大学预科数学三角函数的题目 这道大学数学 高等数学 题目怎么做的这个题目三角函数求导
\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u9898\u76ee(1)
a.b=0
(sin\u03b8,-1).(1,cos\u03b8)=0
sin\u03b8-cos\u03b8=0
tan\u03b8=1
g(\u03b8)
=f(sin\u03b8+cos\u03b8) +2\u221a3(cos\u03b8)^2
=(sin\u03b8+cos\u03b8)^2 +2\u221a3(cos\u03b8)^2
=1+ 2sin\u03b8.cos\u03b8 +2\u221a3(cos\u03b8)^2
=1+2(1/\u221a2)(1/\u221a2) +2\u221a3(1/\u221a2)^2
=1 + 1 + \u221a3
=2+\u221a3
(2)
g(x)
=1+ 2sinx.cosx +2\u221a3(cosx)^2
=1+ sin2x +\u221a3.( 1+cos2x)
=1+\u221a3 + 2\u221a3.sin(2x+\u03c0/3)
max g(x) = 1+3\u221a3
2x+\u03c0/3 = 2k\u03c0 +\u03c0/2
x= k\u03c0 +5\u03c0/12
\u96c6\u5408 = { x| x= k\u03c0 +5\u03c0/12 , k\u2208N }
min g(x) = 1-\u221a3
2x+\u03c0/3 = 2k\u03c0 -\u03c0/2
x = k\u03c0 -\u03c0/12
\u96c6\u5408 = { x| x= k\u03c0 -\u03c0/12 , k\u2208N }
答案在纸上
arctanx=¼π-2arctan½→x=tan(arctanx)=
(tan¼π-tan2arctan½)/(1+tan¼π·tantan2arctan½)=(1-4/3)/(1+4/3)=-1/75sin(2x+φ)=5 tanφ=4/3→2x+φ=2kπ+½π→x=kπ+¼π-½arctan(4/3)
√(6+2√3)sin(x+φ)=1→
x+arctan(2+√3)=2kπ+arcsin(1/√(6+2√3))或x+arctan(2+√3)=(2k+1)π-arcsin(1/√(6+2√3))
x=2kπ+arcsin(1/√(6+2√3))-arctan(2+√3)
x=(2k+1)π-arcsin(1/√(6+2√3))-arctan(2+√3)2sinxcosx/cosx=1→sinx=½→x=2kπ+⅙π 或x=2kπ+⅚π
这个有点难度。
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