Question

What is the proof of the half angle formula?

Answer 1

Hint: In the given question we basically mean to find the formula at half angles using trigonometric functions. We already might be aware of most of the identities that are used of half angles; we just need to prove them.

Complete step by step solution:
We will have to attain the following two terms of the half angle formula, the terms are:
$\begin{align}
  & \cos \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1}{2}(1+\cos \theta )} \\
 & \sin \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1}{2}(1-\cos \theta )} \\
\end{align}$
Now we know that theta is half of twice theta that is why we will put theta equal to twice alpha i.e., $\theta =2\alpha $ in order to make $\alpha =\dfrac{\theta }{2}$ .
Now we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and now if we solve this algebraically, we will get the half angle cosine formula,
We also know that $\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $ .
Now, we have replaced the value of alpha by theta as we have assumed before.
Further after transposing this and exchanging the sides
$\begin{align}
  & \cos \theta +1=2{{\cos }^{2}}\dfrac{\theta }{2} \\
 & \Rightarrow \dfrac{\cos \theta +1}{2}={{\cos }^{2}}\dfrac{\theta }{2} \\
 & \Rightarrow \sqrt{\dfrac{\cos \theta +1}{2}}=\cos \dfrac{\theta }{2} \\
\end{align}$
Now, exchanging sides we get, $\cos \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1}{2}(1+\cos \theta )}$ . This is the half angle formula for the cosine and also, we should know that $\pm $ this sign will depend on the quadrant of the half angle. Any argument theta or alpha can be used as will does not make any change.
Now, for the sine half angle formula we transpose and exchange the sides of $\cos 2\alpha =1-2{{\sin }^{2}}\alpha $ as
$\begin{align}
  & \cos 2\alpha =1-2{{\sin }^{2}}\alpha \\
 & \Rightarrow 2{{\sin }^{2}}\alpha =1-\cos 2\alpha \\
 & \Rightarrow {{\sin }^{2}}\alpha =\dfrac{1-\cos 2\alpha }{2} \\
 & \Rightarrow \sin \alpha =\dfrac{1-\cos 2\alpha }{2} \\
\end{align}$
And now after putting the value of alpha we get $\sin \dfrac{\theta }{2}=\pm \sqrt{\dfrac{1}{2}(1-\cos \theta )}$.

Note: We have studied all these properties before in trigonometry and at various topics of calculus like differentiation and integration in order to obtain the answer. In this we must be very clear with the concept of the quadrant in which the angles lie because that plays a very important role in this. It is not just a change in sign rather it changes the meaning of the trigonometric function completely.