Question

数学题求解！！！！

Answer 1

sinB=-5/13 -π/2<B<0
∴cosb=12/13cos(a-b)=3/5
∵0<A<π/2,-π/2<B<0
∴0<a-b<π/2
∴sin(a-b)=4/5 分解得出sina*cosb-cosa*sinb=4/5 12/13 sina +5/13 cosa=4/55cosa=52/5 -12sina两边平方化简得出 (cos^2a=1-sin^2a)169sin^2a-1248/5 sina +2079/25=0
解方程 sina=（1248/5±78）/（169*2）=(1248±390)/1690
∵0<A<π/2 -π/2<B<0 COS(A-B)=3/5
∴0<A-B<π/2
∴sina=(1248-390)/1690=33/65

追问

其实还是有点乱。。我慢慢看好了。 还能帮我解一下下一题吗？

追答

①∵向量OP*向量OQ
=1/2sin²θ-cos²θ
=1/4*(1-cos2θ)-1/2(1+cos2θ)
=3/4cos2θ-1/2=-1/2
∴cos2θ=0(sin2θ=1)
②｜op｜=√(1/4+cos⁴θ),｜OQ｜=√(sin⁴θ+1)
sin(α+β)=sinαcosβ+cosαsinβ
=[(sin²θ*cos²θ)+1/2]/√(1/4+cos⁴θ)√(1+sin⁴θ)
=1/4sin²2θ+1/2/√[1/4+1/4(1+cos2θ)²]√[1+1/4(1-cos2θ)²]
=1/4+1/2/√(1/2)√(5/4)
=(3/4 )/ √5/8
=3√10/10