Cos double angle identity in terms of Tan function

Formula

$\cos{2\theta}$ $\,=\,$ $\dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$

A mathematical identity that expresses the expansion of cosine of double angle in terms of tan squared of angle is called the cosine of double angle identity in tangent.

Introduction

Let the theta be an angle of a right triangle. The square of tan of angle is written as $\tan^2{\theta}$ and the cosine of double angle is written as the $\cos{2\theta}$ in trigonometric mathematics.

The cosine of double angle is equal to the quotient of the subtraction of square of tangent from one by the sum of one and square of tan function.

$\cos{2\theta}$ $\,=\,$ $\dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$

It is called the cosine of double angle identity in terms of tangent function.

Usage

In trigonometry, the cos double angle identity can be used as a formula in two distinct cases.

Expansion

It is used to expand the cosine of double angle functions as the quotient of the subtraction of tan squared of angle from one by the sum of one and tan squared of angle.

$\implies$ $\cos{2\theta}$ $\,=\,$ $\dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$

Simplified form

It is also used to simplify the quotient of the subtraction of tan squared of angle from one by the sum of one and tan squared of angle as the cosine of double angle.

$\implies$ $\dfrac{1-\tan^2{\theta}}{1+\tan^2{\theta}}$ $\,=\,$ $\cos{2\theta}$

Other forms

In the cosine double angle formula, the angle can be represented by any symbol. Hence, it is also written in two other forms popularly.

$(1). \,\,\,\,\,\,$ $\cos{2x}$ $\,=\,$ $\dfrac{1-\tan^2{x}}{1+\tan^2{x}}$

$(2). \,\,\,\,\,\,$ $\cos{2A}$ $\,=\,$ $\dfrac{1-\tan^2{A}}{1+\tan^2{A}}$

Proof

Learn how to prove the cosine of double angle rule in terms of square of tan function in trigonometry.

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.