求根号2/4sin(π/4-x)+根号6/4cos(π/4-x)的化简,

根号2/4sin(π/4-x)+根号6/4cos(π/4-x)
=(根号2)/2 *[1/2 *sin(π/4-x) + (根号3)/2 *cos(π/4-x)]
=(根号2)/2 *[sin(π/4-x)*cos(π/3) + cos(π/4-x)*sin(π/3)]
=(根号2)/2 *[sin(π/4 -x +π/3)]
=(根号2)/2 *[sin(7π/12 -x)]
=(根号2)/2 *[sin(π - 5π/12 -x)]
=(根号2)/2 *sin(x+ 5π/12)那请问一下 (根号3)/2是哪来的啊，谢谢啦，我不太懂根号6/4=根号2/2 * 根号3/2