三角函数求解Sin(π/4-3x)×cos(π/3-3x)-cos(π/6+3x)×sin(π/4+3x) 不好意思我不
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三角函数求解
Sin(π/4-3x)×cos(π/3-3x)-cos(π/6+3x)×sin(π/4+3x)
不好意思我不会打分号就用斜杠代替
Sin(π/4-3x)×cos(π/3-3x)-cos(π/6+3x)×sin(π/4+3x)
不好意思我不会打分号就用斜杠代替
π/4-3x + π/4+3x = π/2
π/3-3x + π/6+3x = π/2
所以
sin(π/4-3x)*cos(π/3-3x)-cos(π/6+3x)*sin(π/4+3x)
=sin(π/4-3x)*cos(π/3-3x)-cos(π/4-3x)*sin(π/3-3x)
=sin(π/4-3x-(π/3-3x))
=sin(-π/12)
=-sin(π/12)
=-(√6-√2)/4
π/3-3x + π/6+3x = π/2
所以
sin(π/4-3x)*cos(π/3-3x)-cos(π/6+3x)*sin(π/4+3x)
=sin(π/4-3x)*cos(π/3-3x)-cos(π/4-3x)*sin(π/3-3x)
=sin(π/4-3x-(π/3-3x))
=sin(-π/12)
=-sin(π/12)
=-(√6-√2)/4
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