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二直线重合的条件在解几中已有广泛的应用,下面举几个三角方面的例子: 例1 消去θ acosθ+bsinθ=c, acos3θ+bsin3θ=c. 解:设直线ax+by-c=0 ①显然,点(cosθ,sinθ)、(coc3θ,sin3θ)在此直线上,又过这二点的直线方程可写成 (y-sinθ)/(x-cosθ)=(sinθ-sin3θ)/(cosθ-cos3θ),即cos2θ·x+sin2θ·y-cosθ=0 ②由于①、②为同一直线故可得a/cos2θ=b/sin2θ=c/cosθ,∴a~2/cos~22θ=b~/sin~22θ=c~2/cos~2θ,∴(a~2+b~2-2c~2)~2=a~2(a~2+b~2).
The conditions for the two straight lines to overlap are widely used in several solutions. Here are some trigonometric examples: Example 1 Eliminating θ acosθ+bsinθ=c, acos3θ+bsin3θ=c. Solution: Set the line ax+by-c= Obviously, the points (cosθ, sinθ), (coc3θ, sin3θ) are on this straight line, and the straight line equations that pass these two points can be written as (y-sinθ)/(x-cosθ)=(sinθ-sin3θ)/( Cosθ-cos3θ), ie cos2θ·x+sin2θ·y-cosθ=0 2 Since 1 and 2 are the same line, a/cos2θ=b/sin2θ=c/cosθ, ∴a~2/cos~22θ=b ~/sin~22θ=c~2/cos~2θ, ∴(a~2+b~2-2c~2)~2=a~2(a~2+b~2).