Question

已知函数f（x）=cos²x/2－sin²x/2＋sin＋求f（x）最小正周期

当x0∈（0，π/4）且f（x0）=4根号2分之5时，求f（x0+π/6）的值

Answer 1

f（x）=cos²x/2－sin²x/2 + sinx
=cosx + sinx
=根号2(sinxcosπ/4+cosxsinπ/4)
=根号2 sin(x+π/4)
最小正周期2π

x0∈（0，π/4）且f（x0）=4根号2分之5，即：
根号2 sin(x0+π/4)=5/(4根号2)
sin(x0+π/4)=5/8
cos(x0+π/4)=根号(1-5^2/8^2))=根号39 /8
f（x0+π/6）=sin(x0+π/6)=sin[(x0+π/4)-(π/4-π/6)]
=sin(x0+π/4)cos(π/4-π/6) - cos(x0+π/4)sin(π/4-π/6)
=5/8(cosπ/4cosπ/6+sinπ/4sinπ/6)-7/8(sinπ/4cosπ/6-cosπ/4sinπ/6)
=5/8 * 根号2/2 * (根号3/2+1/2) - 根号39 /8* 根号2/2 * (根号3/2-1/2)
=根号2/32 * [ 5(根号3+1) - 根号39(根号3-1)]
=根号2/32 * [ 5+5根号3+根号39- 3根号13)]
= [ 5根号2+5根号6+根号78- 3根号26)]/32