高一数学题求过程。谢谢。
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根据公式cos(α-β)=cosαcosβ+sinαsinβ, cos(α+β)=cosαcosβ-sinαsinβ
sin(45°+a)sin(45°-a)
={cos[(45°+a )-(45°-a)]- cos[(45°+a )+(45°-a)]}/2
=[cos(2a)- cos90°]/2
=cos(2a)/2
=-1/4
(1)∵0°<a<90°
∴2a=120°即a=60°
(2)根据公式sin(α+β)= sinαcosβ+cosαsinβ, sin (α-β)= sinαcosβ- cosαsinβ得
sin(a+10°)(1-(根号3)*tan(a-10°))
=sin70°[1-(根号3)*tan50°]
=sin70°cos50°[cos50°-(根号3)*sin50°]
=2sin70°cos50°[1/2cos50°-(根号3)/2*sin50°]
=[sin(70°+50°)+ sin(70°-50°)] [sin30°cos50°- cos30°sin50°]
=(sin120°+ sin20°)sin(-20°)
=-(根号3)/2*sin20°-1/2sin40°
sin(45°+a)sin(45°-a)
={cos[(45°+a )-(45°-a)]- cos[(45°+a )+(45°-a)]}/2
=[cos(2a)- cos90°]/2
=cos(2a)/2
=-1/4
(1)∵0°<a<90°
∴2a=120°即a=60°
(2)根据公式sin(α+β)= sinαcosβ+cosαsinβ, sin (α-β)= sinαcosβ- cosαsinβ得
sin(a+10°)(1-(根号3)*tan(a-10°))
=sin70°[1-(根号3)*tan50°]
=sin70°cos50°[cos50°-(根号3)*sin50°]
=2sin70°cos50°[1/2cos50°-(根号3)/2*sin50°]
=[sin(70°+50°)+ sin(70°-50°)] [sin30°cos50°- cos30°sin50°]
=(sin120°+ sin20°)sin(-20°)
=-(根号3)/2*sin20°-1/2sin40°
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