Question

How do you find the exact values of tan pi/8 using the half-angle formula?

Answer 1

In the above question, we were asked to find the exact value of ran pi/8 using the half-angle formula. We will use $\tan (\dfrac{\theta }{2})=\dfrac{1-\cos \theta }{\sin \theta }$ , this is the half-angle formula for tan. We will substitute $\dfrac{\theta }{2}$ with $\dfrac{\pi }{8}$. So, let’s see how we can solve this problem.

Complete step by step solution:
The half-angle formula for tangent can be written as: $\tan (\dfrac{\theta }{2})=\dfrac{1-\cos \theta }{\sin \theta }$ and we will use this formula to solve the above problem.
Let, $\theta =\dfrac{\pi }{4}$
$\Rightarrow \tan (\dfrac{\pi }{8})=\dfrac{1-\cos (\dfrac{\pi }{4})}{\sin (\dfrac{\pi }{4})}$
Value of $\cos (\dfrac{\pi }{4})$ and $\sin (\dfrac{\pi }{4})$ is $\dfrac{1}{\sqrt{2}}$. On multiplying both the numerator and denominator with $\sqrt{2}$ we will get $\dfrac{\sqrt{2}}{2}$.
$=\dfrac{1-\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}$
After simplifying the numerator, we get
$=\dfrac{\dfrac{2-\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}$
On solving the numerator and denominator we will get,
$=\dfrac{2-\sqrt{2}}{\sqrt{2}}$

Therefore, $\tan (\dfrac{\pi }{8})=\dfrac{2-\sqrt{2}}{\sqrt{2}}$.

Additional Information:
In the above solution, we have used a half-angle formula for the tangent. There is a half-angle formula for sin and cos as well. $\sin (\dfrac{\theta }{2})=\pm \sqrt{\dfrac{1-\cos \theta }{2}}$ and $\cos (\dfrac{\theta }{2})=\pm \sqrt{\dfrac{1+\cos \theta }{2}}$ . Also, $\cos (2\theta )={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta =2{{\cos }^{2}}\theta -1$, $\sin 2\theta =2\sin \theta \cos \theta$ and $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$. All these formulas are very useful and sometimes they are converted according to the problem statement. We will study these in the coming lectures.

Note:
In the above solution, we have used the formula $\tan (\dfrac{\theta }{2})=\dfrac{1-\cos \theta }{\sin \theta }$. Let’s see how it is derived. Formula of $\tan (\dfrac{\theta }{2})=\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}$.
$\Rightarrow \tan (\dfrac{\theta }{2})=\dfrac{\sin \theta }{1+\cos \theta }$
$\Rightarrow \tan (\dfrac{\theta }{2})=\dfrac{1-\cos \theta }{sin\theta }$
This is how we get this formula.