求 tan 31pi/3 的值
三角学是数学中的一个领域,它处理三角形中边的比率及其角度。通过使用三角形的边及其对应角的各种关系和恒等式,可以更精确、更轻松地解决和计算许多问题(距离、高度等)。三角形的角和边之间有几个标准比率或关系,有助于解决一些基本问题和复杂问题。
三角比
三角比定义为直角三角形的边与角的比例。标准三角比可以作为直角三角形的边与任一锐角的比值来获得。正弦、余弦、正切等一些标准三角比的一些定义如下,
- 正弦函数是以角度 θ 为参数的函数,角度 θ 是直角三角形中的任一锐角,定义为直角三角形对边的长度与斜边的比值。用技术术语来说,它可以写成如下,
sin(θ) = opposite side / hypotenuse
- 余弦函数是以角度 θ 为参数的函数,角度 θ 是直角三角形中的任一锐角,定义为直角三角形相邻边的长度与斜边的比值。用技术术语来说,它可以写成如下,
cos(θ) = adjacent side / hypotenuse
- 正切函数是以角度 θ 为参数的函数,它是直角三角形中的锐角之一,定义为直角三角形的对边与相邻边的长度之比.用技术术语来说,它可以写成如下,
tan(θ) = opposite side / adjacent side
这些三角比使用一些三角恒等式和公式相互关联,
tan(θ) = sin(θ) / cos(θ)
罪2 (θ) + cos 2 (θ) = 1
每个三角比都有其他三个导出的三角比,这些三角比是通过取各自比率的倒数来推导出的。其他三个三角比是余割、正割和余切,在数学上用作 cosec、sec 和 cot。这些与主要三角比率有关,如下所示,
cosec(θ) = 1 / sin(θ)
sec(θ) = 1 / cos(θ)
cot(θ) = 1 / tan(θ) = cos(θ) / sin(θ)
下面是一些与标准三角比和派生三角比相关的恒等式,
tan 2 (θ) + 1 = sec 2 (θ)
婴儿床2 (θ) + 1 = cosec 2 (θ)
三角表
下表列出了一些常用角度和基本三角比。三角函数中每个角度的值是固定的且已知的,但提到的更常见且最常用,Ratio\Angle(θ) 0° 30° 45° 60° 90° sin(θ) 0 1/2 1/√2 √3/2 1 cosθ) 1 √3/2 1/√2 1/2 0 tan(θ) 0 1/√3 1 √3 ∞ cosec(θ) ∞ 2 √2 2/√3 1 sec(θ) 1 2/√3 √2 2 ∞ cot(θ) ∞ √3 1 1/√3 0
除了直角三角形之外,还有一些其他的三角比率可以应用:
sin(-θ) = – sin(θ)
cos(-θ) = cos(θ)
tan(-θ) = – tan(θ)
补角和补角
为补角和补角定义了某些公式。互补角是加起来形成90°或π/2 弧度的角度。我们可以形成这样的角度,并根据三角比找到等效的角度。
补角是加起来形成180°或π 弧度的角度。我们可以形成这样的角度,并根据三角比找到等效的角度。
从 90° (π/2 弧度) 中减去一个适当的角度或在 180° (π 弧度) 上加上一个角度以获得互补角。在 90° (π/2 弧度) 上加上一个适当的角度或从 180° (π 弧度) 中减去一个角度以获得一对补角。实际角度可以在三角比的函数中调整,以形成互补角或补角,然后根据下面给出的公式列表评估推导的三角比。补角和补角有一些三角比,
- sin(nπ/2 + θ) = cos(θ) 或 sin(n90° + θ) = cos(θ)
- sin(nπ/2 – θ) = cos(θ) 或 sin(n × 90° – θ) = cos(θ)
- cos(nπ/2 + θ) = -sin(θ) 或 cos(n × 90°+ θ) = -sin(θ)
- cos(nπ/2 – θ) = sin(θ) 或 cos(n × 90° – θ) = sin(θ)
- tan(nπ/2 + θ) = -cot(θ) 或 tan(n × 90° + θ) = -cot(θ)
- tan(nπ/2 – θ) = cot(θ) 或 tan(n × 90° – θ) = cot(θ)
- sin(nπ + θ) = -sin(θ) 或 sin(n × 180° + θ) = -sin(θ)
- sin(nπ – θ) = sin(θ) 或 sin(n × 180° – θ) = sin(θ)
- cos(nπ + θ) = -cos(θ) 或 cos(n × 180° + θ) = -cos(θ)
- cos(nπ – θ) = -cos(θ) 或 cos(n × 180° – θ) = -cos(θ)
- tan(nπ + θ) = tan(θ) 或 tan(n × 180° + θ) = tan(θ)
- tan(nπ – θ) = -tan(θ) 或 sin(n × 180° – θ) = -tan(θ)
- sin(3nπ/2 + θ) = -cos(θ) 或 sin(n × 270° + θ) = -cos(θ)
- sin(3nπ/2 – θ) = -cos(θ) 或 sin(n × 270° – θ) = -cos(θ)
- cos(3nπ/2 + θ) = sin(θ) 或 cos(n × 270° + θ) = sin(θ)
- cos(3nπ/2 – θ) = -sin(θ) 或 cos(n × 270° – θ) = -sin(θ)
- tan(3nπ/2 – θ) = cot(θ) 或 tan(n × 270° + θ) = cot(θ)
- tan(3nπ/2 + θ) = -cot(θ) 或 tan(n × 270° – θ) = -cot(θ)
- sin(2nπ + θ) = sin(θ) 或 sin(n × 360° + θ) = sin(θ)
- sin(2nπ – θ) = -sin(θ) 或 sin(n × 360° – θ) = -sin(θ)
- cos(2nπ + θ) = cos(θ) 或 cos(n × 360°+ θ) = cos(θ)
- cos(2nπ – θ) = cos(θ) 或 cos(n × 360° – θ) = cos(θ)
- tan(2nπ + θ) = tan(θ) 或 tan(n × 360°+ θ) = tan(θ)
- tan(2nπ – θ) = -tan(θ) 或 tan(n × 360° – θ) = -tan(θ)
一些更重要的公式,
- tan(A + B) = [tan(A) + tan(B)] / [1 – (tan(A)tan(B))]
- tan(A – B) = [tan(A) – tan(B)] / [1 + (tan(A)tan(B))]
求 tan 31π/3 的值
方法一
tan(31π/3)
It can be written 31π/3 as (10π + π/3),
Thus,
tan(31π/3) = tan(10π + π/3)
= tan [ 5(2π) + π/3 ]
tan(2nπ + θ) = tan(θ)
Thus,
tan(31π/3) = tan [ 5(2π) + π/3 ]
= tan(π/3)
= √3
Therefore,
tan(31π/3) = √3
方法二
tan(31π/3)
It can be written, 31π/3 as (21π/2 – π/6),
Thus,
tan(31π/3) = tan [ 21π/2 – π/6 ]
= tan [ 7(3π/2) – π/6 ]
tan(3nπ/2 – θ ) = cot ( θ )
Thus,
tan(31π/2) = tan [ 7(3π/2) – π/6 ]
= cot ( π/6 )
= √3
Therefore,
tan(31π/3) = √3
方法三
31π/3 can be written as 14π/3 + 17π/3,
Now,
14π/3 = 9π/2 + π/6 and,
17π/3 = 11π/2 + π/6
Therefore,
tan(31π/3) = tan [ (9π/2 + π/6 ) + (11π/2 + π/6) ]
tan(A + B) = [tan(A) + tan(B)] / [1 – (tan(A).tan(B))]
Here, A= 9π/2 + π/6 and B = 11π/2 + π/6,
Thus, tan(A) = tan(9π/2 + π/6 ) = tan(3(3π/2) + π/6 )
tan(B) = tan(11π/2 + π/6) = tan(11(π/2) + π/6 )
tan(3nπ/2 + θ) = -cot(θ)
tan(nπ/2 + θ) = -cot(θ)
Therefore,
tan(A) = tan((9π/2 + π/6) = -cot(π/6) = -√3
tan(B) = tan((11π/2 + π/6)= -cot(π/6) = -√3
Now,
tan(31π/3) = tan [ (9π/2 + π/6 ) + (11π/2 + π/6) ]
= [ tan(9π/2 + π/6 ) + tan(11π/2 + π/6) ] / [ 1 – (tan((9π/2 + π/6).tan(11π/2 + π/6) ]
= [ (-√3) + (-√3) ] / [ 1 – (-√3).(-√3) ]
= [ (-2√3 ] / [ 1 – 3 ]
= [ -2√3 ] / [ -2 ]
= √3
Therefore,
tan(31π/3) = √3
因此 tan(31π/3) 的值是 √3,几乎是 1.732……
类似问题
问题一:求tan(5π/4)的值
解决方案:
5π/4 can be written as π + π/4
Thus, tan(5π/4) = tan(π + π/4)
tan(π + θ) = tan(θ) Thus,
tan(5π/4) = tan( π + π/4)
= tan( π/4)
=1
tan(5π/4) = 1
问题 2:求 tan(π/12) 的值
解决方案:
π/12 can be written as π/4 – π/6,
Thus, tan(π/12) = tan (π/4 – π/6)
tan(A + B) = [tan(A) + tan(B)] / [1 – (tan(A).tan(B))]
Here,
A = π/4 and B = π/6 Thus,
tan(π/12) = tan (π/4 – π/6)
= [tan (π/4) – tan(π/6)] / [1 – (tan(π/4).tan(π/6))]
= [1 – 1/√3] / [1 – ((1).(1/√3))]
= [√3 -1] / [(√3 -1) / √3]
= (√3 -1) √3 / (√3 – 1)
= (3 -√3) / (√3 -1)
Therefore,
tan(π/12) = (3 – √3) / (√3 – 1)
问题 3:求 tan(300°)
解决方案:
We can write 300° = 270° + 30°
Note Here, angles are used in degree and not in radians
tan(300°) = tan(270° + 30°)
tan(270° + θ) = -cot(θ) Thus,
tan(300°) = tan(270° + 30°)
= -cot(30°)
= -√3
Thus,
tan(300°) = -√3