求证:(1)2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2 (2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]求证:(1) 2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2(2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]
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![求证:(1)2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2 (2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]求证:(1) 2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2(2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]](/uploads/image/z/8883506-2-6.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%892+%EF%BC%881-sinA%EF%BC%89%281%2BcosA%29%3D%281-sinA%2BcosA%29%5E2+%282%29%5BtanAsinA%5D%2F%5BtanA-sinA%5D%3D%5BtanA%2BsinA%5D%5BtanAsinA%5D%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89+2+%EF%BC%881-sinA%EF%BC%89%281%2BcosA%29%3D%281-sinA%2BcosA%29%5E2%282%29%5BtanAsinA%5D%2F%5BtanA-sinA%5D%3D%5BtanA%2BsinA%5D%5BtanAsinA%5D)
求证:(1)2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2 (2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]求证:(1) 2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2(2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]
求证:(1)2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2 (2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]
求证:(1) 2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2
(2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]
求证:(1)2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2 (2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]求证:(1) 2 (1-sinA)(1+cosA)=(1-sinA+cosA)^2(2)[tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA]
(1)(1-sinA+cosA)^2=(1-sinA)^2+2(1-sinA)cosA+(cosA)^2
=(1-sinA)^2+2(1-sinA)cosA+1-(sinA)^2
=(1-sinA)^2+2(1-sinA)cosA+(1-sinA)(1+sinA)
=(1-sinA)(1-sinA+2cosA+1+sinA)
=2(1-sinA)(1+cosA)
(2) [tanAsinA]/[tanA-sinA]=[tanA+sinA][tanAsinA] 等价于(tanAsinA)^2=(tanA)^2-(sinA)^2
(tanA)^2-(sinA)^2=(tanA)^2-(tanAcosA)^2=(tanA)^2[1-(cosA)^2]=(tanAsinA)^2