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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Factor

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

1 - cos^{2} (x)

Factor out common term

- 1

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Cancel the common factor:

cos (x) + 1

= - (cos (x) - 1)

Negate

- (cos (x) - 1) = - cos (x) + 1

= - cos (x) + 1

= 1 - cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin (2 θ) + 2 sin (θ) - cos (θ) = 2

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

\frac{csc ( x ) + 1}{cot ( x ) + cos ( x )} = sec (x)

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