求和与两个角度之差的三角函数

三角学是数学的一个分支，涉及三角形的角度，长度和高度以及它们之间的关系。它在计算复杂函数或大距离方面起着重要作用，而如果没有三角函数则无法计算。在解决三角学问题时，我们遇到了许多情况，我们必须计算角度和或角度差的三角解。例如

这里， $\frac{BC}{AB} = \frac{Opposite \hspace{0.1cm}side} {Adjacent\hspace{0.1cm} side}$

这是正切三角比，其角度与BC相反。

tan(θ+Φ)= $\frac{BC}{AB}$

如果θ= 30°和Φ= 45°。我们知道45°和30°的三角角，但是我们不知道(45°+ 30°= 75°)的三角角。因此，以简化这些类型的问题。我们将学习三角公式或两个角度之和与差的恒等式，这将使事情变得容易。

在继续前进之前，我们将在四个象限中看到三角函数的符号。这些符号在三角学中起着重要的作用。

三角恒等式

现在我们要找到 三角恒等式。我们知道

sin(-x)= – sin x

cos(-x)= cos x

因为在第四象限中只有cos和sec是正数。因此，现在我们证明一些有关角度之和和差的结果：

Let’s consider a unit circle (having radius as 1) with centre at the origin. Let x be the ∠DOA and y be the ∠AOB. Then (x + y) is the ∠DOB. Also let (– y) be the ∠DOC. 

Therefore, the coordinates of A, B, C and D are

A = (cos x, sin x)

B = [cos (x + y), sin (x + y)]

C = [cos (– y), sin (– y)] 

D = (1, 0).

As, ∠AOB = ∠COD

Adding, ∠BOC both side, we get

∠AOB + ∠BOC = ∠COD + ∠BOC

∠AOC = ∠BOD

In △ AOC and △ BOD

OA = OB (radius of circle)

∠AOC = ∠BOD (Proved earlier)

OC = OD (radius of circle)

△ AOC ≅ △ BOD by SAS congruency.

By using distance formula, for

AC² = [cos x – cos (– y)]² + [sin x – sin(–y]²

AC² = 2 – 2 (cos x cos y – sin x sin y) …………….(i)

And, now

Similarly, using distance formula, we get

BD² = [1 – cos (x + y)]² + [0 – sin (x + y)]²

BD² = 2 – 2 cos (x + y) …………….(ii)

As, △ AOC ≅ △ BOD

AC = BD, So AC² = BD²

From eq(i) and eq(ii), we get

2 – 2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y)

So, 

cos (x + y) = cos x cos y – sin x sin y

Take y = -y, we get

cos (x + (-y)) = cos x cos (-y) – sin x sin (-y)

cos (x – y) = cos x cos y + sin x sin y

Now, taking 

cos ($\frac{\pi}{2}$-(x + y)) = cos (($\frac{\pi}{2}$-x) – y) (cos ($\frac{\pi}{2}$-θ) = sin θ)

sin (x – y) = sin x cos y – cos x sin y

take y = -y, we get

sin (x – (-y)) = sin x cos (-y) – cos x sin (-y)

sin (x + y) = sin x cos y + cos x sin y

复合角的三角比的导出公式如下：

sin (A + B) = sin A cos B + cos A sin B     ………………..(1)

sin (A – B) = sin A cos B – cos A sin B       ………………..(2)

cos (A + B) = cos A cos B – sin A sin B     ..………………(3)

cos (A – B) = cos A cos B + sin A sin B     ………………..(4)

通过使用以下公式，我们可以获得一些重要且最常用的形式：

(1)取A = $\mathbf{\frac{\pi}{2}}$

In eq(1) and (3), we get

sin ($\frac{\pi}{2}$+B) = cos B

cos ($\frac{\pi}{2}$+B) = – sin A

(2)取A =π

In eq(1), (2), (3) and (4) we get

sin (π + B) = – sin B

sin (π – B) = sin B

cos (π ± B) = – cos B

(3)取A =2π

In eq(2) and (4) we get

sin (2π – B) = – sin B

cos (2π – B) = cos B

婴儿床A，棕褐色A，秒A和Cosec A同样

(4) $\mathbf{tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}}$

Here, A, B, and (A + B) is not an odd multiple of π/2, so, cosA, cosB and cos(A + B) are non-zero

tan(A + B) = sin(A + B)/cos(A + B)

From eq(1) and (3), we get

tan(A + B) = sin A cos B + cos A sin B/cos A cos B – sin A sin B

Now divide the numerator and denominator by cos A cos B we get

tan(A + B) = $\frac{\frac{sin A cos B}{cosAcosB} + \frac{cos A sin B}{cosAcosB}}{\frac{cos A cos B}{cosAcosB} - \frac{sin A sin B}{cosAcosB}}$

$tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}$

(5) $\mathbf{tan(A - B) = \frac{tan A - tan B}{1+tan A tan B}}$

As we know that 

$tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}$

So, on putting B = -B, we get

$tan(A - B) = \frac{tan A - tan B}{1+tan A tan B}$

(6) $\mathbf{cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}}$

Here, A, B, and (A + B) is not a multiple of π, so, sinA, sinB and sin(A + B) are non-zero

cot(A + B) = cos(A + B)/sin(A + B)

From eq(1) and (3), we get

cot(A + B) = cos A cos B – sin A sin B/sin A cos B + cos A sin B

Now divide the numerator and denominator by sin A sin B we get

cot(A + B) = $\frac{\frac{cos A cos B}{sinAsinB} - \frac{sin A sin B}{sinAsinB}}{\frac{sin A cos B}{sinAsinB} + \frac{cos A sin B}{sinAsinB}}$

$cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}$

(7) $\mathbf{cot(A - B) = \frac{cotAcot B+1}{cot A - cot B}}$

As we know that 

$cot(A + B) = \frac{cotAcot B-1}{cot A + cot B}$

So, on putting B = -B, we get

$cot(A - B) = \frac{cotAcot B+1}{cot A - cot B}$

在这里，我们将建立两组转换公式：因式分解和去因式公式。

分解公式

在三角学中，分解是指将乘积转换为总和或差。分解公式为：

(1)2罪A cos B =罪(A + B)+罪(A – B)

证明：

As we know that 

sin (A + B) = sin A cos B + cos A sin B     ………………………(1)

sin (A – B) = sin A cos B – cos A sin B      ………………………(2)

By adding eq(1) and (2), we get

2 sin A cos B = sin (A + B) + sin (A – B)

(2)2 cos A sin B = sin(A + B)– sin(A – B)

证明：

As we know that 

sin (A + B) = sin A cos B + cos A sin B     ………………………(1)

sin (A – B) = sin A cos B – cos A sin B      ………………………(2)

By subtracting eq(2) from (1), we get

2 cos A sin B = sin (A + B) – sin (A – B)

(3)2个cos A cos B = cos(A + B)+ cos(A – B)

证明：

As we know that 

cos (A + B) = cos A cos B – sin A sin B     ………………………(1)

cos (A – B) = cos A cos B + sin A sin B     ………………………(2)

By adding eq(1) and (2), we get

2 cos A cos B = cos (A + B) + cos (A – B)

(4)2 sin A sin B = cos(A – B)– cos(A + B)

证明：

cos (A + B) = cos A cos B – sin A sin B     ………………………(1)

cos (A – B) = cos A cos B + sin A sin B     ………………………(2)

By subtracting eq(3) from (4), we get

2 sin A sin B = cos (A – B) – cos (A + B)  

示例1.将以下乘积中的每一个转换为总和或差额。

(i)2正弦40°cos 30°

(ii)2 sin 75°sin 15°

(iii)cos 75°cos 15°

解决方案：

(i) Given: A = 40° and B = 30°

Now put all these values in the formula, 

2 sin A cos B = sin (A + B) + sin (A – B)

We get

2 sin 40° cos 30° = sin (40 + 30) + sin (40 – 30) 

= sin (70°) + sin (10°) 

(ii) Given: A = 75° and B = 15°

Now put all these values in the formula, 

2 sin A sin B = cos (A – B) – cos (A + B)

We get

2 sin 75° sin 15° = cos (75-15) – cos (75+15)

= cos (60°) – cos (90°)

(iii) Given: A = 75° and B = 15°

Now put all these values in the formula, 

2 cos A cos B = cos (A + B) + cos (A – B)

We get

cos 75° cos 15° = 1/2(cos (75+15) + cos (75-15))

= 1/2 (cos (90°) + cos (60°))

例子2.解决$2cos \frac{\pi}{13}cos \frac{9\pi}{13}+cos \frac{3\pi}{13}+cos \frac{5\pi}{13}$

解决方案：

Using the formula

2 cos A cos B = cos (A + B) + cos (A – B) 

$2cos \frac{\pi}{13}cos \frac{9\pi}{13} =cos (\frac{9\pi}{13}+\frac{\pi}{13})+cos (\frac{9\pi}{13}-\frac{\pi}{13}) + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}$ 

= $cos \frac{10\pi}{13}+cos \frac{8\pi}{13} + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}$

= $cos (\pi-\frac{3\pi}{13})+cos (\pi-\frac{5\pi}{13}) + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}$

= $-cos \frac{3\pi}{13}-cos \frac{5\pi}{13} + cos\frac{3\pi}{13}+cos\frac{5\pi}{13}$

Hence, 

$2cos \frac{\pi}{13}cos \frac{9\pi}{13}+cos \frac{3\pi}{13}+cos \frac{5\pi}{13}$ = 0

因式分解公式

在三角学中，因式分解是指将总和或差转换为乘积。分解公式为：

(1)罪(C)+罪(D)= 2罪$(\mathbf{\frac{C+D}{2}})$ cos $(\mathbf{\frac{C-D}{2}})$

证明：

We have 

2 sin A cos B = sin (A + B) + sin (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

Now put all these values in eq(1), we get

2 sin ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$) = sin (C) + sin (D)

Or 

sin (C) + sin (D) = 2 sin ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$)

(2)罪(C)–罪(D)= 2 cos $(\mathbf{\frac{C+D}{2}})$罪$(\mathbf{\frac{C-D}{2}})$

证明：

We have 

2 cos A sin B = sin (A + B) – sin (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

Now put all these values in eq(1), we get

2 cos ($\frac{C+D}{2}$) sin ($\frac{C-D}{2}$) = sin (C) – sin (D)

Or 

sin (C) – sin (D) = 2 cos ($\frac{C+D}{2}$) sin ($\frac{C-D}{2}$)

(3)cos(C)+ cos(D)= 2 cos $(\mathbf{\frac{C+D}{2}})$ cos $(\mathbf{\frac{C-D}{2}})$

证明：

We have 

2 cos A cos B = cos (A + B) + cos (A – B) ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

Now put all these values in eq(1), we get

2 cos ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$) = cos (C) + cos (D)

Or 

cos (C) + cos (D) = 2 cos ($\frac{C+D}{2}$) cos ($\frac{C-D}{2}$)

(4)cos(C)– cos(D)= 2罪$(\mathbf{\frac{C+D}{2}})$罪$(\mathbf{\frac{D-C}{2}})$

证明：

We have 

2 sin A sin B = cos (A – B) – cos (A + B)  ………………………(1)

So now, we are taking 

A + B = C and A – B = D

Then, A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

Now put all these values in eq(1), we get

2 sin ($\frac{C+D}{2}$) sin ($\frac{C-D}{2}$) = cos (C) – cos (D)

Or 

cos (C) – cos (D) = 2 sin ($\frac{C+D}{2}$) sin ($\frac{C-D}{2}$)

说明1.将以下各项表示为产品

(i)正弦40°+正弦20°

(ii)正弦60°–正弦20°

(iii)cos 40°+ cos 80°

解决方案：

(i) Given: C = 40° and D = 20°

Now put all these values in the formula, 

sin (C) + sin (D) = 2 sin $(\frac{C+D}{2})$cos $(\frac{C-D}{2})$ 

We get

sin 40° + sin 20° = 2 sin $(\frac{40+20}{2})$cos $(\frac{40-20}{2})$

= 2 sin $(\frac{60}{2})$ cos $(\frac{20}{2})$

= 2 sin 30° cos 10°

(ii) Given: C = 60° and D = 20°

Now put all these values in the formula, 

sin (C) – sin (D) = 2 cos $(\frac{C+D}{2})$sin $(\frac{C-D}{2})$

We get

sin 60° – sin 20° = 2 cos $(\frac{60+20}{2})$sin $(\frac{60-20}{2})$

= 2 cos $(\frac{80}{2})$ sin $(\frac{40}{2})$

= 2 cos 40° sin 20°

(iii) Given: C = 80° and D = 40°

 Now put all these values in the formula, 

cos (C) + cos (D) = 2 cos $(\frac{C+D}{2})$cos $(\frac{C-D}{2})$

We get

cos 40° + cos 80° = 2 cos $(\frac{40+80}{2})$ cos $(\frac{80-40}{2})$

= 2 cos $(\frac{120}{2})$ cos $(\frac{40}{2})$

= 2 cos 60° cos 20°

示例2.证明：1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x

解决方案：

Lets take LHS

1 + cos 2x + cos 4x + cos 6x

Here, cos 0x = 1

So,

(cos 0x + cos 2x) + (cos 4x + cos 6x)

Using formula

cos (C) + cos (D) = 2 cos $(\frac{C+D}{2})$cos $(\frac{C-D}{2})$ 

We get

(2 cos $(\frac{0x+2x}{2})$cos $(\frac{2x-0x}{2})$) + (2 cos $(\frac{4x+6x}{2})$ cos $(\frac{6x-4x}{2})$)

(2 cos x cos x) + (2 cos 5x cos x)

Taking 2 cos x common, we have

2 cos x (cos x + cos 5x)

Again using the forumla 

cos (C) + cos (D) = 2 cos $(\frac{C+D}{2})$cos $(\frac{C-D}{2})$

We get

2 cos x (2 cos $(\frac{x+5x}{2})$cos $(\frac{5x-x}{2})$)

2 cos x (2 cos 3x cos 2x)

4 cos x cos 2x cos 3x

LHS = RHS

Hence proved

关于角度A的多个角度(2A)的三角比

直角三角形中某个角度的三角比例定义了该角度与其边长之间的关系。 sin 2x或cos 2x等也就是这样的一个三角公式，也称为双角公式，因为它具有双角。

(1)罪2A = 2罪A cos A

证明：

As we know that 

sin (A + B) = sin A cos B + cos A sin B ………………..(1)

Now taking B = A, in eq(1), we get

sin (A + A) = sin A cos A + cos A sin A

sin 2A = 2 sin A cos A  

(2)cos 2A = cos ² A – sin ² A

证明：

As we know that 

cos (A + B) = cos A cos B – sin A sin B ………………..(1)

Now taking B = A, in eq(1), we get

cos (A + A) = cos A cos A + sin A sin A

cos 2A = cos² A – sin² A

(3)cos 2A = 2cos ² A – 1

证明：

As we know that 

cos 2A = cos² A – sin² A ………………..(1)

We also know that 

sin² A + cos² A = 1

So, sin² A = 1 – cos² A

Now put the value of sin² A in eq(1), we get

cos 2A = cos² A – (1 – cos² A)

cos 2A = cos² A – 1 + cos² A

cos 2A = 2cos² A – 1

(4)cos 2A = 1 – 2sin ² A

证明：

As we know that 

cos 2A = 2cos² A – 1 ………………..(1)

We also know that 

sin² A + cos² A = 1

So, cos² A = 1 – sin² A

Now put the value of sin² A in eq(1), we get

cos 2A = 2(1 – sin² A) – 1

cos 2A = 2 – 2sin² A) – 1

cos 2A = 1 – 2sin² A 

(5)cos 2A = $\mathbf{\frac{1-tan^2A}{1+tan^2A}}$

证明：

As we know that 

cos 2A = cos² A – sin² A 

So, now dividing, by sin² A + cos² A = 1, we get

cos 2A = $\frac{cos^2 A - sin^2 A}{sin^2 A + cos^2 A}$

Again dividing the numerator and denominator by cos² A, we get

cos 2A = $\frac{\frac{cos^2 A - sin^2 A}{cos^2A}}{\frac{sin^2 A + cos^2 A}{cos^2A}}$

cos 2A = $\frac{1-tan^2A}{1+tan^2A}$

(6)罪2A = $\mathbf{\frac{2tanA}{1+tan^2A}}$

证明：

As we know that 

sin (A + B) = sin A cos B + cos A sin B ………………..(1)

Now taking B = A, in eq(1), we get

sin (A + A) = sin A cos A + cos A sin A

sin 2A = 2 sin A cos A  

As we also know that sin² A + cos² A = 1

So, now dividing, by sin² A + cos² A = 1, we get

sin 2A = $\frac{2 sin A cos A}{cos^2A + sin^2A}$

Now, on dividing the numerator and denominator by cos² A, we get

sin 2A = $\frac{2 tanA}{1+tan^2A}$

(7)棕褐色2A = $\mathbf{\frac{2 tanA}{1-tan^2A}}$

证明：

As we know that 

$tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}$ ………………..(1)

Now taking B = A, in eq(1), we get

tan(A + A) = $\frac{tan A + tan A}{1-tan A tan A}$

tan 2A = $\frac{2tan A}{1-tan^2 A}$

示例：证明

(一世) $\mathbf{\frac{sin 2\theta}{1+cos 2\theta}}$ = tanθ

(ii) $\mathbf{\frac{sin 2\theta}{1-cos 2\theta}}$ =婴儿床θ

(iii)cos 4x = 1 – 8 sin ² x cos ² x

解决方案：

(i) sin 2θ = 2 sin θ cos θ ………..(from identity 1)

and, 1 + cos 2θ = 2cos²θ  ………..(from identity 3)

$\frac{sin 2\theta}{1+cos 2\theta} = \frac{2 sin \theta cos \theta}{2cos^2\theta}$

= $\frac{sin \theta }{cos\theta}$

= tan θ 

Hence Proved

(ii) sin 2θ = 2 sin θ cos θ ………..(from identity 1)

and, 1 – cos 2θ = 2sin²θ  ………..(from identity 4)

$\frac{sin 2\theta}{1-cos 2\theta} = \frac{2 sin \theta cos \theta}{2sin^2\theta}$

= $\frac{cos \theta }{sin\theta}$

= cot θ

Hence Proved

(iii) cos 4x = cos 2(2x)

= 1 – 2sin²(2x) (using 16)

= 1 – 2(sin(2x))²

= 1 – 2(2 sin x cos x)²    (using identity 1)

= 1 – 2(4 sin² x cos² x)

cos 4x = 1 – 8 sin² x cos² x

Hence Proved

关于角度A的多个角度(3A)的三角比

直角三角形中某个角度的三角比例定义了该角度与其边长之间的关系。 sin 3x或cos 3x等也都是这样的三角公式，也称为三重角度公式，因为它具有三重角度。

(1)sin 3A = 3sin A – 4 sin ³ A

证明：

Let’s take LHS

sin 3A = sin(2A + A)

Using identity 

sin (A + B) = sin A cos B + cos A sin B

We get

sin 3A = sin 2A cos A + cos 2A sin A

= 2sin A cos A cos A + (1 – 2 sin²A)sin A

= 2sin A(1 – sin²A) + sin A – 2 sin³A

= 2sin A – 2sin³A + sin A – 2 sin³A

sin 3A = 3sin A – 4 sin³A

(2)cos 3A = 4 cos ³ A – 3cos A

证明：

Let’s take LHS

sin 3A = sin(2A + A)

Using identity 

cos (A + B) = cos A cos B – sin A sin B

We get

cos 3A = cos 2A cos A – sin 2A sin A

= (2cos²A – 1)cos A – 2sin A cos A sin A

= (2cos²A – 1)cos A – 2cos A(1 – cos²A)

= 2cos³A – cos A – 2cos A + 2cos³A)

cos 3A = 4 cos³A – 3cos A 

(3)棕褐色3A = $\mathbf{\frac{3tan A-tan^3A}{1-3tan^2A}}$

证明：

Let’s take LHS

tan 3A = tan(2A + A)

Using identity 

$tan(A + B) = \frac{tan A + tan B}{1-tan A tan B}$

We get

tan 3A = $\frac{tan 2A + tan A}{1-tan 2A tan A}$

= $\frac{\frac{tan 2A}{1-tan^2A} + tan A}{1-\frac{tan 2A tan A}{1-tan^2A}}$

= $\frac{2tan A + tan A-tan^3A}{1-tan^2A-2 tan^2 A}$

= $\frac{3tan A-tan^3A}{1-3tan^2A}$

示例1.解决2sin3xsinx。

解决方案：

We have 2sin3xsinx

We also write as y = y1 . y2  ….(1)

Here, y1 = 2sin3x

y2 = sinx

So let’s solve y1 = 2sin3x

Using identity 

sin 3A = 3sin A – 4 sin³A

We get

y1 = 2(sin x – 4 sin³x)

= 2sin x – 8 sin³x

Now put these values in eq(1), we get

y = (2sin x – 8 sin³x)(sinx)

= 2sin² x – 8 sin⁴x

示例2.解决2tan3xtanx。

解决方案：

We have 2tan3xtanx

We also write as y = y1 . y2  ….(1)

Here, y1 = 2tan3x

y2 = tanx

So let’s solve y1 = 2tan3x

Using identity 

tan 3A = $\frac{3tan A-tan^3A}{1-3tan^2A}$

We get

y1 = 2($\frac{3tan x-tan^3x}{1-3tan^2x}$)

= $\frac{6tan x-2tan^3x}{1-3tan^2x}$

Now put these values in eq(1), we get

y = ($\frac{6tan x-2tan^3x}{1-3tan^2x}$)(tanx)

= $\frac{6tan^2 x-2tan^4x}{1-3tan^2x}$