Is cot^2(x)=(1+cos(2x))/(1-cos(2x)) ?

The answer to whether cot^2(x)=(1+cos(2x))/(1-cos(2x)) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cot^2(x)=(1+cos(2x))/(1-cos(2x))

prove

cot^{2} (x) = \frac{1 + cos ( 2 x )}{1 - cos ( 2 x )}

True

Solution steps

cot^{2} (x) = \frac{1 + cos ( 2 x )}{1 - cos ( 2 x )}

Manipulating right side

\frac{1 + cos ( 2 x )}{1 - cos ( 2 x )}

Rewrite using trig identities

\frac{1 + cos ( 2 x )}{1 - cos ( 2 x )}

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= \frac{1 + 1 - 2 sin ^{2} ( x )}{1 - ( 1 - 2 sin ^{2} ( x ) )}

Simplify

\frac{1 + 1 - 2 sin ^{2} ( x )}{1 - ( 1 - 2 sin ^{2} ( x ) )} : \frac{- sin ^{2} ( x ) + 1}{sin ^{2} ( x )}

\frac{1 + 1 - 2 sin ^{2} ( x )}{1 - ( 1 - 2 sin ^{2} ( x ) )}

Add the numbers:

1 + 1 = 2

= \frac{- 2 sin ^{2} ( x ) + 2}{1 - ( - 2 sin ^{2} ( x ) + 1 )}

Expand

1 - (1 - 2 sin^{2} (x)) : 2 sin^{2} (x)

1 - (1 - 2 sin^{2} (x))

- (1 - 2 sin^{2} (x)) : - 1 + 2 sin^{2} (x)

- (1 - 2 sin^{2} (x))

Distribute parentheses

= - (1) - (- 2 sin^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (x)

= 1 - 1 + 2 sin^{2} (x)

1 - 1 = 0

= 2 sin^{2} (x)

= \frac{- 2 sin ^{2} ( x ) + 2}{2 sin ^{2} ( x )}

Factor

- 2 sin^{2} (x) + 2 : 2 (- sin^{2} (x) + 1)

- 2 sin^{2} (x) + 2

Rewrite as

= - 2 sin^{2} (x) + 2 \cdot 1

Factor out common term

2

= 2 (- sin^{2} (x) + 1)

= \frac{2 ( - sin ^{2} ( x ) + 1 )}{2 sin ^{2} ( x )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{- sin ^{2} ( x ) + 1}{sin ^{2} ( x )}

= \frac{- sin ^{2} ( x ) + 1}{sin ^{2} ( x )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

= \frac{cos ( x )}{sin ( x )} \cdot \frac{cos ( x )}{sin ( x )}

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

\frac{cos ( x ) cot ( x )}{sin ( x )}

= cot (x) \frac{cos ( x )}{sin ( x )}

Use the basic trigonometric identity:

\frac{cos ( x )}{sin ( x )} = cot (x)

cot (x) cot (x)

Simplify

cot (x) cot (x) : cot^{2} (x)

cot (x) cot (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (x) cot (x) = cot^{1 + 1} (x)

= cot^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cot^{2} (x)

cot^{2} (x)

cot^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (π + x) = cos (x)

sec (2 x) = \frac{( sec ^{2} ( x ) )}{1 - tan ^{2} ( x )}

(sin^{2} (x) - 1)^{2} = cos (2 x) + sin^{4} (x)

sec (x) - cos (x) = sin^{2} (x) sec (x)

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) + 1

Is cot^2(x)=(1+cos(2x))/(1-cos(2x)) ?
The answer to whether cot^2(x)=(1+cos(2x))/(1-cos(2x)) is True