Question

同角的三角函数解题思路

Answer 1

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.
解：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.