如果 Sin A = 3/4，计算 cos A 和 tan A

三角学是数学的分支之一。它是对三角形性质的研究，就像它的边长和它们之间的角度之间的关系一样。三角问题中涉及的比率称为三角比率。三角函数有助于理解任何涉及距离、角度、波浪问题等的主题。三角函数（sin、cos 和 tan）是研究科学、技术和工程的关键。

上面给出的问题很简单，可以使用三角比和毕达哥拉斯定理来解决。

三角比

三角比是三角学中最开始学习的东西。这些比率来自理解这一数学分支所需的基本基础。这是我们正在谈论的所有三角比，

直角三角形

sin(θ) = 垂直/斜边

cos(θ) = 底边/斜边

tan(θ) = 垂直/底

cot (θ) = 1/tan (θ) = 底/垂直

sec (θ) = 1/cos (θ) = 斜边/底

cosec (θ) = 1/sin (θ) = 斜边/垂直

毕达哥拉斯定理

毕达哥拉斯定理指出，如果三角形是直角三角形，则三角形的垂线和底边的平方和等于三角形斜边的平方，即

如果在直角三角形中，AB = 垂直，BC = 底边，AC = 斜边

那么，（垂直） ² +（底） ² =（斜边） ²

因此，AB ² + BC ² = AC ²

这被称为毕达哥拉斯定理。

如果 Sin A = 3/4，计算 cos A 和 tan A。

解决方案：

Given that, Sin A = 3/4  i.e.

For a given problem, let Base = X 

Given: AB = Perpendicular = 3 ,  AC = Hypotenuse = 4

Sin(A) = Perpendicular/Hypotenuse = 3/4

Using above Trigonometric ratios:

Cos(A) = Base/Hypotenuse= X/4  ⇢ (equation 1)

Tan(A) = Perpendicular/Base = 3/X  ⇢ (equation 2)

Now to find X we need to apply Pythagoras theorem

Therefore, AB² + BC² = AC²

3² + X² = 4²   

9 + X² = 16

X² = 16 – 9

X ²= 7

X = √ 7.

Now substituting X in equation 1 and 2 we get:

Cos(A) = Base/Hypotenuse=  √7/4  

Tan(A) = Perpendicular/Base = 3/√7  

类似问题

问题 1：如果 Sin A = 4/5，计算 Cot A 和 Cosec A。

解决方案：

Given that, Sin A = 4/5  i.e.

AB = Perpendicular = 4,  AC = Hypotenuse = 5

For a given problem, let Base = X

Sin(A) = Perpendicular/Hypotenuse = 4/5

Using Trigonometric ratios:

(1) cot (A) = Base/Perpendicular = X/4

Now to find X we need to apply Pythagoras theorem

Therefore, AB² + BC² = AC²

4² + BC² = 5²

BC² = 25 – 16

BC² = 9

BC = √9

BC = 3

Therefore, cot (A) = Base/Perpendicular = 3/4

(2) cosec (A) = 1/Sin (A) = Hypotenuse/Perpendicular = 5/4

问题 2：如果 Tan = √7/3，计算 Sec A 和 Cos A。

解决方案：

Perpendicular =  √ 7 , Base = 3 , Hypotenuse = ?

Given that Tan A = √7/3

Tan A = Perpendicular/Base

So, Perpendicular = √7, Base = 3

Let Hypotenuse be X

Using Pythagoras theorem

√7² + 3² = X²

7² + 9 = X²

16 = X²

Therefore, X = 4

Hypotenuse = X = 4

Perpendicular =  √ 7 , Base = 3 , Hypotenuse = 4

(1) Sec A = Hypotenuse/Base = 4/3

(2) Cos A = 1/Sec A = 3/4

问题 3：如果 Cot A = √3，证明 Tan A + Sin A = (2+√3) / 2√3。

解决方案：

Given Cot A = √3

Cot A = Base/Perpendicular

So, Base = √3, Perpendicular = 1

Let Hypotenuse be X,

Perpendicular =  1 , Base =  √ 3 , Hypotenuse = ?

Using Pythagoras theorem 

1² + (√3)²  = X²

1 + 3 = X²

4 = X²

Therefore, X= 2

Hypotenuse = X = 2

Tan A + Sin A = ?

Tan A =1/Cot A = 1/√ 3

Sin A = Perpendicular/Hypotenuse = 1/2

Tan A + Sin A =  1/√3 + 1/2 

= (2 + √ 3) / 2√3

Therefore, Tan A + Sin A =  (2 + √3) / 2√3.

问题 4：如果 cot ² A = 3 (sec ² A – tan ² A) ，那么求 3(sin A)？

解决方案：

Given:  cot² A = 3 (sec² A -tan² A)

sec² A – tan² A = 1

Therefore,  cot² A = 3(1)

So, Cot  A = √3

Cot A =  Base/Perpendicular = √3/1

 Base = √3, Perpendicular = 1

Let Hypotenuse be X,

Perpendicular =  1 , Base =  √ 3 , Hypotenuse = X

Using Pythagoras theorem 

1² + (√3)² = X²

1 + 3 = X²

4 = X²

Therefore, X = 2

Hypotenuse = X = 2

3(Sin A) = ?

Sin A = Perpendicular/Hypotenuse = 3(Perpendicular/Hypotenuse)

Therefore, Sin A = 3(1/2) = 3/2

问题 5：如果 sin ² A sec A + cos A = 5/4，那么求 tan A？

解决方案：

Given: sin²A sec A + cos A = 5/4

Multiply and Divide L.H.S by Cos A ,

Sin²A Sec A + cos A × (Cos A/Cos A) = 5/4

(Sin² A × Sec A × Cos A + Cos² A)/Cos A = 5/4

Sec A = 1/Cos A

So, (Sin² A × 1/ Cos A × Cos A + Cos² A)/Cos A = 5/4

(Sin²A + Cos² A)/Cos A = 5/4

1/Cos A = 5/4

Therefore, Cos A = 4/5

Cos A = Base/Hypotenuse = 4/5

Let Perpendicular be X

Base = 4 , Hypotenuse = 5 , Perpendicular = X

Using Pythagoras theorem, 

4² + X² = 5²

16 + X²  = 25

X² = 25 – 16

X² = 9

X = √9

X = 3

Tan A = ?

Hence, Tan A = Perpendicular/Base = 3/4