sin(x+π/3)=sin(x)cos(π/3)+cos(x)sin(π/3)=(1/2)sin(x)+(√3/2)cos(x)
cos(x+π/6)=cos(x)cos(π/6)-sin(x)sin(π/6)=(√3/2)cos(x)-(1/2)sin(x)
f(x)=(1/2)sin(x)+(√3/2)cos(x)+√3cos(x)-sin(x)=(-1/2)sin(x)+(3√3/2)cos(x)
sinとcosの合成より、f(x)=Asin(x+α) ※A=√((-1/2)²+(3√3/2)²)=√7、α=tan⁻¹((3√3/2)/(-1/2))
-1≦sin(x+α)≦1なので、-√7≦f(x)≦√7
Answers
Were you able to resolve your confusion?
Users viewing this question
are also looking at these questions 😉
Recommended
詳説【数学Ⅰ】第一章 数と式~整式・実数・不等式~
8991
117
数学ⅠA公式集
5737
20
詳説【数学Ⅰ】第三章 図形と計量(前半)~鋭角鈍角の三角比~
4580
11
【セ対】三角比 基礎〜センター約8割レベル
985
3