Question

高中数学19题~求详细过程~

Answer 1

（1）先化简f（x）=1+cos2x+cos2x*cosπ/3-sin2x*sinπ/3=1+cos2x+3/2cos2x-根号3/2sin2x 统一函数名f（x）=1-根号3* sin(2x-π/3 ） f（a）=1- 根号3* sin(2a-π/3 ）=根号3/3+1 所以sin（2a-π/3 ）= -1/3 cos（2a-π/3 ）=根号8/3 sin2a=sin[(2a-π/3 )+π/3 ]=（根号24-1）/6
(2)f(A)=1-根号3* sin(2A-π/3 ）=-1/2 所以A=π/6=30° sinA=1/2 用cosA那个余玄定理

Answer 2

（1）解析：f(x)=2cos^2x+cos(2x+π/3)=cos2x+1+1/2cos2x-√3/2sin2x

=3/2cos2x-√3/2sin2x+1=√3cos(2x+π/6)+1

f(α)= √3cos(2α+π/6)+1=√3/3+1

==> cos(2α+π/6)= √3/2cos2α-1/2sin2α=1/3

==>3√3cos2α-3sin2α=2

与(cos2α)^2+(sin2α)^2=1联立

解得sin2α=(-1+2√6)/6

(2)解析：f(A)= √3cos(2A+π/6)+1=-1/2

==> cos(2A+π/6)=-√3/2==>2A+π/6=5π/6==>A=π/6；

S=1/2bcsinA=3b/4=3√3==>b=4√3

∴a=√(b^2+c^2-2bccosA)= √21