√2sina=-√3cosa,求√2cos(2a-π/4)
问题描述：
√2sina=-√3cosa,求√2cos(2a-π/4)
答
√2sina=-√3cosa
sina=[-√(3/2)]cosa
sin²a+cos²a=1
(3/2)cos²a+cos²a=1
cos²a=2/5
sin(2a)=2sinacosa=2[-(√3/2)cosa]cosa=-√6cos²a=-2√6/5
cos(2a)=2cos²a-1=4/5 -1=-1/5
√2cos(2a-π/4)
=√2[cos(2a)cos(π/4)+sin(2a)sin(π/4)]
=√2[cos(2a)(√2/2)+sin(2a)(√/2/2)]
=cos(2a)+sin(2a)
=-1/5-2√6/5
=-(1+2√6)/5