Is cot(2β)=(cot^2(β)-1)/(2cot(β)) ?

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

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prove cot(2β)=(cot^2(β)-1)/(2cot(β))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

2 \cdot \frac{cos ( β )}{sin ( β )}

Multiply fractions:

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

= \frac{cos ( β ) \cdot 2}{sin ( β )}

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot sin^{2} (β) = sin^{2} (β)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (β)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

= \frac{cos ( 2 β )}{sin ( 2 β )}

= \frac{cos ( 2 β )}{sin ( 2 β )}

Use the basic trigonometric identity:

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

= cot (2 β)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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