(15pi/4) 的 cos 和 sin 是什么?
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。使用三角函数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。它可以定义为直角三角形的边的简单三角比,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
每个三角角的值是固定的和定义的,但是为了方便计算,引入了一些非常常见的角及其值,下表是一些常见的角和基本三角比。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,
- sin(-θ) = – sin(θ)
- cos(-θ) = cos(θ)
- tan(-θ) = – tan(θ)
并且,这些关系在笛卡尔坐标系中使用,这些也将用于解决给出的问题陈述,
- sin(nπ/2 + θ) = cos(θ), sin(nπ/2 – θ) = cos(θ)
- cos(nπ/2 + θ) = -sin(θ), cos(nπ/2 – θ) = sin(θ)
- sin(nπ + θ) = -sin(θ), sin(nπ – θ) = sin(θ)
- cos(nπ + θ) = -cos(θ), cos(nπ – θ) = -cos(θ)
- sin(3nπ/2 + θ) = -cos(θ), sin(3nπ/2 – θ) = -cos(θ)
- cos(3nπ/2 + θ) = sin(θ), cos(3nπ/2 – θ) = -sin(θ)
- sin(2nπ + θ) = sin(θ), sin(2nπ – θ) = -sin(θ)
- cos(2nπ + θ) = cos(θ),cos(2nπ – θ) = cos(θ)
切线函数还有一些特殊的三角公式,
- cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
- cos (A – B) = [cos(A) × cos(B)] + [sin(A) × sin(B)]
- sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]
- sin (A – B) = [sin(A) × cos(B)] – [sin(B) × cos(A)]
(15pi/4) 的 cos 和 sin 是什么?
方法一:
利用一些复合角公式来计算 cos(15π/4) 的值。这里使用以下恒等式或公式,
cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]
解决方案:
cos(15π/4)
Write (15π/4) as (2π + 7π/4) So,
cos(15π/4) = cos(2π + 7π/4)
cos (A + B) = [cos(A) × cos(B)] – [sin(A) × sin(B)]
Here, A = 2π and B = 7π/4
cos(15π/4) = cos(2π + 7π/4)
= [cos(2π) × cos(7π/4)] – [sin(2π) × sin(7π/4)]
= [(1) × cos(2π – (π/4)) – [(0) × sin(2π – (π/4))]
= [cos(π/4)] – [0]
= 1/√2
sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]
Here, A = 2π and B = 7π/4
sin(15π/4) = sin(2π + 7π/4)
= [sin(2π) × cos(7π/4)] + [cos(2π) × sin(7π/4)]
= [(0) × cos(2π – (π/4))] + [(1).sin(2π – (π/4))]
= [0] + [-sin(π/4)]
= -1/√2
Therefore,
sin(15π/4) = -1/√2
cos(15π/4) = 1/√2
方法二:
利用简单的三角恒等式计算 cos(15π/4) 和 sin(15π/4) 的值。这里使用以下标识,
sin(2π – θ) = – sin(θ)
cos(2π – θ) = cos(θ)
解决方案:
(15π/4)
Write (15π/4) as (4π/ – π/4), So,
sin(15π/4) = cos(4π – π/4)
sin(2π – θ) = – sin(θ)
Here, θ = π/4 = 45°
sin(15π/4) = sin(4π – π/4)
= sin [ (2(2π)) – (π/4) ]
= – sin (π/4)
= -1/√2
Similarly,
cos(2π – θ) = cos(θ)
here, θ = π/4
cos(15π/4) = cos(4π – π/4)
= cos [ (2(2π)) – (π/4) ]
= cos (π/4)
= 1/√2
Therefore,
sin(15π/4) = -1/√2
cos(15π/4) = 1/√2
方法三
利用三角恒等式和复合角度公式计算 sin(15π/4) 和 cos(15π/4) 的值,
解决方案:
cos(15π/4)
Write (15π/4) as (3π + 3π/4), So,
cos(15π/4) = cos(3π + 3π/4)
The compound angle formulae,
cos (A + B) = (cos(A) × cos(B)) – (sin(A) × sin(B)),
Here, A = 3π and B = 3π/4
Therefore,
cos(15π/4) = cos( 3π + 3π/4)
= [ cos(3π).cos(3π/4) ] – [ sin(3π).sin(3π/4) ]
= [ (0).cos(π/2+(π/4)) ] – [ (-1).sin(π/2 + (π/4]) ]
= [ -sin(π/4).(0) ] – [ -cos(π/4) ]
= 0 – (-1/√2)
= 1/√2
Now,
sin (A + B) = [sin(A) × cos(B)] + [sin(B) × cos(A)]
sin(15π/4) = sin(3π + 3π/4)
= [sin(3π) × cos(3π/4)] – [cos(3π) × sin(3π/4)]
= [(-1) × cos(π/2 + (π/4))] – [(0) × sin(π/2 + (π/4])]
= [-sin(π/4)] – [(0) × (cos(π/4)]
= (-1/√2) – (0)
= -1/√2
Therefore,
sin(15π/4) = -1/√2
cos(15π/4) = 1/√2
因此通过上述方法,计算出 cos(15π/4) 和 sin(15π/4) 的值分别为 1/√2 和 -1/√2。
示例问题
问题一:求cos(3π/4)的值
解决方案:
Write (3π/4) as (π/2 + π/4) So,
cos(3π/4) = cos (π/2 + π/4)
cos(nπ/2 + θ) = – sin(θ)
Here, θ = π/4
cos(3π/4) = – sin(π/4)
= – 1/√2
Therefore,
cos(3π/4) = -1/√2
问题2:求sin(19π/6)的值
解决方案:
Write (19π/6) as (3π + π/6) So,
sin(19π/6) = sin(3π + π/6)
sin(nπ + θ) = – sin(θ)
Here, θ = π/6
sin(19π/6) = – sin(π/6)
= -1/2
= -0.5
Therefore,
sin(19π/6) = -0.5
问题 3:求 (7π/6) 的 cosec 和 sec
解决方案:
Write (7π/6) as (π + π/6) Therefore,
sin(7π/6) = sin(π + π/6)
Since, sin(nπ + θ) = – sin(θ)
sin(7π/6) = sin(π + π/6)
= – sin(π/6)
= -1/2
Since, cosec(θ) = 1 / sin(θ)
Therefore,
cosec(7π/6) = 1/ sin(7π/6)
= 1 / (-1/2)
= -2
Now, cos(7π/6) = cos(π + π/6)
Since, cos(nπ + θ) = – cos(θ)
cos(7π/6) = cos(π + π/6]
= – cos(π/6)
= – √3/2
Since, sec(θ) = 1/cos(θ)
sec(7π/6) = 1 / cos(7π/6)
= 1 / (-√3/2)
= -2/√3
Thus,
cosec(7π/6) = -2
sec(7π/6) = -2/√3