1.
已知函數(shù)f(x)=cos(2x-π/3) 2sin(x-π/4)sin(x π/4)
(1)求函數(shù)f(x)的最小正周期和圖像的對(duì)稱軸方程
（2）求函數(shù)f(x)在區(qū)間[-π/12,π/2]上的值域
⑴f(x)=cos(2x-π/3) 2sin(x-π/4)sin(x π/4)
=cos(2x-π/3) √2sin(x-π/4)√2sin(x π/4)
=cos(2x-π/3) (sinx-cosx)(sinx cosx)
=cos(2x-π/3) (sinx^2-cosx^2)
=cos(2x-π/3) cos2x
=1/2cos2x-√3/2sin2x cos2x
=3/2cos2x-√3/2sin2x
=√3(√3/2cos2x-1/2sin2x)
=√3sin(2x-π/3)
∴函數(shù)f(x)的最小正周期為π,圖像的對(duì)稱軸方程 x=kπ/2 π/6
（2）函數(shù)f(x)在區(qū)間[-π/12,π/2]上的值域 [-√3,√3]
2.
(1)函數(shù)y=T3*cos2x+sin2x+m (T為根號(hào))
=2(T3/2*cos2x+1/2*sin2x)+m
=2[sin(2x+派/3)]+m
所以最小正周期為 2派/2=派
(2)x∈[0.3.14/2] 2x∈[0 ,派]
2x+派/3∈[派/3,4派/3]
當(dāng) 2x+派/3=4派/3時(shí) 有最小值-T3+m=5
所以 m=5+T3
3.
因?yàn)橄蛄縜//向量b,得
-sinx-3/2cosx=0,得-sinx=3/2cosx
tanx=-3/2,
再化簡(jiǎn)2cos^2x-sin2x,用降冪公式得：1+cos2x=2cos^x
=1+cos2x-sin2x
再用萬(wàn)能公式
cos2x=(1-tan^2x)/(1+tan^x)=(1-9/4)/(1+9/4)=-5/13
sin2x=2tanx/(1+tan^x)=2*(-3/2)/（1+9/4)=-12/13,
最后將sin2x,cos2x代人,得
1-5/13+12/13=20/13