同角三角函数的化简求解题思路
同角三角函数的化简求解题思路
我同角三角函数的倒数关系,商数关系,平方关系都会背了 但是三角函数的化简总是 摸不清解题的方法 不能灵活运用公式 怎么样才能理清思路?比如：sin^4α+sin^2αcos^2α+cos^2α 这题很虽然知道各个函数可以怎么化但是却无法将之联系起来 希望有人帮我解答
换元法.
同角三角函数问题,实质上是代数问题.
= = = = = = = = =
例1.化简 (sin α)^4 +(sin α)^2 (cos α)^2 +(cos α)^2.
令 x =(sin α)^2,
则 (cos α)^2 =1-x.
所以 原式=x^2 +x (1-x) +(1-x)
=x^2 +(x -x^2) +(1-x)
=1.
= = = = = = = = =
用这种换元法,注意次数问题.
x =(sin α)^2,
则 (sin α)^4 =x^2,而不是 x^4.
= = = = = = = = =
例2.求证：(tan x)^2 -(sin x)^2 =(tan x)^2 (sin x)^2.
证明：令 u =(sin x)^2,
则 (cos x)^2 =1-u,
(tan x)^2 =u /(1-u).
所以 左边= ...
右边= ...
......
= = = = = = = = =
例3.求证：1 +3 (sin x)^2 (sec x)^4 =(sec x)^6 -(tan x)^6.
证明：令 u =(sin x)^2,
则 (cos x)^2 =1-u,
(sec x)^2 =1/(1-u),
(tan x)^2 =u/(1-u).
所以 左边= 1 +3u/(1-u)^2
= (1+u+u^2) /(1-u)^2.
右边= 1/(1-u)^3 -(u^3)/(1-u)^3
=(1 -u^3) /(1-u)^3
=(1-u) (1+u+u^2) / (1-u)^3
= (1+u+u^2) /(1-u)^2.
所以 左边 =右边.
所以 1 +3 (sin x)^2 (sec x)^4 =(sec x)^6 -(tan x)^6.
= = = = = = = = =
注意次数.
(sec x)^4 =1/(1-u)^2,
(sec x)^6 =1/(1-u)^3,
(tan x)^6 =(u^3) /(1-u)^3.