Is cot(2u)=(cot^2(u)-1)/(2cot(u)) ?

The answer to whether cot(2u)=(cot^2(u)-1)/(2cot(u)) is True

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Popular Trigonometry >

prove cot(2u)=(cot^2(u)-1)/(2cot(u))

Solution

prove

cot (2 u) = \frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

Solution

True

Solution steps

cot (2 u) = \frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

Manipulating right side

\frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( u ) + cos ^{2} ( u )}{sin ^{2} ( u )}

Multiply:

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

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

= \frac{\frac{c o s ^{2} ( u ) - s i n ^{2} ( u )}{s i n ^{2} ( u )} sin ( u )}{2 cos ( u )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (u)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( u ) - sin ^{2} ( u )}{2 cos ( u ) sin ( u )}

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

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

Use the basic trigonometric identity:

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

= cot (2 u)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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