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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (b)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

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

Use the basic trigonometric identity:

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

= cot (2 b)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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