Question

Find $\tan \left(22.5^{\circ }\right)$ using the half-angle formula.

Answer 1

Find the value using half-angle formula

We know that $\tan \left(2x\right)=\frac{2\tan x}{1-{\left(\tan \left(x\right)\right)}^2}$ and

Let $y=\tan \left(x\right)$ where $x={\left(22.5\right)}^{\circ }$

So $2x={\left(45\right)}^{\circ }$

We know that $\tan \left({45}^{\circ }\right)=1$

So $\frac{2y}{1-y^2}=1$

$$\begin{gathered} \Rightarrow 1-y^2=2y \\ \Rightarrow y^2+2y-1=0 \end{gathered}$$

We know if $\alpha$ and $\beta$ are the roots of a quadratic equation of the form $a{x}^2+bx+c=0$ then the roots are

$$\begin{gathered} \alpha =\frac{-b+\sqrt{b^2-4ac}}{2a} \\ \beta =\frac{-b-\sqrt{b^2-4ac}}{2a} \end{gathered}$$

Therefore the roots are $\sqrt{2}-1$ and $-\sqrt{2}-1$

Since, $\tan \left(22.5^{\circ }\right)$ lie in first quadrant so we will take the positive value.

Hence, the value of $\tan \left(22.5^{\circ }\right)$ is $\sqrt{2}-1$