oneloner
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设arcsin(3/5)=α∈(0,π/2)
,则sinα=3/5,cosα=4/5
设arcsin(3/10）=β∈（0,π/2）
则sinβ=3/10,cosβ=√91/10
∴4*sin(arcsin(3/5)-arcsin(3/10))
=4sin(α-β)
=4(sinαcosβ-cosαsinβ)
=4(3/5*√91/10-4/5*3/10)
=4(3√91-12)/50
=(6√91-24)/25
1年前
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