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三角恒等交换的题目
已知x+y=√2sin(a+n/4),x-y=√2sin(a-n/4),求证x^2+y^2=1
人气:377 ℃ 时间:2020-04-27 20:07:58
解答
x+y=√2sin(a+n/4)
(x+y)^2=2sin(a+n/4)
x^2+2xy+y^2=2sin(a+n/4)(1)
x-y=√2sin(a-n/4)
(x-y)^2=2sin(a-n/4)
x^2-2xy+y^2=2sin(a-n/4)(2)
(1)+(2),并化简得,x^2+y^2=2sina*cos(n/4)
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