1. ## Half angle Identities

Hi
This may take a while and a few posts so be pateint please!!
I can't understand when to use which formula and how to recognise half angle problems.
Can I use either of the formulae below to start solving?

If I am told for example tan theta = 4 and asked to find cos x/2 which is more beneficial?
Thanks in advance, as I say this may take a few post

2. Originally Posted by 200001
Hi
This may take a while and a few posts so be pateint please!!
I can't understand when to use which formula and how to recognise half angle problems.
Can I use either of the formulae below to start solving?

If I am told for example tan theta = 4 and asked to find cos x/2 which is more beneficial?
Thanks in advance, as I say this may take a few post
The second formula is not valid: it should be $\displaystyle \cos\theta =2\cos^2\frac{\theta}{2}-1$. Solving for $\displaystyle \cos\frac{\theta}{2}$ in this improved version of the second formula gives the first formula. So these two formulas are not fundamentally different, and therefore there is hardly any difference in their applicability to this or any other problem...

Now, since $\displaystyle \tan\theta=4$ is given, you need to be able to express $\displaystyle \cos\theta$ in this formula for $\displaystyle \cos\frac{\theta}{2}$ in terms of $\displaystyle \tan\theta$.
And how do you do that? Tipp: sketch a right triangle with acute angle $\displaystyle \theta$, adjacent leg $\displaystyle 1$ and opposite leg $\displaystyle \tan\theta$. By Pyhtagoras's theorem you have that the length of the hypotenuse of this triangle is $\displaystyle \sqrt{1+\tan^2\theta}$.
Therefore, by definition, we have $\displaystyle \cos\theta =\frac{1}{\sqrt{1+\tan^2\theta}}$. This is valid only for acute angles $\displaystyle \theta$. In the general case, you need to write this as $\displaystyle \cos\theta=\pm\frac{1}{\sqrt{1+\tan^2\theta}}$. The sign depends on the quadrant in which the point that corresponds to $\displaystyle \theta$ lies on the unit circle.

To sum it up you get:
$\displaystyle \cos\frac{\theta}{2}=\pm\sqrt{\frac{1\pm \frac{1}{\sqrt{1+\tan^2\theta}}}{2}}$
If you are allowed to assume that $\displaystyle \theta$ is an acute angle, then you only take the $\displaystyle +$ case of $\displaystyle \pm$. Now plug in the given value of $\displaystyle \tan\theta=4$...

3. Awesome
Thanks for the lengthy insight.
I have applied that to the ones I have and its helped. I am struggling to find examples to do after a google.
Any links to some exercises would be fantastic

Thanks once again for your help

4. Originally Posted by 200001
Awesome
Thanks for the lengthy insight.
I have applied that to the ones I have and its helped. I am struggling to find examples to do after a google.
Any links to some exercises would be fantastic

Thanks once again for your help
Dear 200001,

There are many trignometric books and worksheets in the link below.