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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{cos ( u )}{sin ( u )} \cdot \frac{1}{2}

Multiply fractions:

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

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

Multiply:

cos (u) \cdot 1 = cos (u)

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

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

\frac{1}{2} \cdot \frac{sin ( u )}{cos ( u )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (u) = sin (u)

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

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

Least Common Multiplier of

sin (u) 2, 2 cos (u) : 2 sin (u) cos (u)

sin (u) \cdot 2, 2 cos (u)

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

sin (u) 2

2 cos (u)

= 2 sin (u) cos (u)

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

2 sin (u) cos (u)

For

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

multiply the denominator and numerator by

cos (u)

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

For

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

multiply the denominator and numerator by

sin (u)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Therefore

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

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