求解两道三角恒等变换题1.sin(36°+α)sin(54°-α)- cos(36°+α)cos(54°-α)2.α为第三象限角,sinα=-3/5,求sin(37π/6-α)的值已知cosα=-3/5,π<α<2π,求cos(π/6-α)的值.

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求解两道三角恒等变换题
1.sin(36°+α)sin(54°-α)- cos(36°+α)cos(54°-α)
2.α为第三象限角,sinα=-3/5,求sin(37π/6-α)的值已知cosα=-3/5,π<α<2π,求cos(π/6-α)的值.

1,
sin(36°+x)sin(54°-x) -cos(36°+x)cos(54°-x)
=-cos[(36°+x)+(54°-x)]
=cos90°=0.
2
α为第三象限角,且sinα=-3/5 则cosa=-4/5
即sin2a=-24/25 cos2a=7/25
sin(37π/6-2α)的值=sin(π/6-2α)
=0.5*cos2a-根号3/2*sin2a
代进去就是7/50+(12根号3)/25
cos(a - π/6) = cos(a)cos(π/6) + sin(a)sin(π/6) = 3/5 * (根3)/2 + 4/5 * 1/2 = (4 + 3*根3)/10