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求根号2/4sin(π/4-x)+根号6/4cos(π/4-x)的化简,
题目详情
求根号2/4sin(π/4-x)+根号6/4cos(π/4-x)的化简,
▼优质解答
答案和解析
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)
=(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)
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