如果 x sin3θ + y cos3θ = sin θ cos θ 且 x sin θ – y cos θ = 0，则证明 x2 + y2 = 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

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.

三角比

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

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 it 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 & 2

x = cos θ & y = sin θ   

x² = cos² θ    & y ²  = sin² θ 

So now          

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

x² + y²  =  1

Hence proved 

类似问题

问题 1：如果 cos θ + sin θ = √2 cos θ，证明 cos θ – sin θ = √2 sin θ？

解决方案：

Here we have

cos θ + sin θ = √2 cos θ

Squaring both the sides

(cos θ + sin θ)² = (√2 cos θ)²

cos² θ + sin² θ + 2 sin θcos θ = 2 cos ²θ 

cos² θ – sin² θ = 2 sin θcos θ 

(cos θ + sin θ)(cos θ – sin θ) = 2 sin θcos θ 

(cos θ – sin θ) = 2 sin θcos θ/(cos θ + sin θ)

cos θ – sin θ = 2 sin θcos θ/√2 cos θ               {cos θ + sin θ = √2 cos θ}

cos θ – sin θ = √2 sin θ 

Hence proved

问题 2：从等式中消除 theta：tan θ – cot θ = a 和 cos θ + sin θ = b

解决方案：

tan θ – cot θ = a ………. (1)

cos θ + sin θ = b ………. (2)  

Squaring both sides of (eq2) we get,  

(cos θ + sin θ)² = (b)²

cos² θ + sin² θ + 2cos θ sin θ = b²                                        {sin² θ  + cos² θ = 1}

1 + 2 cos θ sin θ  = b²

2 cos θ sin θ = b² – 1 ………. (3)  

Again, from (1)

We get, (sin θ/cos θ) – (cos θ/sin θ) = a  

(sin² θ – cos² θ)/(cos θ sin θ) = a  

sin²θ – cos²θ = a sin θ cos θ

(sin θ + cos θ) (sin θ – cos θ) = a ∙ (b² – 1)/2                       ……….          [by (3)]

b(sin θ – cos θ) = (½) a (b² – 1)                                       [by (2)]  

b² (sin θ – cos θ)² = (1/4) a² (b² – 1)2                            [Squaring both the sides]  

b² [(sin θ + cos θ)² – 4 sinθ cos θ] = (1/4) a² (b² – 1)2

b² [b² – 2 ∙ (b² – 1)] = (1/4) a² (b² – 1)2                                  [from (2) and (3)]  

4b² (2 – b²) = a² (b² – 1)2

which is the required θ-eliminate.