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例1 若acosθ+bsinθ=c(1) dcosθ+esinθ=f(2)求证(ce-bf)~2+(af-ed)~2=(ae-bd)~2(3) 其中ae-bd≠0。对于此题,欲证(3)成立,只要从(1)、(2)中消去参数θ即可。具体作法是 (1)×d-(2)×a得 sinθ=af-ed/ae-bd, (1)×e-(2)×b得 cosθ=af-ed/ae-bd代入恒等式Sin~2θ+COS~2θ=1,即得(3)。这种方法是众所周知的,而有时要想从关于f(sinθ,cosθ)的条件等式中,直接解出sinθ,Cosθ,然后利用sin~2θ+cos~2θ=1去消参就相当困难,甚至是不可能的,因此必须另辟途
Example 1 If acosθ+bsinθ=c(1) dcosθ+esinθ=f(2) Verification (ce-bf)~2+(af-ed)~2=(ae-bd)~2(3) where ae-bd ≠ 0. For this question, if the testimony (3) holds, it is sufficient to eliminate the parameter θ from (1) and (2). The specific method is (1) × d - (2) × a sin θ = af-ed / ae-bd, (1) × e- (2) × b cos θ = af-ed / ae-bd substituted into the identity Sin ~ 2θ+COS~2θ=1, that is, (3). This method is well-known, and sometimes it is quite difficult to solve the conditional equations for f(sinθ,cosθ) directly from sinθ, Cosθ, and then use sin~2θ+cos~2θ=1 to eliminate the parameters. It is not even possible and therefore must be provided