1.cos2x=1-2*(sinx)^2 故 f(sinx)=3-cos2x=2+2*(sinx)^2 即 f(x)=2+2*x^2 所以 f(cosx)=2+2*(cosx)^2=2+1+[2*(cosx)^2-1]=3+cos2x 得C
2.sin(7π/6 +a)=sin(2π/3 +a+π/2)=cos(2π/3 +a) = -1/4 ； cos(5π/6 -a)=cos(-2π/3 -a+3π/2)=cos(-2π/3 -a-π/2)= - sin(2π/3 +a) ,cos^2(5π/6 -a)=sin^2(2π/3 +a) =1-cos^2(2π/3 +a)=15/16 ,故 -1/4 + 15/16 =11/16
3.[sin(π/2 +a)-cos(π-a)]/[sin(π/2 -a)-sin(π-a)]= (cosa+cosa)/(cosa-sina)=2/(1-tan a) =-2
即化簡后上下同除以 cosa
以后有問題找同學(xué)或者老師吧……你看這樣多累