sin*2π/8+cos*27π/24+sinπ/8cos7π/24
人气:475 ℃ 时间:2020-10-01 19:18:01
解答
sin²π/8+cos²7π/24+sinπ/8cos7π/24
=(1-cosπ/4)/2+(1+cos7π/12)/2+1/2[sin(π/8+7π/24)+sin(π/8-7π/24)]
=1-√2/4+1/2*cos7π/12+1/2[sin5π/12-sinπ/6]
=1-√2/4-1/4+1/2(sin5π/12-cos5π/12)
=3/4-√2/4+√2/2(sin5π/12*√2/2-cos5π/12*√2/2)
=3/4-√2/4+√2/2sin(5π/12-π/4)
=3/4-√2/4+√2/2sinπ/6
=3/4=1-√2/4+1/2*cos7π/12+1/2[sin5π/12-sinπ/6]=1-√2/4-1/4+1/2(sin5π/12-cos5π/12)为什么???=1-√2/4+1/2*cos7π/12+1/2[sin5π/12-sinπ/6]=1-√2/4+1/2*cos(π-5π/12)+1/2*sin5π/12-1/2*1/2=1-√2/2-1/4-1/2cos5π/12+/2*sin5π/12=1-√2/4-1/4+1/2(sin5π/12-cos5π/12) cos7π/12=cos(π-5π/12)=-cos5π/12
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