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

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

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

prove

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

True

Solution steps

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

Let:

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

We showed that the two sides could take the same form

\Rightarrow True

2 cot (θ) = csc^{2} (θ) sin (2 θ)

cos (3 t) = cos^{3} (t) - 3 sin^{2} (t) cos (t)

sin^{2} (x) sec (x) csc (x) = tan (x)

4 sin (x) cos (x) + 1 = 2 sin (2 x) + 1

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

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