求 cos (31π/3) 的值
三角学是标准化数学的一个分支,它处理长度、高度和角度之间的关系。三角学是数学的一个分支,它处理三角形的边和角之间的比率和关系。使用 Trigonometry cab 可以计算连接到三角形的各种测量值。定义了一些标准比率,以便于计算与直角三角形边的长度和角度有关的一些常见问题。
三角比
三角比是直角三角形中任何一个锐角的边的比例。一个简单的三角比可以定义为直角三角形的边,即斜边、底边和垂直边。有三个简单的三角比wiz。正弦、余弦和正切。
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,
sin(θ) = 对边 / 斜边
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,
cos(θ) = 邻边 / 斜边
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,
tan(θ) = 对边 / 邻边
这些三角比使用一些三角恒等式和公式相互关联,这些关系是三角学的基础,并且在许多计算中用于简化,例如,
- 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)
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 cos(θ) 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可以应用,比如负号如果出现在角的前面,可以取出来,也是奇函数的一个例子,所以如果 sin(-θ) = – sin(θ),表示sin是奇函数,但是cos不允许带出符号,是偶函数,
- sin(-θ) = – sin(θ)
- cos(-θ) = cos(θ)
- tan(-θ) = – tan(θ)
如果从 90° 加上或减去一个角度,三角比之间定义了关系,它被称为互补函数的三角比,让我们看看这些方程的广义形式,
- sin(nπ/2 + θ) = cos(θ), sin(nπ/2 – θ) = cos(θ)
- cos(nπ/2 + θ) = -sin(θ), cos(nπ/2 – θ) = sin(θ)
- tan(nπ/2 + θ) = -cot(θ), tan(nπ/2 – θ) = cot(θ)
如果从 180° 增加或减去一个角度,则三角比之间定义了关系,它被称为补充函数的三角比,让我们看看这些方程的广义形式,
- sin(nπ + θ) = -sin(θ), sin(nπ – θ) = sin(θ)
- cos(nπ + θ) = -cos(θ), sin(nπ – θ) = sin(θ)
- tan(nπ + θ) = tan(θ), tan(nπ – θ) = -tan(θ)
正切函数有特殊的三角公式,下面提到的公式对于问题陈述也很重要,
- cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
- cos (A – B) = [cos(A) × cos(B)] + [sin(A) × sin(B)]
求 cos (31π/3) 的值
方法一:
cos(31π/3)
It can be written as (10π + π/3),
Thus,
cos(31π/3) = cos(10π + π/3)
= cos [ 5(2π) + π/3 ]
It is known,
cos(2nπ + θ) = cos(θ)
Thus,
cos(31π/3) = cos [ 5(2π) + π/3 ]
= cos(π/3)
= 1/2
Therefore,
cos(31π/3) = 1/2
方法二:
cos(31π/3)
Write, 31π/3 as (21π/2 – π/6),
Thus,
cos(31π/3) = cos [ 21π/2 – π/6 ]
= cos [ 21(π/2) – π/6 ]
It is known,
cos(nπ/2 – θ ) = sin ( θ )
Thus,
cos(31π/2) = cos [ 21(π/2) – π/6 ]
= sin ( π/6 )
= 1/2
Therefore,
cos(31π/3) = 1/2
方法三:
cos(31π/3)
Write 31π/3 as 14π/3 + 17π/3,
Now,
14π/3 = 9π/2 + π/6
and,
17π/3 = 11π/2 + π/6
Therefore,
cos(31π/3) = cos [ (9π/2 + π/6 ) + (11π/2 + π/6) ]
It is known,
cos (A + B) = [cos(A) .cos(B)] – [sin(A).sin(B)]
Here, A= 9π/2 + π/6 and B = 11π/2 + π/6,
Thus,
cos(A) = tan(9π/2 + π/6 ) = cos(3(3π/2) + π/6 )
cos(B) = tan(11π/2 + π/6) = cos(11(π/2) + π/6 )
It is known,
cos(3nπ/2 + θ) = sin(θ)
cos(nπ/2 + θ) = -sin(θ)
Therefore,
cos(A) = cos((9π/2 + π/6) = sin(π/6) = 1/2
cos(B) = cos((11π/2 + π/6)= -sin(π/6) = -1/2
Now,
cos(31π/3) = cos [ (9π/2 + π/6 ) + (11π/2 + π/6) ]
= [cos(9π/2 + π/6 ) .cos(11π/2 + π/6)] – [sin((9π/2 + π/6 )).sin(11π/2 + π/6)]
= [ (1/2) + (-1/2) ] – [ (1/2).(-1/2) ]
= [ 0 ] + [ 1/2 ]
= 1/2
Therefore,
cos(31π/3) = 1/2
因此,我们能够找到 cos(31π/3) 的值为 0.5。
类似问题
问题一:求cos(4π/3)的值
解决方案:
cos(4π/3)
Write 4π/3 as π + π/3
Thus,
cos(4π/3) = cos(π + π/3)
It is known,
cos(π + θ) = – cos(θ)
So, cos(4π/3) = cos ( π + π/3 )
= – cos (π/3)
= – 1/2
Thus,
cos(4π/3) = -1/2
问题2:求cos(5π/12)的值
解决方案:
cos(5π/12)
We can write (5π/12) as (π/6 + π/4)
So, cos(5π/12) = cos (π/6 + π/4)
cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
here, A = π/6 and B= π/4
So,
cos(5π/12) = cos (π/6 + π/4)
= [cos(π/6) × cos(π/4)] – [sin(π/6) × sin(π/4)]
= [(√3/2) × (1/√2)] – [(1/2) × (1/√2)]
= (√3/2√2) – ( 1/2√2)
= (√(3)-1 ) / (2√2)
= 0.258819.
Therefore,
cos(5π/12) = (√(3)-1) / (2√2)
= 0.258819.
问题 3:求 cos(330 ° )。
解决方案:
Write 330° as 360° – 30°
Note: We are using degrees here and not radians
So,
cos(330°) = cos(360° – 30°)
It is known,
cos(360 – θ) = cos(θ)
Thus,
cos(330°) = cos(360° – 30°)
= cos(30°)
= √3/2
Thus,
cos(330°) = √3/2