证明 cos θ/(1 + sin θ) = (1 + cos θ – sin θ)/(1 + cos θ + sin θ)

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、正割或余割。

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°.

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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.

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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

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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

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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

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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

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证明 cos θ/(1 + sin θ) = (1 + cos θ – sin θ)/(1 + cos θ + sin θ)

解决方案：

We have cos θ/(1 + sin θ) = (1 + cos θ – sin θ)/(1 + cos θ + sin θ)

First we will take LHS = cos θ/(1 + sin θ)

= {[cos θ/(1 + sin θ)] × [(1-sin θ)/(1-sin θ)]}

= [cos θ(1-sin θ)]/(1- sin²θ)

= [cos θ(1-sin θ)]/ cos² θ { 1- sin²θ = cos²θ }

= (1-sin θ) / cos θ

= (1/ cos θ) – (sin θ/cos θ)

LHS = sec θ – tan θ

Now

RHS = (1 + cos θ – sin θ)/(1 + cos θ + sin θ)

= [(1 + cos θ – sin θ)/(1 + cos θ + sin θ) × (1 + cos θ – sin θ)/(1 + cos θ – sin θ)]

= [(1 + cos θ – sin θ)²] /[(1 + cos θ)² + (sin θ)²]

= [(1 + cos θ)² + sin²θ – 2 (1 + cos θ)(sin θ) ]/(1 + cos²θ + 2cos θ – sin² θ)

= [(1 + cos ²θ + sin² θ + 2 cos θ – 2 (1+cos θ)(sin θ)]/[1+ cos² θ + 2cos θ -(1-cos ²θ)] {… 1- cos²θ = sin²θ}

= [2 +2 cos θ – 2 (sin θ) (1 + cos θ) ]/[1 +cos² θ + 2cos θ – 1 +cos ²θ] {sin²θ + cos²θ = 1}

= [2 + 2cos θ – 2sin θ (1 + cos θ)]/[2cos ²θ + 2cos θ]

= [2 + 2cos θ – 2sin θ (1 + cos θ)] / [2 cos θ (cos θ +1)]

= [2 (1 + cos θ) – 2 sin θ(1 + cos θ)] / [2 cos θ (cos θ +1)]

= [2 (1 + cos θ) (1 – sin θ)]/[2 cos θ (cos θ +1)]

= (1 – sin θ) / cos θ

= (1/ cos θ) – (sin θ/ cos θ)

RHS = sec θ – tan θ

Therefore LHS = RHS

cos θ/(1 + sin θ) = (1 + cos θ – sin θ)/(1 + cos θ + sin θ)

Hence proved

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类似问题

问题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，（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 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 & y² = sin² θ

so now x²+ y² = cos² θ + sin²θ { cos²θ + sin ²θ = 1 }

x² + y² = 1

Hence proved

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问题 2：评估 (Sin 30° – Sin 90° + 2 Cos 0°) / Tan 30° Tan 60°？

解决方案：

Here we have (Sin 45° – Sin 90° + 2 Cos 0°) / Tan 45° Tan 60°

As per the trigonometric values

(Sin 45° – Sin 90° +2 Cos 0°) / Tan 45° Tan 60°

= (1/√2 – 1 + 2 × 1) / 1 × √3

= (1/√2 – 1 + 2) / √3

= (1/√2 + 1) / √3

= (1+√2 / √2) / √3

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