Question

第二题，求详细过程！

Answer 1

(1)
sinA=3/5, cosA=√[1-(3/5)²]=4/5sinB=5/13, cosB=√[1-(5/13)²]=12/13
13sin(A-B) = sinAcosB-cosAsinB=(3/5)(12/13)-(4/5)(5/13)=36/65-20/65=16/65
(2)
sinx=(√5-1)/2, sin²x=(3-√5)/2, 2sin²x=(3-√5),
sin2(x-π/4)=sin(2x-π/2)=-cos2x=-1+2sin²x=-1+(3-√5)=2-√5

(3)
tan12°+tan33°=tan(12°+33°)(1-tan12°tan33°)=1-tan12°tan33°+tan12°tan33°=1
(4)
2sin75°sin15°
=2sin75°sin(90°-75°)
=2sin75°cos75°
=sin(2x75°)
=sin150°
=sin(180°-30°)
=sin30°
=1/2

Answer 2

嗯