题目
设sin(π/4-a/2)cos(π/4-a/2)=1/6,求sin2a
提问时间：2021-03-13
答案
sin(π/4-a/2)cos(π/4-a/2)
=(sinπ/4cosa/2-cosπ/4sina/2)(cosπ/4cosa/2+sinπ/4sina/2)
=√ 2/2 (cosa/2-sina/2) × √ 2/2 (cosa/2+sina/2)
=1/2 (cosa/2-sina/2)(cosa/2+sina/2)
=1/2(cos²a/2-sin²a/2)
=1/2 cosa
=1/6
=> cosa=1/3 =>sina=±(2√2)/3
sin2a=2sinacosa=±(4√2)/9
=(sinπ/4cosa/2-cosπ/4sina/2)(cosπ/4cosa/2+sinπ/4sina/2)
=√ 2/2 (cosa/2-sina/2) × √ 2/2 (cosa/2+sina/2)
=1/2 (cosa/2-sina/2)(cosa/2+sina/2)
=1/2(cos²a/2-sin²a/2)
=1/2 cosa
=1/6
=> cosa=1/3 =>sina=±(2√2)/3
sin2a=2sinacosa=±(4√2)/9
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