Is ((1))/((2))(cot(θ)-tan(θ))=cot(2θ) ?

The answer to whether ((1))/((2))(cot(θ)-tan(θ))=cot(2θ) is True

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prove ((1))/((2))(cot(θ)-tan(θ))=cot(2θ)

Solution

prove

\frac{( 1 )}{( 2 )} (cot (θ) - tan (θ)) = cot (2 θ)

Solution

True

Solution steps

\frac{1}{2} (cot (θ) - tan (θ)) = cot (2 θ)

Manipulating left side

\frac{1}{2} (cot (θ) - tan (θ))

Express with sin, cos

(cot (θ) - tan (θ)) \frac{1}{2}

Use the basic trigonometric identity:

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

= (\frac{cos ( θ )}{sin ( θ )} - tan (θ)) \frac{1}{2}

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}) \frac{1}{2}

Multiply fractions:

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

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

1 \cdot (\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}) = \frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}

1 \cdot (\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )})

Multiply:

1 \cdot (\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}) = (\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )})

= (\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )})

Remove parentheses:

(a) = a

= \frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{s i n ( θ )}{c o s ( θ )}}{2}

Join

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

\frac{cos ( θ )}{sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}

Least Common Multiplier of

sin (θ), cos (θ) : sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (θ) cos (θ)

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

\frac{cos ( θ )}{sin ( θ )} = \frac{cos ( θ ) cos ( θ )}{sin ( θ ) cos ( θ )} = \frac{cos ^{2} ( θ )}{sin ( θ ) cos ( θ )}

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

\frac{sin ( θ )}{cos ( θ )} = \frac{sin ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = \frac{sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{c o s ^{2} ( θ ) - s i n ^{2} ( θ )}{s i n ( θ ) c o s ( θ )}}{2}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

Use the basic trigonometric identity:

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

= cot (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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