Is cot(x/2)=(1+cos(x))/(sin(x)) ?

The answer to whether cot(x/2)=(1+cos(x))/(sin(x)) is True

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prove cot(x/2)=(1+cos(x))/(sin(x))

prove

cot (\frac{x}{2}) = \frac{1 + cos ( x )}{sin ( x )}

True

Solution steps

cot (\frac{x}{2}) = \frac{1 + cos ( x )}{sin ( x )}

Let:

u = \frac{x}{2}

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

Prove

cot (u) = \frac{1 + cos ( 2 u )}{sin ( 2 u )} :

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

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

Simplify

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

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

1 + 2 cos^{2} (u) - 1 = 2 cos^{2} (u)

1 + 2 cos^{2} (u) - 1

Group like terms

= 2 cos^{2} (u) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (u)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (u)

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

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

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

Use the basic trigonometric identity:

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

= cot (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

cot (\frac{x}{2}) = \frac{1 + cos ( x )}{sin ( x )}

We showed that the two sides could take the same form

\Rightarrow True

(sin (x)) (csc (x)) = 1

1 + cos (2 t) = cot (t) sin (2 t)

sin (α) = \frac{2 tan ( \frac{α}{2} )}{1 + tan ^{2} ( \frac{α}{2} )}

(1 + sin (x)) (1 - sin (x)) = \frac{1}{sec ^{2} ( x )}

sin (3 x) cos (2 x) - cos (3 x) sin (2 x) = sin (x)

Is cot(x/2)=(1+cos(x))/(sin(x)) ?
The answer to whether cot(x/2)=(1+cos(x))/(sin(x)) is True