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有些代数问题,如用纯代数方法求解往往比较困难,但通过适当的换元,变成三角问题求解,不但可以简化书写过程,而且能使数量系明朗化,从而化难为易,找到解决问题的途经。代数问题进行三角代换,关键在于熟悉三角函数的性质和一些重要大系式。下面归类举例说明: 一形如x~2+y~2=1,x+y=1(x,y为正数),可设x=sina,y=cosa 或者x=sin~2a,y=cos~2a。例1 已知a~2+b~2=1,c~2+d~2=1,求证|2abd+(a~2-b~3)c|≤1 证明:因为 a~2+b~2=1,c~2+a~2=1,故可设=sina,则b=±cosa,又令C=sinβ,则d=±cosβ而有 |2abd+(a~2-b~2)c|=|2sina(±cosa)(±cosB)
Some algebra problems, such as using pure algebraic methods, are often difficult to solve, but through appropriate substitutions, triangulation problems can be solved, not only can simplify the writing process, but also can make the number of systems clear, thus making it difficult to find solutions to problems. via. The key to the trigonometric substitution of algebraic problems lies in the familiarity with the nature of the trigonometric functions and some important major systems. The following categorization examples illustrate: A shape such as x~2+y~2=1, x+y=1 (x, y is a positive number), can be set x=sina,y=cosa or x=sin~2a,y =cos~2a. Example 1 It is known that a~2+b~2=1, c~2+d~2=1. Prove that |2abd+(a~2-b~3)c|≤1 prove: because a~2+b~2 =1,c~2+a~2=1, so we can set =sina, b=±cosa, and let C=sinβ, then d=±cosβ and |2abd+(a~2-b~2)c |=|2sina(±cosa)(±cosB)