f(x)=sin(x+5π/4)+cos(x+3π/4) ①求f(x)的单调递增区间 ②若f(x)=(2√2)/3 sinα/1+tanα值
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![f(x)=sin(x+5π/4)+cos(x+3π/4) ①求f(x)的单调递增区间 ②若f(x)=(2√2)/3 sinα/1+tanα值](/uploads/image/z/8620252-52-2.jpg?t=f%EF%BC%88x%EF%BC%89%3Dsin%EF%BC%88x%2B5%CF%80%2F4%29%2Bcos%EF%BC%88x%2B3%CF%80%2F4%EF%BC%89+%E2%91%A0%E6%B1%82f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4+%E2%91%A1%E8%8B%A5f%EF%BC%88x%EF%BC%89%3D%EF%BC%882%E2%88%9A2%EF%BC%89%2F3+sin%CE%B1%2F1%2Btan%CE%B1%E5%80%BC)
f(x)=sin(x+5π/4)+cos(x+3π/4) ①求f(x)的单调递增区间 ②若f(x)=(2√2)/3 sinα/1+tanα值
f(x)=sin(x+5π/4)+cos(x+3π/4) ①求f(x)的单调递增区间 ②若f(x)=(2√2)/3 sinα/1+tanα值
f(x)=sin(x+5π/4)+cos(x+3π/4) ①求f(x)的单调递增区间 ②若f(x)=(2√2)/3 sinα/1+tanα值
f(x)=sin(x+5π/4)+cos(x+3π/4),①求f(x)的单调递增区间; ②若f(x)=(2√2)/3,
求 sinx/(1+tanx)的值.
f(x)=sin[π+(x+π/4)]+cos[π+(x-π/4)]=-sin(x+π/4)-cos(x-π/4)=-(√2/2)[(sinx+cosx)+(cosx+sinx)]
=-(√2)(sinx+cosx)=-2sin(x+π/4)
由π/2+2kπ≦x+π/4≦3π/2+2kπ,得单增区间为π/4+2kπ≦x≦5π/4+2kπ,k∈Z.
若f(x)=-2sin(x+π/4)=-(√2)(sinx+cosx)=(2√2)/3,则sinx+cosx=-2/3
故此时sinx/(1+tanx)=sinxcosx/(sinx+cosx)=2sinxcosx/[2(sinx+cosx)]
=[(sinx+cosx)²-1]/[2(sinx+cosx)]=[(-2/3)²-1]/[2(-2/3)]=(5/9)/(4/3)=5/12.