三角学的创始人是谁？

三角学是一门数学学科，研究直角三角形的边长和角之间的关系。三角函数，也称为测角函数、角函数或圆函数，是建立角度与直角三角形的两条边之比之间关系的函数。六个主要的三角函数是正弦、余弦、正切、余切、正割和余割。

Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.

如上图中的直角三角形所示：

• 斜边：与直角相对的边是斜边，它是直角三角形中最长的边，与90°角相对。
• 底：角 C 所在的一侧称为底。
• 垂直：考虑角度 C 的对边。

三角函数

三角函数有 6 个基本的三角函数，它们是正弦、余弦、正切、余割、正割和余切。现在让我们看看三角函数。六个三角函数如下，

sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ

cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ

tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ

cosecant: It is the reciprocal of sin θ and is represented as cosec θ.

secant: It is the reciprocal of cos θ and is represented as sec θ.

cotangent: It is the reciprocal of tan θ and is represented as cot θ.

根据上图，三角比是

Sin θ = Perpendicular / Hypotenuse = AB/AC

Cosine θ = Base / Hypotenuse = BC/AC

Tangent θ = Perpendicular / Base = AB/BC

Cosecant θ = Hypotenuse / Perpendicular = AC/AB

Secant θ = Hypotenuse / Base = AC/BC

Cotangent θ = Base / Perpendicular = BC/AB

互惠身份

Sin θ = 1/ Cosec θ                    OR        Cosec θ = 1/ Sin θ

Cos θ = 1/ Sec θ                       OR        Sec θ = 1 / Cos θ

Cot θ = 1 / Tan θ                     OR         Tan θ = 1 / Cot θ

Cot θ = Cos θ / Sin θ               OR         Tan θ = Sin θ / Cos θ

Tan θ.Cot θ = 1

三角比值

	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	Not Defined
Cosec θ	Not Defined	2	√2	2/√3	1
Sec θ	1	2/√3	√2	2	Not Defined
Cot θ	Not Defined	√3	1	1/√3	0

补角和补角的三角恒等式

• 互补角：和等于90°的一对角
• 补角：和等于 180° 的一对角

互补角的恒等式是

sin (90° – θ) = cos θ

cos (90° – θ) = sin θ

tan (90° – θ) = cot θ

cot (90° – θ) = tan θ

sec (90° – θ) = cosec θ

cosec (90° – θ) = sec θ

补角的恒等式

sin (180° – θ) = sin θ

cos (180° – θ) = – cos θ

tan (180° – θ) = – tan θ

cot (180° – θ) = – cot θ

sec (180° – θ) = – sec θ

cosec (180° – θ) = – cosec θ

三角学的创始人是谁？

解决方案：

The first trigonometric table was apparently formed by Hipparchus, who is consequently now known as “the father of trigonometry”.

Hipparchus of Nicaea,  Hipparkhos; c. 190 – c. 120 BC was a Greek astronomer, geographer, and mathematician. 

He is considered the founder of trigonometry, but is most famous for his incidental discovery of precession of the equinoxes.

 Hipparchus was born in Nicaea, Bithynia, and probably died on the island of Rhodes, Greece. 

He is known to have been a working astronomer between 162 and 127 BC.

He is the first who created the first trigonometric table and solved several problems of spherical trigonometry by representing the corresponding values of arc and chord for a series of angles such as 0°, 30°, 45°, 60° and 90°, etc.

示例问题

问题1：若x sin ³ θ + y cos ³ θ = sin θ cos θ 且x sin θ – y cos θ = 0，则证明x ² + y ² = 1，（其中，sin θ ≠ 0 且cos θ ≠ 0)。

解决方案：

Here we have,    

x sin³ θ + y cos³ θ = sin θ cos θ

Given:             

x sin³ θ + y cos³ θ = sin θ cos θ

⇒ (x sin θ) sin² θ + (y cos θ) cos² θ = sin θ cos θ

⇒ (x sin θ) sin² θ + (x sin θ) cos² θ = sin θ cos θ             (∵  y cos θ = x sin θ)

⇒ x sin θ(sin² θ + cos² θ) = sin θ cos θ                           (sin² θ + cos² θ = 1)

⇒ x sin θ = sin θ cos θ

⇒ x = cos θ         ….(eq. 1)

now another trigono eq we have  x sin θ – y cos θ = 0

we can write it as                                         

x sin θ  = y cos θ

from eq. 1 we have   x = cos θ   

so put in above eq.  x sin θ  = y cos θ

So, x sin θ  = y cos θ

cos θ sin θ  = y cos θ

y   = sin θ                    eq. 2

Now  by squaring and adding both the equation 1 & 

x = cos θ    &     y   = sin θ  

x²  = cos² θ    &    y²  = sin² θ

So now          

x² + y² = cos² θ + sin ²θ        { cos² θ + sin ²θ = 1 }

x² + y²  =  1

Hence proved 

问题 2：如果 tan A = 4 且 tan B = 2，求 tan(A – B) 的值。

解决方案：

As per the formula

tan(A + B) = (tanA-tanB)/(1+tanAtanB)                

                 = (4 – 2)/(1 + 4 × 2)                 

                 = 2/9

Therefore, the value of tan(A – B) is 2/19.

问题 3：如果 sin(y) = 10/29 那么求 csc(-y) 的值

解决方案：

Here we have

sin(y) = P/B = 10/29

So               

cosec(-y) = 1/sin(-y)

= 1/(-siny)

= – (1/siny)                                                    

= – cosec(y)                     {Cosecant θ = Hypotenuse / Perpendicular = AC/AB}

= -{1/(10/29)}

= -29/10

问题4：执行指示的操作并简化结果。 {sec X}/{csc X} + {csc X}/{sec x}

解决方案：

We have

{sec X}/{csc X} + {csc X}/{sec x}

here we can write csc x = 1/sin x and sec x = 1/cos x

= {(1/cos x )/ (1/ sin x) } + {(1 /sin x) / ( 1/cos x)}

= (sin x / cos x ) + ( cos x / sin x )

= tan x + cot x          {Tan x = Sin x / Cos x and Cot x = Cos x / Sin x}

Therefore, {sec X}/{csc X} + {csc X}/{sec x} = tan x + cot x