反三角恒等式
在数学中,反三角函数也称为弧函数或反三角函数。反三角函数是基本三角函数的反函数,即正弦、余弦、正切、余割、正割和余切。它用于找到具有任何三角比的角度。反三角函数一般用于几何、工程等领域。反三角函数的表示形式有:
如果 a = f(b),则反函数为
b = f-1(a)
反三角函数的例子有 sin -1 x、cos -1 x、tan -1 x 等。
下表显示了一些三角函数及其域和范围。
Function | Domain | Range |
y = sin-1 x | [-1, 1] | [-π/2, π/2] |
y = cos-1 x | [-1, 1] | [0 , π] |
y = cosec-1 x | R – (-1,1 ) | [-π/2, π/2] – {0} |
y = sec-1 x | R – (-1, 1) | [0 , π] – {π/2} |
y = tan-1 x | R | (-π/2, π/2) |
y = cot-1 x | R | (0 , π) |
反三角函数的性质
以下是反三角函数的性质:
属性一:
- sin -1 (1/x) = cosec -1 x,对于 x ≥ 1 或 x ≤ -1
- cos -1 (1/x) = sec -1 x,对于 x ≥ 1 或 x ≤ -1
- tan -1 (1/x) = cot -1 x,对于 x > 0
属性 2:
- sin -1 (-x) = -sin -1 x, 对于 x ∈ [-1 , 1]
- tan -1 (-x) = -tan -1 x,对于 x ∈ R
- cosec -1 (-x) = -cosec -1 x,对于 |x| ≥1
财产 3
- cos -1 (-x) = π – cos -1 x, 对于 x ∈ [-1 , 1]
- 秒-1 (-x) = π – 秒-1 x,对于 |x| ≥1
- cot -1 (-x) = π – cot -1 x,对于 x ∈ R
属性 4
- sin -1 x + cos -1 x = π /2,对于 x ∈ [-1,1]
- tan -1 x + cot -1 x = π /2,对于 x ∈ R
- cosec -1 x + sec -1 x = π /2 , 对于 |x| ≥1
财产 5
- tan -1 x + tan -1 y = tan -1 ( x + y )/(1 – xy),对于 xy < 1
- tan -1 x – tan -1 y = tan -1 (x – y)/(1 + xy),对于 xy > -1
- tan -1 x + tan -1 y = π + tan -1 (x + y)/(1 – xy),对于 xy >1 ; x, y >0
财产 6
- 2tan -1 x = sin -1 (2x)/(1 + x 2 ),对于 |x| ≤ 1
- 2tan -1 x = cos -1 (1 – x 2 )/(1 + x 2 ),对于 x ≥ 0
- 2tan -1 x = tan -1 (2x)/(1 – x 2 ),对于 -1 < x <1
函数的恒等式
以下是反三角函数的恒等式:
- sin -1 (sin x) = x 提供 – π /2 ≤ x ≤ π /2
- cos -1 (cos x) = x 提供 0 ≤ x ≤ π
- tan -1 (tan x) = x 提供 – π /2 < x < π /2
- sin(sin -1 x) = x 提供 -1 ≤ x ≤ 1
- cos(cos -1 x) = x 提供 -1 ≤ x ≤ 1
- tan(tan -1 x) = x 假设 x ∈ R
- cosec(cosec -1 x) = x 提供 -1 ≤ x ≤ ∞ 或 -∞ < x ≤ 1
- sec(sec -1 x) = x 提供 1 ≤ x ≤ ∞ 或 -∞ < x ≤ 1
- cot(cot -1 x) = x 提供 -∞ < x < ∞
- 2cos -1 x = cos -1 (2x 2 – 1)
- 2sin -1 x = sin -1 2x√(1 – x 2 )
- 3sin -1 x = sin -1 (3x – 4x 3 )
- 3cos -1 x = cos -1 (4x 3 – 3x)
- 3tan -1 x = tan -1 ((3x – x 3 /1 – 3x 2 ))
- sin -1 x + sin -1 y = sin -1 { x√(1 – y 2 ) + y√(1 – x 2 )}
- sin -1 x – sin -1 y = sin -1 { x√(1 – y 2 ) – y√(1 – x 2 )}
- cos -1 x + cos -1 y = cos -1 [xy – √{(1 – x 2 )(1 – y 2 )}]
- cos -1 x – cos -1 y = cos -1 [xy + √{(1 – x 2 )(1 – y 2 )}
- 棕褐色-1 x + 棕褐色-1 y = 棕褐色-1 (x + y/1 – xy)
- 棕褐色-1 x – 棕褐色-1 y = 棕褐色-1 (x – y/1 + xy)
- tan -1 x + tan -1 y +tan -1 z = tan -1 (x + y + z – xyz)/(1 – xy – yz – zx)
示例问题
问题 1:证明 sin -1 x = sec -1 1/√(1-x 2 )
解决方案:
Let sin-1 x = y
⇒ sin y = x , (since sin y = perpendicular/hypotenuse ⇒ cos y = √(1- perpendicular2 )/hypotenuse )
⇒ cos y = √(1 – x2), here hypotenuse = 1
⇒ sec y = 1/cos y
⇒ sec y = 1/√(1 – x2)
⇒ y = sec-1 1/√(1 – x2)
⇒ sin-1 x = sec-1 1/√(1 – x2)
Hence, proved.
问题 2:证明 tan -1 x = cosec -1 √(1 + x 2 )/x
解决方案:
Let tan-1 x = y
⇒ tan y = x , perpendicular = x and base = 1
⇒ sin y = x/√(x2 + 1) , (since hypotenuse = √(perpendicular2 + base2 ) )
⇒ cosec y = 1/sin y
⇒ cosec y = √(x2 + 1)/x
⇒ y = cosec-1 √(x2 + 1)/x
⇒ tan-1 x = cosec-1 √(x2 + 1)/x
Hence, proved.
问题 3:计算 tan(cos -1 x)
解决方案:
Let cos-1 x = y
⇒ cos y = x , base = x and hypotenuse = 1 therefore sin y = √(1 – x2)/1
⇒ tan y = sin y/ cos y
⇒ tan y = √(1 – x2)/x
⇒ y = tan-1 √(1 – x2)/x
⇒ cos-1 x = tan-1 √(1 – x2)/x
Therefore, tan(cos-1 x) = tan(tan-1 √(1 – x2)/x ) = √(1 – x2)/x.
问题 4:tan -1 √(sin x) + cot -1 √(sin x) = y。找到 cos y。
解决方案:
We know that tan-1 x + cot-1 x = /2 therefore comparing this identity with the equation given in the question we get y = π/2
Thus, cos y = cos π/2 = 0.
问题 5:tan -1 (1 – x)/(1 + x) = (1/2)tan -1 x, x > 0。求解 x。
解决方案:
tan-1 (1 – x)/(1 + x) = (1/2)tan-1 x
⇒ 2tan-1 (1 – x)/(1 + x) = tan-1 x …(1)
We know that, 2tan-1 x = tan-1 2x/(1 – x2).
Therefore, LHS of equation (1) can be written as
tan-1 [ { 2(1 – x)/(1 + x)}/{ 1 – [(1 – x)(1 + x)]2}]
= tan-1 [ {2(1 – x)(1 + x)} / { (1 + x)2 – (1 – x)2 }]
= tan-1 [ 2(1 – x2)/(4x)]
= tan-1 (1 – x2)/(2x)
Since, LHS = RHS therefore
tan-1 (1 – x2)/(2x) = tan-1 x
⇒ (1 – x2)/2x = x
⇒ 1 – x2 = 2x2
⇒ 3x2 = 1
⇒ x = ± 1/√3
Since, x must be greater than 0 therefore x = 1/√3 is the acceptable answer.
问题 6:证明 tan -1 √x = (1/2)cos -1 (1 – x)/(1 + x)
解决方案:
Let tan-1 √x = y
⇒ tan y = √x
⇒ tan2 y = x
Therefore,
RHS = (1/2)cos-1 ( 1- tan2 y)/(1 + tan2 y)
= (1/2)cos-1 (cos2 y – sin2 y)/(cos2 y + sin2 y)
= (1/2)cos-1 (cos2 y – sin2 y)
= (1/2)cos-1 (cos 2y)
= (1/2)(2y)
= y
= tan-1 √x
= LHS
Hence, proved.
问题 7:tan -1 (2x)/(1 – x 2 ) + cot -1 (1 – x 2 )/(2x) = π /2, -1 < x < 1。求解 x。
解决方案:
tan-1 (2x)/(1 – x2) + cot-1 (1 – x2)/(2x) = π/2
⇒ tan-1 (2x)/(1 – x2) + tan-1 (2x)/(1 – x2) = π/2
⇒ 2tan-1 (2x)/(1 – x2) = ∏/2
⇒ tan-1 (2x)/(1 – x2) = ∏/4
⇒ (2x)/(1 – x2) = tan ∏/4
⇒ (2x)/(1 – x2) = 1
⇒ 2x = 1 – x2
⇒ x2 + 2x -1 = 0
⇒ x = [-2 ± √(22 – 4(1)(-1))] / 2
⇒ x = [-2 ± √8] / 2
⇒ x = -1 ± √2
⇒ x = -1 + √2 or x = -1 – √2
But according to the question x ∈ (-1, 1) therefore for the given equation the solution set is x ∈ ∅.
问题 8:tan -1 1/(1 + 1.2) + tan -1 1/(1 + 2.3) + … + tan -1 1/(1 + n(n + 1)) = tan -1 x。求解 x。
解决方案:
tan-1 1/(1 + 1.2) + tan-1 1/(1 + 2.3) + … + tan-1 1/(1 + n(n + 1)) = tan-1 x
⇒ tan-1 (2 – 1)/(1 + 1.2) + tan-1 (3 – 2)/(1 + 2.3) + … + tan-1 (n + 1 – n)/(1 + n(n + 1)) = tan-1 x
⇒ (tan-1 2 – tan-1 1) + (tan-1 3 – tan-1 2) + … + (tan-1 (n + 1) – tan-1 n) = tan-1 x
⇒ tan-1 (n + 1) – tan-1 1 = tan-1 x
⇒ tan-1 n/(1 + (n + 1).1) = tan-1 x
⇒ tan-1 n/(n + 2) = tan-1 x
⇒ x = n/(n + 2)
问题 9:如果 2tan -1 (sin x) = tan -1 (2sec x) 那么求解 x。
解决方案:
2tan-1 (sin x) = tan-1 (2sec x)
⇒ tan-1 (2sin x)/(1 – sin2 x) = tan-1 (2/cos x)
⇒ (2sin x)/(1 – sin2 x) = 2/cos x
⇒ sin x/cos2 x = 1/cos x
⇒ sin x cos x = cos2 x
⇒ sin x cos x – cos2 x = 0
⇒ cos x(sin x – cos x) = 0
⇒ cos x = 0 or sin x – cos x = 0
⇒ cos x = cos π/2 or tan x = tan π/4
⇒ x = π/2 or x = π/4
But at x = π/2 the given equation does not exist hence x = π/4 is the only solution.
问题 10:证明 cot -1 [ {√(1 + sin x) + √(1 – sin x)}/{√(1 + sin x) – √(1 – sin x)}] = x/2, x ∈ (0, π /4)
解决方案:
Let x = 2y therefore
LHS = cot-1 [{√(1+sin 2y) + √(1-sin 2y)}/{√(1+sin 2y) – √(1-sin 2y)}]
= cot-1 [{√(cos2 y + sin2 y + 2sin y cos y) + √(cos2 y + sin2 y – 2sin y cos y)}/{√(cos2 y + sin2 y + 2sin y cos y) – √(cos2 y + sin2 y – 2sin y cos y)} ]
= cot-1 [{√(cos y + sin y)2 + √(cos y – sin y)2} / {√(cos y + sin y)2 – √(cos y – sin y)2}]
= cot-1 [( cos y + sin y + cos y – sin y )/(cos y + sin y – cos y + sin y)]
= cot-1 (2cos y)/(2sin y)
= cot-1 (cot y)
= y
= x/2.