求 cos(11(pi)/6) 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,
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 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率可以应用:
sin(-θ) = – sin(θ)
cos(-θ) = cos(θ)
tan(-θ) = – tan(θ)
正切函数有特殊的三角公式,
cos (A + B) = [cos(A) .cos(B)] – [sin(A).sin(B)]
cos (A – B) = [cos(A) .cos(B)] + [sin(A).sin(B)]
求 cos(11(pi)/6) 的值。
方法一
利用简单的三角恒等式计算 cos(11π/6) 的值。在这里,我们使用以下恒等式或公式,
cos(2π – θ) = cos(θ)
解决方案:
cos(11π/6)
(11π/6) as (2π – π/6)
So, cos(11π/6) = cos(2π – π/6)
cos(2π – θ) = cos(θ)
Here, θ = π/6 = 30°
cos(11π/6) = cos(π/6)
= √3/2
Therefore, cos(11π/6) = √3/2
= 0.866025.
方法二
利用简单的三角恒等式计算 cos(11(pi)/6) 的值。在这里,我们使用以下标识,
cos(3π/2 + θ) = sin(θ)
解决方案:
cos(11π/6)
(11π/6) as (3π/2 + π/3),
So, cos(11π/6) = cos(3π/2 – π/3)
cos(3π/2 + θ) = sin(θ)
Here, θ = π/3 = 60°
cos(11π/6) = sin(π/3)
= √3/2
Therefore, cos(11π/6) = √3/2
= 0.866025.
方法三
利用复合角度公式计算 cos(11(pi)/6) 的值。这里使用了以下恒等式和公式,
cos(A + B) = (cos(A).cos(B)) – (sin(A).sin(B))
解决方案:
cos(11π/6)
(11π/6) as (2π/6 + 9π/6),
So, cos(11π/6) = cos(2π/6 – 9π/6)
The comp[ound angle formulae,
cos (A + B) = (cos(A).cos(B)) – (sin(A).sin(B)),
here, A = 2π/6 and B = 9π/6
Therefore, cos(11π/6) = cos( 2π/6 + 9π/6)
= [cos(2π/6).cos(9π/6)] – [sin(2π/6).sin(9π/6)]
= [cos(π/3).cos(3π/2)] – [sin(π/3).sin(3π/2)]
= [(1/2).(0)] – (√3/2).(-1)]
= 0 – (-√3/2)
= √3/2
Therefore,cos(11π/6) = √3/2
= 0.866025.
因此通过上述方法,我们能够计算出 cos(11(pi)/6) 的值 cos(330°) 为 √3/2 或大约 0.86602。
类似问题
问题1:求cos(5π/6)的值
解决方案:
cos(5π/6)
(5π/6) as ( π/2 + π/3)
So, cos(5π/6) = cos (π/2 + π/3)
cos(π/2 + θ) = -sin(θ)
So, cos(5π/6) = cos (π/2 + π/3)
= – sin(π/3)
= – √3/2
Therefore,
cos(5π/6) = – √3/2
问题2:求cos(5π/12)的值
解决方案:
cos(5π/12)
(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:如果 A = 3π/4,求所有三角比的值。
解决方案:
A = 3π/4,
Find the value of all trigonometric ratios
That is, sin(A), cos(A), tan(A), cosec(A), sec(A) and cot(A)
(3π/4) as (π – π/4) can be written,
So, sin(A) = sin(3π/4)
= sin (π – π/4)
sin(π – θ) = sin(θ)
Therefore, sin(A) = sin (π – π/4)
= sin( π/4)
= 1 / √2
Thus, sin(3π/4) = 1 / √2
Similarly, cos(A) = cos (3π/4)
= cos (π -π/4)
= – cos (π/4)
= – 1 / √2
Thus, cos(3π/4) = -1 / √2
Now, tan(A) = sin(A) / cos(A)
Thus, tan(A) = (1 / √2) / (- 1/√2)
tan(A) = -1
Now, cosec(A) = 1/sin(A) = 1 / (1/√2) = √2
sec(A) = 1/ cos(A) = 1 / (-1 / √2) = -√2
cot(A) = 1/ tan(A) = 1/ (-1) = -1
Therefore,
sin(A) = 1/√2, cosec(A) = √2
cos(A) = – 1/√2, sec(A) = – √2
tan(A) = -1, cot(A) = -1