三角函数的主要功能是什么？

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

三角学象限

三角函数的主要功能是什么？

解决方案：

The primary function of trigonometry are 

Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,

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.

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

示例问题

问题 1：如果 Sin θ = 4/5 那么找到所有的三角函数值？

解决方案：

Here we have 

sin θ = 4/5

therefore Sin θ = Perpendicular / Hypotenuse = AB/AC

so we have perpendicular (p)= 4 and hypotenuse(h) = 5

So as per the pythagoras theorem 

Lets find out the value of base (b)

h² = b²+p²

5² = b² + 4²

25 = b² + 16

25 -16 = b²

or b²  = 9

b = 3

So now 

As per the trigonometric functions

We have 

Sin θ = Perpendicular/Hypotenuse = AB/AC = 4/5

Cosine θ = Base/Hypotenuse = BC/AC = 3/5

Tangent θ = Perpendicular/Base = AB/BC = 4/3

Cosecant θ = Hypotenuse/Perpendicular = AC/AB = 5/4

Secant θ = Hypotenuse/Base = AC/BC = 5/3

Cotangent θ = Base/Perpendicular = BC/AB = 3/4

问题 2：在B 处的直角三角形中，如果 sin C =3/5，求 cos C 和 tan C？

解决方案：

Here we have sin C = 3/5

Therefore sin C = 3/5 and  Sin θ = Perpendicular / Hypotenuse = AB/AC

AB (p)= 3 , AC (h) = 5

Now for cos C = Cosine θ = Base / Hypotenuse = BC / AC

here AC = 5 and base BC = ?

for this we will use pythagoras theorem 

AC² = AB² + BC²

5² = 3² + BC²

25 = 9 + BC²

BC²= 25-9

BC² = 16

BC = 4

Therefore

 cos C = Cosine θ = Base/Hypotenuse = BC/AC = 4/5 and

tan C = Tangent θ = Perpendicular/Base = AB/BC = 3/4