高一三角函数两道解答题求大神解答要详细过程

Question

如题所述

Answer 1

二
1)f(x)=3sinxcosx-3√3(cosx)^2+3√3/2
=3/2sin2x-3√3/2(1+cos2x)+3√3/2
=3/2sin2x-3√3/2cos2x
=3(1/2sin2x-√3/2cos2x)
=3sin(2x-π/3)
sin(2x-π/3)最大值=1, 2x-π/3=2kπ+π/2
f(x)=3sin(2x-π/3)最大值=3,x=kπ+π/12
当,x=kπ+π/12时,f(x)最大值=3
2)f(2π/3-a/2)=3sin(π-a)=9/5
sina=3/5
f(b/2+5π/12)=3sin(π/2+b)=-36/13
cosb=-12/13

所以,sina=3/5 ,cosa=4/5
sinb=5/13 ,cosa=-12/13
cos(a+b)=cosacosb-sinasinb=-4/5*12/13-3/5*5/13=-63/65
cos(a+b)=-63/65
三

f(x)=2sinxcosx-2(cosx)^2+1
=sin2x-cos2x
=√2(√2/2sin2x-√2/2cos2x)
=√2sin(2x-π/4)
2x-π/4在[2kπ-π/2,2k+π/2]是单调递增
x在[kπ-π/8,kπ+3π/8]是单调递增
f(a/2+π/8)=√2sina=3√2/5
sina=3/5
cosa=4/5
tana=sina/cosa=3/4
tan(a+π/4)=(1+tana)/(1-tana)=(1+3/4)/(1-3/4)=7
tan(a+π/4)=7

Answer 2

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