**Introduction:**

Learn in online is very easy task in now a days. Instead of online, we refer any books (or) any teacher to study about our subjects. But It takes lot of time. How means, searching related book in library from thousand number of books. But if we use to search in online for learningany subject, it takes fraction of a second to get exact content what weare looking for. So we save our time. In online, Many tutorials of different math topic are available. Tutorial is used to share information about topic. In Tutorial, we will know definition, proofs and example problems of a topic. Now we will study about half angle identities in tutorial.

There are three half angle identities used in trigonometry. They are sine , cosine and tangent. Now we will see each proof .

From double angle identities for cosine , we can derive half angle for cosine.

Cosine Double angle identities is

**Cos 2θ = 2cos ^{2} θ – 1**

Let θ be `theta/2`.

Cos 2( `theta/2`) = 2cos^{2} `theta/2` – 1

Cos θ = 2cos^{2} `theta/2` – 1

Now we need to solve for cos `theta/2`, for that we rearrange the above one.

2cos^{2 }`theta/2` = 1 + cos θ

Divide by 2 each side.

Cos^{2} `theta/2` = `(1+cos theta)/2`

Take square root on each side.

**Cos `theta/2`**** = ± **`sqrt((1+cos theta)/2)`

If we get positive answer, half angle is in first (or) fourth quadrants.

If we get negative answer, half angle is in second (or) third quadrants.

**Prepare Example Problem:**

Find the value of cos `theta/2` if cos `theta` = `12/13` where 0° < `theta` < 90°.

**Solution:**

Half-angle formulae for cosine:

Cos `theta/2` = ± `sqrt((1+cos theta)/2)`

The given angle is in between first quadrants, so we use positive formulae.

Cos `(theta)/2` = + `sqrt((1+(12/13))/2)`

Cos `(theta)/2` = `sqrt(((13+12)/13)/2)`

Cos `(theta)/2` = `sqrt((25)/26)`

Cos `(theta)/2` = `5/sqrt(26)`

**Cos `(theta)/2` ** **= 0.9803 **** **

**Half angle identities – sine:**

From double angle identities for cosine , we can derive half angle for sine.

Cosine Double angle formulas is

**Cos 2θ = 1 – 2sin ^{2} θ**

Let θ be `theta/2`.

Cos 2(`theta/2`) = 1 – 2sin^{2} `theta/2`

Cos θ = 1 – 2sin^{2} `theta/2`

Now we need to solve for sin `theta/2`, for that we rearrange the above one.

2sin^{2} `theta/2` = 1 – cos θ

Divide by 2 each side.

Sin^{2} `theta/2` = `(1-cos theta)/2`

Take square root on each side.

**Sin `theta/2`**** = ± **`sqrt((1-cos theta)/2)`

If we get positive answer, half angle is in first (or) second quadrants.

If we get negative answer, half angle is in third (or) fourth quadrants.

**Prepare Example Problem:**

Find the value of sin `theta/2` if cos `theta` = `12/13` where 0° < `theta` < 90°.

**Solution:**

Half-angle identities for sine:

Sin `theta/2` = ± `sqrt((1-cos theta)/2)`

The given angle is in between first quadrants, so we use positive formulae.

Sin `(theta)/2` = + `sqrt((1-(12/13))/2)`

Sin `(theta)/2` = `sqrt(((13-12)/13)/2)`

Sin `(theta)/2` = `sqrt((1)/26)`

Sin `(theta)/2` = `1/sqrt(26)`

**Sin `(theta)/2` ** **= 0.1961 **

**Half angle identities – tangent:**

The tangent half angle identites can be derived from trignometric ratio,

Tan θ = `(sin theta)/(cos theta)`

Tan `theta/2` = `(sin theta/2)/(cos theta/2)`

Now we apply sine and cosine half angle formulae into above one.

Tan `theta/2` = `(+-sqrt((1-cos theta)/2))/(+-sqrt((1+cos theta)/2))`

Now we rearrange the above one.

Tan `theta/2` = ± `sqrt((1-cos theta)/(1+cos theta))`

Other forms of tangent half angle formulas are written as

Tan `theta/2` = `(1-cos theta)/(sin theta)` = `(sin theta)/(1+cos theta)`

**Prepare Example Problem:**

Find the value of tan `theta/2` if cos `theta` = `12/13` where 0° < `theta` < 90°.

**Solution:**

Half-angle formulae for cosine:

Tan `theta/2` = ± `sqrt((1+cos theta)/(1-cos theta))`

The given angle is in between first quadrants, so we use positive formulae.

Tan `(theta)/2` = + `sqrt((1+(12/13))/(1-(12/13)))`

Tan `(theta)/2` = `sqrt(((13+12)/13)/((13-12)/13))`

Tan `(theta)/2` = `sqrt((25)/1)`

**Tan `(theta)/2` = ** `5`