Is cot^2(θ)(1-cos^2(θ))=cos^2(θ) ?

The answer to whether cot^2(θ)(1-cos^2(θ))=cos^2(θ) 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(θ)(1-cos^2(θ))=cos^2(θ)

prove

cot^{2} (θ) (1 - cos^{2} (θ)) = cos^{2} (θ)

True

Solution steps

cot^{2} (θ) (1 - cos^{2} (θ)) = cos^{2} (θ)

Manipulating left side

cot^{2} (θ) (1 - cos^{2} (θ))

Express with sin, cos

(1 - cos^{2} (θ)) cot^{2} (θ)

Use the basic trigonometric identity:

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

= (1 - cos^{2} (θ)) (\frac{cos ( θ )}{sin ( θ )})^{2}

Simplify

(1 - cos^{2} (θ)) (\frac{cos ( θ )}{sin ( θ )})^{2} : \frac{cos ^{2} ( θ ) ( 1 - cos ^{2} ( θ ) )}{sin ^{2} ( θ )}

(1 - cos^{2} (θ)) (\frac{cos ( θ )}{sin ( θ )})^{2}

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ^{2} ( θ ) ( 1 - cos ^{2} ( θ ) )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) ( 1 - cos ^{2} ( θ ) )}{sin ^{2} ( θ )}

= \frac{( 1 - cos ^{2} ( θ ) ) cos ^{2} ( θ )}{sin ^{2} ( θ )}

Rewrite using trig identities

\frac{( 1 - cos ^{2} ( θ ) ) cos ^{2} ( θ )}{sin ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin^{2} (θ)

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{π}{3} + x) - cos (\frac{π}{6} + x) = sin (x)

csc (2 x) - cot (2 x) = tan (x)

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

tan^{2} (θ) = csc^{2} (θ) tan^{2} (θ) - 1

cos (x + \frac{π}{6}) + sin (x - \frac{π}{3}) = 0

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