如果 un = cosnθ + sinnθ，则证明 2u6 – 3u4 + 1 = 0

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。

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

直角三角形

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，

1. sine: the ratio of perpendicular and hypotenuse is defined as sine and It is represented as sin θ
2. cosine: the ratio of base and hypotenuse is defined as cosine  and it is represented as cos θ
3. tangent: the ratio of sine and cosine of an angle is defined as tangent. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
4. cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
5. secant: It is the reciprocal of cos θ and is represented as sec θ.
6. cotangent: It is the reciprocal of tan θ and is represented as cot θ.

根据上图，三角比是

1. sin θ = 垂直 / 斜边 = AB/AC
2. 余弦 θ = 底 / 斜边 = BC / AC
3. 切线 θ = 垂直 / 底边 = AB / BC
4. 余割 θ = 斜边 / 垂直 = AC/AB
5. 割线 θ = 斜边 / 底 = AC/BC
6. 余切 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
Sec θ	NOT DEFINED	2	√2	2/√3	1
Cosec 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。

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：和等于 180° 的一对角。

补角的恒等式是：

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

问题：如果 u _n = cos ⁿ θ + sin ⁿ θ 那么证明，2u ₆ – 3u ₄ + 1 = 0。

解决方案：

Here we have:

u_n  = cosⁿθ+sinⁿθ 

To prove : 2u₆ – 3u₄ + 1 = 0

Lets n = 6

So, u_n  = cosⁿθ+sinⁿθ 

u₆ = cos⁶θ+sin⁶θ 

u₆ = (cos²θ)³+ (sin²θ) ³

u₆ = (cos²θ +sin²θ)³− 3(cos ²θ)(sin ²θ)(cos ²θ+sin ²θ) 

u₆ =1−3cos²θ sin²θ                                                                                         ……. (1)                                                   { cos²θ +sin²θ = 1 }

Now let n = 4      

u₄ = cos⁴θ + sin⁴θ

u₄​  = (cos²θ)² + (sin²θ)²

u₄ = (cos²θ + sin²θ) ²−2(cos ²θ)(sin ²θ)                                                                                                                      { cos²θ +sin²θ = 1 }

u_4​  = 1− 2cos²θsin²θ                                                                                 ……. (2)                                                   

Now to prove 

2u₆ – 3u₄ + 1 = 0

Put values from equations (1) and (2).

So,

2u₆ – 3u₄ + 1 = 0

= 2(1−3cos ²θ sin ²θ) − 3(1−2cos ²θsin ²θ)+1

= 2 − 6cos²θ sin²θ − 3 + 6cos²θ sin²θ)+1

= 2 -3 +1 

= 0 

Therefore  2u₆ – 3u₄ + 1 = 0

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：证明 (cos θ sec θ)/cot θ = tan θ。

解决方案：

Here we have   

cos θ sec θ / cot θ = tan θ

Therefore,

{ cos θ sec θ }/ cot θ = tan θ

By taking L.H.S

cos θ sec θ / cot θ

we can write cos θ sec θ as 1  

= (cos θ sec θ )/cot θ

= 1/cot θ                                                                                                                              {cos θ = 1/ sec θ   therefore Cos θ Sec θ  = 1}

= tan θ                                                                                                                                 {tan θ = 1 / cot θ }

Therefore LHS = RHS

{cos θ sec θ}/ cot θ = tan θ

Hence Proved

问题 3：证明 {(cot A + cosec A – 1) / (cot A – cosec A + 1)} = {(1 + cos A) / sin A} ？

解决方案：

We have 

To prove : {(cot A + cosec A – 1) / (cot A – cosec A + 1)} = {(1 + cos A) / sin A}

First LHS = {(cot A + cosec A – 1) / (cot A – cosec A + 1)}

              = {(cot A + cosec A ) – ( cosec² A – cot²A ) } / {(cot A – cosec A) + 1}          { cosec² A – cot²A = 1 }

              =  {( cosec A + cot A ) –  ( cosec A + cot A )( cosec A – cot A ) } / (cot A – cosec A + 1 )

              =  {( cosec A + cot A ) [ 1 – ( cosec A – cot A ) ] } / (cot A – cosec A + 1 )

              =  {( cosec A + cot A ) ( cot A – cosec A  + 1 )} / (cot A – cosec A + 1 )

              =    cosec A + cot A

              =   (1 / sin A) + (cos A / sin A)

              =  (1 + cos A) / sin A

              =  RHS
Hence Proved

问题 4：如果 sin θ + cos θ = √2 ，则证明 tan θ + cot θ = 2

解决方案：

We have sin θ + cos θ  = √2 

Squaring both sides

(sin θ + cos θ)²  = (√2 )²

sin² θ + cos² θ  + 2 sin θcos θ = 2 

1 + 2 sin θcos θ = 2                                                                                                                                             {sin² θ + cos² θ = 1}

2 sin θcos θ = 2 – 1

2 sin θcos θ = 1

2 sin θcos θ = sin² θ + cos² θ

Divide both sides by sin θcos θ

2 sin θcos θ / sin θcos θ = (sin² θ + cos² θ) / sin θcos θ

2 = (sin² θ / sin θcos θ) + (cos² θ / sin θcos θ)

2 = (sin θ / cos θ) + (cos θ / sin θ)

2 = tan θ  + cot θ

Hence proved, tan θ  + cot θ  = 2.