两点间距离公式证明余弦定理(d^2)=((acos(α)-bcos(β))^2)+((asin(α)-bsin(β))^2)=(a^2)((cos(α))^2)+(b^2)((cos(β))^2)+(a^2)((sin(α))^2)+(b^2)((sin(β))^2)-2abcos(α)cos(β)-2absin(α)sin(β)=(a^2)+(b^2)-2ab(cosαcosβ+sinαsinβ) 可
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![两点间距离公式证明余弦定理(d^2)=((acos(α)-bcos(β))^2)+((asin(α)-bsin(β))^2)=(a^2)((cos(α))^2)+(b^2)((cos(β))^2)+(a^2)((sin(α))^2)+(b^2)((sin(β))^2)-2abcos(α)cos(β)-2absin(α)sin(β)=(a^2)+(b^2)-2ab(cosαcosβ+sinαsinβ) 可](/uploads/image/z/1609316-44-6.jpg?t=%E4%B8%A4%E7%82%B9%E9%97%B4%E8%B7%9D%E7%A6%BB%E5%85%AC%E5%BC%8F%E8%AF%81%E6%98%8E%E4%BD%99%E5%BC%A6%E5%AE%9A%E7%90%86%28d%5E2%29%3D%28%28acos%28%CE%B1%29-bcos%28%CE%B2%29%29%5E2%29%2B%28%28asin%28%CE%B1%29-bsin%28%CE%B2%29%29%5E2%29%3D%28a%5E2%29%28%28cos%28%CE%B1%29%29%5E2%29%2B%28b%5E2%29%28%28cos%28%CE%B2%29%29%5E2%29%2B%28a%5E2%29%28%28sin%28%CE%B1%29%29%5E2%29%2B%28b%5E2%29%28%28sin%28%CE%B2%29%29%5E2%29-2abcos%28%CE%B1%29cos%28%CE%B2%29-2absin%28%CE%B1%29sin%28%CE%B2%29%3D%28a%5E2%29%2B%28b%5E2%29-2ab%EF%BC%88cos%CE%B1cos%CE%B2%2Bsin%CE%B1sin%CE%B2%29+%E5%8F%AF)
两点间距离公式证明余弦定理(d^2)=((acos(α)-bcos(β))^2)+((asin(α)-bsin(β))^2)=(a^2)((cos(α))^2)+(b^2)((cos(β))^2)+(a^2)((sin(α))^2)+(b^2)((sin(β))^2)-2abcos(α)cos(β)-2absin(α)sin(β)=(a^2)+(b^2)-2ab(cosαcosβ+sinαsinβ) 可
两点间距离公式证明余弦定理
(d^2)=((acos(α)-bcos(β))^2)+((asin(α)-bsin(β))^2)
=(a^2)((cos(α))^2)+(b^2)((cos(β))^2)+(a^2)((sin(α))^2)+(b^2)((sin(β))^2)-2abcos(α)cos(β)-2absin(α)sin(β)
=(a^2)+(b^2)-2ab(cosαcosβ+sinαsinβ)
可证到此步骤,以下该怎样证?注明详细步骤.
两点间距离公式证明余弦定理(d^2)=((acos(α)-bcos(β))^2)+((asin(α)-bsin(β))^2)=(a^2)((cos(α))^2)+(b^2)((cos(β))^2)+(a^2)((sin(α))^2)+(b^2)((sin(β))^2)-2abcos(α)cos(β)-2absin(α)sin(β)=(a^2)+(b^2)-2ab(cosαcosβ+sinαsinβ) 可
你的条件错了.
如果α=∠AOY,
应该是A(asinα,acosα)(你写反了)
d²=(asinα-bcosβ)²+(acosα-bsinβ)²
=a²sin²α-2absinαcosβ+b²cos²β+a²cos²α-2abcosαsinβ+b²sin²β
=a²+b²-2ab(sinαcosβ+cosαsinβ)
=a²+b²-2absin(α+β)
=a²+b²-2absin(90°-θ)(∵α+β+θ=90°)
=a²+b²-2abcosθ.