4A)(1-cos4B)的值.不是,
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![1.求函数y=sinx+cosx+2sinxcosx+2的最大值和最小值,又若x∈[0,π]呢?2.若4cosAcosB=√6,4sinAsinB=√2,求(1-cos4A)(1-cos4B)的值.不是,](/uploads/image/z/616052-20-2.jpg?t=1.%E6%B1%82%E5%87%BD%E6%95%B0y%3Dsinx%2Bcosx%2B2sinxcosx%2B2%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%2C%E5%8F%88%E8%8B%A5x%E2%88%88%5B0%2C%CF%80%5D%E5%91%A2%3F2.%E8%8B%A54cosAcosB%3D%E2%88%9A6%2C4sinAsinB%3D%E2%88%9A2%2C%E6%B1%82%EF%BC%881-cos4A%EF%BC%89%EF%BC%881-cos4B%EF%BC%89%E7%9A%84%E5%80%BC.%E4%B8%8D%E6%98%AF%2C)
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1.求函数y=sinx+cosx+2sinxcosx+2的最大值和最小值,又若x∈[0,π]呢?2.若4cosAcosB=√6,4sinAsinB=√2,求(1-cos4A)(1-cos4B)的值.不是,
1.求函数y=sinx+cosx+2sinxcosx+2的最大值和最小值,又若x∈[0,π]呢?
2.若4cosAcosB=√6,4sinAsinB=√2,
求(1-cos4A)(1-cos4B)的值.
不是,
1.求函数y=sinx+cosx+2sinxcosx+2的最大值和最小值,又若x∈[0,π]呢?2.若4cosAcosB=√6,4sinAsinB=√2,求(1-cos4A)(1-cos4B)的值.不是,
1.y=(sinx+cosx)+(sinx+cosx)^2+1
然后你知道了吧.
2.cos4A=(cos2A)^2-(sin2A)^2
1-cos4A=2[sin(2A)]^2
同理
1-cos4B=2[sin(2B)]^2
所以:
(1-cos4A)(1-cos4B)
=4[sin2A*sin2B]^2
=4[4sinAcosA*4sinBcosB]^2
剩下的把题目条件带入咯