已知x+y=√2 sin(θ+π/4),x—y=√2 sin(θ—π/4),求证:x²+y²=1.
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已知x+y=√2sin(θ+π/4),x—y=√2sin(θ—π/4),求证:x²+y²=1.已知x+y=√2sin(θ+π/4),x—y=√2sin(θ—π/4),求证:x
已知x+y=√2 sin(θ+π/4),x—y=√2 sin(θ—π/4),求证:x²+y²=1.
已知x+y=√2 sin(θ+π/4),x—y=√2 sin(θ—π/4),求证:x²+y²=1.
已知x+y=√2 sin(θ+π/4),x—y=√2 sin(θ—π/4),求证:x²+y²=1.
显然sin(θ+π/4)=sinθ*sinπ/4 +cosθ*cosπ/4,
sin(θ-π/4)=sinθ*cosπ/4 -cosθ*sinπ/4,
而sinπ/4=cosπ/4= √2/2,
故x+y=√2 sin(θ+π/4)=sinθ+cosθ,
x-y=√2 sin(θ-π/4)=sinθ -cosθ,
两式相加,得到x= sinθ,
两式相减,得到y=cosθ,
显然sin²θ+cos²θ=1,
故x²+y²=1
已知x+y=√2 sin(θ+π/4),x—y=√2 sin(θ—π/4),求证:x²+y²=1.
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