已知函数f(x)=sin(π/2-x)+sinx(1)求函数y=f(x)的单调递增区间(2)若f(a-π/4)=根号2/3,求f(2a+π/4)的值
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![已知函数f(x)=sin(π/2-x)+sinx(1)求函数y=f(x)的单调递增区间(2)若f(a-π/4)=根号2/3,求f(2a+π/4)的值](/uploads/image/z/52294-22-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%28%CF%80%2F2-x%29%2Bsinx%281%29%E6%B1%82%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%282%29%E8%8B%A5f%28a-%CF%80%2F4%29%3D%E6%A0%B9%E5%8F%B72%2F3%2C%E6%B1%82f%282a%2B%CF%80%2F4%29%E7%9A%84%E5%80%BC)
已知函数f(x)=sin(π/2-x)+sinx(1)求函数y=f(x)的单调递增区间(2)若f(a-π/4)=根号2/3,求f(2a+π/4)的值
已知函数f(x)=sin(π/2-x)+sinx
(1)求函数y=f(x)的单调递增区间
(2)若f(a-π/4)=根号2/3,求f(2a+π/4)的值
已知函数f(x)=sin(π/2-x)+sinx(1)求函数y=f(x)的单调递增区间(2)若f(a-π/4)=根号2/3,求f(2a+π/4)的值
f(x)=cosx+sinx
f(x)=√2sin(x+π/4)
(1)递增区间:2kπ-π/2≤x+π/4≤2kπ+π/2
得:2kπ-3/4π≤x≤2kπ+π/4
递增区间是:[2kπ-3π/4,2kπ+π/4],其中k∈Z
(2)f(a-π/4)=√2sina=√2/3
则:sina=1/3
f(2a+π/4)=√2sin(2a+π/2)=√2cos2a=√2[1-2sin²a]=(7/9)√2
(1)f(x)=cosx+sinx=根号2乘以sin(x+π/4)
由-π/2+2Kπ<=x+π/4<=π/2+2Kπ解得单调区间[-π3/4+2Kπ,π/4+2Kπ]
(2)f(x)=cosx+sinx=根号2乘以sin(x+π/4)
f(a-π/4)=根号2乘以sin(a)=根号2/3,所以sina=1/3
f(2a+π/4)=根号2乘以sin(2a+π/2)=sin2a=2sinacosa
∵f(x)=sin[(π/2)-x]+sinx
=√2[(√2/2)cosx+(√2/2)sinx]
=√2[sin(π/4)cosx+cos(π/4)sinx]
=√2sin[x+(π/4)]
又∵y=sinx在[-π/2,π/2]上单调递增,
即:-π/2≤x...
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∵f(x)=sin[(π/2)-x]+sinx
=√2[(√2/2)cosx+(√2/2)sinx]
=√2[sin(π/4)cosx+cos(π/4)sinx]
=√2sin[x+(π/4)]
又∵y=sinx在[-π/2,π/2]上单调递增,
即:-π/2≤x≤π/2
∴-π/2≤x+(π/4)≤π/2
整理得:-3π/4≤x≤π/4
∴f(x)在2kπ-(3π/4)≤x≤2kπ+(π/4)(k∈Z)上单调递增;
同理,∵sinx在[π/2,3π/2]上单调递减;
∴π/2≤x+(π/4)≤3π/2
整理得:π/4≤x≤5π/4
∴f(x)在2kπ+(π/4)≤x≤2kπ+(5π/4)(k∈Z)上单调递减;
∵f(a-π/4)=√2/3
∴f(a-π/4)=√2sin[(a-π/4)+(π/4)
=√2sina
即√2sina=√2/3
∴sina=1/3
sina^2=1/9
cosa^2=1-(1/3)^2
=8/9
f(2a+π/4)=√2sin[(2a+π/4)+(π/4)]
=√2sin(2a+π/2]
=-√2cos2a
=-√2(cosa^2-sina^2)
=-√2[(8/9)-(1/9)]
=-7√2/9
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