以知兀/4<α<3兀/4,0<β<兀/4,cos(兀/4+α)=-3/5,sin(3兀/4+β)=5/13,求sin(β+α)
人气：203 ℃ 时间：2020-03-29 18:19:03
解答
cos(兀/4+α)=-3/5.cosα-sinα=-3√2/5.平方得,sinαcosα=7/50.配方得,cosα+sinα=4√2/5.∴cosα=√2/10.sinα=7√2/10.同理,从sin(3兀/4+β)=5/13,展开,平方,配方得.cosβ=17√2/26.sinβ=√2/26.sin(β+α)＝7...
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