设f(x)=2sinxcos^2P/2+cosxsinP-sinx(0

题目

设f(x)=2sinxcos^2P/2+cosxsinP-sinx(0(1)求P的值
(2)在三角形ABC中,abc分别是角ABC的对边,已知a=1,b=根号2,f(A)=根号3/2,求角C

答案

f(x) = 2sinxcos^2(P/2)+cosxsinP-sinx
= sinx(1+cosP)+cosxsinP-sinx
= sinxx + sinxcosP + cosxsinP - sinx
= sinxcosP+cosxsinP
= sin(x+P)
在x=π处取最小值,0则x+P=3π/2
P=π/2
f(x) = sin(x+π/2) = cosx
f(A)=√3/2
A=π/6
a/sinA=b/sinB
sinB=bsinA/a = √2×1/2 /1 = √2/2
B=π/4,或3π/4
C=π-A-B=7π/12,或π/12