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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= 1 - \frac{2}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}

Simplify

1 - \frac{2}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}} : \frac{- sin ^{2} ( x ) + cos ^{2} ( x )}{sin ^{2} ( x ) + cos ^{2} ( x )}

1 - \frac{2}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}

\frac{2}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}} = \frac{2 sin ^{2} ( x )}{sin ^{2} ( x ) + cos ^{2} ( x )}

\frac{2}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}

Apply exponent rule:

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

= \frac{2}{1 + \frac{c o s ^{2} ( x )}{s i n ^{2} ( x )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

= \frac{2}{\frac{s i n ^{2} ( x ) + c o s ^{2} ( x )}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

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

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

Add similar elements:

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

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

= \frac{cos ( 2 x )}{1}

Apply rule

\frac{a}{1} = a

= cos (2 x)

= cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) - 1 = cot (x)

sin^{2} (x) = \frac{sec ( x ) sin ( x )}{tan ( x ) + cot ( x )}

sin (x + y) + sin (x - y) = 2 cos (x) sin (y)

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

(cos^{2} (x) + 1) (sec^{2} (x)) = tan^{2} (x) + 2

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