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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

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 sin ( u ) cos ( u )}{2 cos ^{2} ( u )}

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (u)

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

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

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

Use the basic trigonometric identity:

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

= tan (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{csc ( x )}{cos ( x )} = tan (x) + cot (x)

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

cot (x) \cdot sin (x) = cos (x)

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )} = 1

\frac{cot ( y )}{sec ( y )} = csc (y) - sin (y)

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