Is cot(2x)=(csc^2(x)-2)/(2cot(x)) ?

The answer to whether cot(2x)=(csc^2(x)-2)/(2cot(x)) is True

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Popular Trigonometry >

prove cot(2x)=(csc^2(x)-2)/(2cot(x))

Solution

prove

cot (2 x) = \frac{csc ^{2} ( x ) - 2}{2 cot ( x )}

Solution

True

Solution steps

cot (2 x) = \frac{csc ^{2} ( x ) - 2}{2 cot ( x )}

Manipulating right side

\frac{csc ^{2} ( x ) - 2}{2 cot ( x )}

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

2 \cdot \frac{cos ( x )}{sin ( x )} : \frac{2 cos ( x )}{sin ( x )}

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

Multiply fractions:

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

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

Use the basic trigonometric identity:

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

= cot (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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