计算 tan (150° + 45°) 的准确值
三角学是数学的一个分支,它使用比率将三角形的边和角联系起来。三角函数有助于计算连接到三角形的各种测量值。在三角学中,定义了标准比率,以便于计算与直角三角形边的长度和角度相关的一些常见问题。
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。有三个标准的三角比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
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 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)
对于这个问题,请查看某些特定于切线比的公式和与简单易懂的事物的关系。查看正切函数的补角和补角,
补角和补角
互补角是一对角加起来形成 90° 或 π/2 弧度。可以形成这样的角度并根据三角比找到等效角度。
补角是一对角加起来形成 180° 或 π 弧度。可以形成这样的角度并根据三角比找到等效角度。
从 90° 中减去一个角,得到一对互补角,同样,可以将一个角相加到 90°,形成一个互补角对。换句话说,可以在三角比的函数中调整实际角度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出的三角比。
- 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)
- tan(nπ + θ) = tan(θ) 或 tan(n × 180° + θ) = tan(θ)
- tan(nπ – θ) = -tan(θ) 或 sin(n × 180° – θ) = -tan(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)
- tan(2nπ + θ) = tan(θ) 或 tan(n × 360°+ θ) = tan(θ)
- tan(2nπ – θ) = -tan(θ) 或 tan(n × 360° – θ) = -tan(θ)
三角函数有复合角公式。
- 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(150° + 45°) 的值
解决方案:
方法一
tan(150° + 45°)
Use the compound angles formula for tangent function,
tan(A + B) = [tan(A) + tan(B)] / [1 – (tan(A)tan(B))]
So, here, A = 150° and B = 45°,
tan(150° + 45°) = (tan(150°) + tan(45°)) / [1 – (tan(150).tan(45))]
Now, write 150° as (180° – 30°),
So, tan(150°) = tan(180° – 30°)
sin(n × 180° – θ) = -tan(θ)
Here n = 1 and θ = 30
So,
tan(150°) = tan(180° – 30°)
= -tan(30°)
= – 1/√3
Thus,
tan(150° + 45°) = ( tan(150°) + tan(45°) ) / [1 – (tan(150).tan(45))]
tan(150°) = -1/√3 and tan(45°) = 1,
= ((-1/√3) + 1) / [1 – ((-1/√3).(1))]
= [(-1 + √3)/√3] / [1 + 1/√3]
= [(√3 – 1)/√3] / [(√3 + 1)/√3]
= [√3 – 1] / [√3 +1]
Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)
tan(150° + 45°) = [(√3 – √1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]
= [(√3 – 1)2] / [ (√32) – (12)]
= [3 – 2√3 + 1 ] / [3 – 1]
= [4 – 2√3] / 2
= 2 – √3
~ 0.268
Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268
方法二
tan(150° + 45°)
= tan(195°)
Now, write 195° as (180° + 15°),
So, tan(195°) = tan(180° + 45°)
sin(n × 180° + θ) = tan(θ)
Here, n = 1 and θ = 15
tan(150° + 45°) = tan(195°) = tan(180° +15°)
= tan(15°)
Now, 15° can be written as (45° – 30°)
Thus,
tan(150° + 45°) = tan(15°)
= tan(45° – 30°)
The compound angles formula for tangent function,
tan(A – B) = [tan(A) – tan(B)] / [1 + (tan(A)tan(B))]
Here, A = 45° and B = 30°
tan(45° – 30°) = (tan(45°) – tan(30°)) / [1 + (tan(45).tan(30))]
tan(45°) = -1 and tan(30°) = 1/√3,
= (1 – (1/√3)) / [1 + ((1).(1/√3))]
= [(1 – √3)/√3] / [1 + 1/√3]
= [(√3 – 1)/√3] / [(√3 + 1)/√3]
= [√3 – 1] / [√3 +1]
tan(150° + 45°) = [√3 – 1] / [√3 +1]
Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)
tan(150° + 45°) = tan(15°)
= [(√3 – √1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]
= [(√3 – 1)2] / [ (√32) – (12)]
= [3 – 2√3 + 1] / [3 – 1]
= [4 – 2√3] / 2
= 2 – √3
~ 0.268
Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268
方法三
tan(150° + 45°)
= tan(195°)
Now, write 195° as (240° – 45°),
So, tan(195°) = tan(240° – 45°)
Use the compound angles formula for tangent function,
tan(A – B) = [tan(A) – tan(B)] / [1 + (tan(A)tan(B))]
Here. A = 240° and B = 45°,
tan(240° – 45°) = (tan(240°) – tan(45°)) / [1 + (tan(240).tan(45))]
Now, write 240° as (270° – 30°),
So, tan(240°) = tan(270° – 30°)
sin(n × 270° – θ) = cot(θ)
Here n = 1 and θ = 30,
So, tan(240°) = tan(270° – 30°)
= cot(30°)
= √3
Hence , tan(240°) = √3
tan(150° + 45°) = tan(195°) = tan(240° – 45°)
tan(240° – 45°) = (tan(240°) – tan(45°)) / [1 – (tan(240).tan(45))]
Here, tan(240°) = 3 and tan(45°) = 1
= (tan(240°) – tan(45°)) / [1 + (tan(240°).tan(45°))]
= (3 – 1) / (1 + (3).(1))
tan(150° + 45°) = [√3-1] / [√3 +1]
Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)
= [(√3 – 1)2] / [ (√32) – (12)]
= [3 – 2√3 + 1] / [3 – 1]
= [4 – 2√3] / 2
= 2 – √3
~ 0.268
Thus the value of tan(150° + 45°) = 2 – √3 ~ 0.268
示例问题
问题一:求tan(135°)的值
解决方案:
tan(135°)
135° as (90° + 45°)
So,
tan(135°) = tan(90° + 45°)
tan(n × 90° + θ) = -cot(θ)
Here, n = 1 and θ = 45°
Thus,
tan(135°) = tan(90° + 45°)
= -cot(45)
= -1
Thus, the value of tan(135°) = -1
问题2:求tan(210°)的值
解决方案:
tan(210°)
210° as (180° + 30°)
So,
tan(210°) = tan(180° + 30°)
tan(n × 180° + θ) = tan(θ)
Here, n = 1 and θ = 30°,
Thus,
tan(210°) = tan(180° + 30°)
= tan(30°)
= 1/√3
Thus,
tan(210°) = 1/√3
Thus, the value of tan(210°) = 1/√3 ~ 0.5773
问题3:求tan(135° + 30°)的值
解决方案:
tan(135° + 30°)
Use the compound angles formula for tangent function,
tan(A – B) = [tan(A) – tan(B)] / [1 + (tan(A)tan(B))]
Here. A = 135° and B = 30°,
tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]
To find, tan(135°),
Write 135° as (90° + 45°)
So, tan(135°) = tan(90° + 45°)
tan(n × 90° + θ) = -cot(θ)
Here, n = 1 and θ = 45°
So, tan(135°) = tan(90° + 45°)
= -cot(45°)
= -1
tan(135°) = -1
Thus,
tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]
Here, tan(135°) = -1 and tan(30°) = 1/√3,
tan(135° + 30°) = [tan(135°) + tan(30°)] / [1 – ((tan(135°).(tan(30°))]
= [-1 + (1/√3)] / [1 – ((-1).(1/√3))]
= [(-√3 +1) / √3] / [1 + (1/√3)]
= [(-√3 + 1) / √3] / [(√3 + 1) / √3]
= [-√3 + 1] / [√3 +1]
tan(135° + 30°) = [-√3 + 1] / [√3 + 1]
Rationalizing the denominator (multiplying and dividing by the denominator’s conjugate)
= [(-√3 + 1) × (√3 – 1)] / [(√3 + 1) × (√3 – 1)]
= [-√3 × √3 + √3 + √3 – 1] / [ (√32) – (12) ]
= [-3 + 2√3 – 1] / [3 – 1]
= [-4 + 2√3 ] /2
= -2 + √3
~ -0.268
Thus, the value of tan(135° + 30°) = -2 + √3 ~ -0.268