Why is $t=\frac{1}{2}$ a root for $\tan 4\theta= \frac{4t-4t^3}{1-6t^2+t^4}=\frac{-24}{7}$, where $t=\tan \theta$

Question

Show that $(2+i)^4=-7+24i$

$$\cos 4\theta = \cos^4 \theta - 6\cos^2 \theta \sin^2 \theta + \sin^4 \theta$$

$$\sin 4\theta= 4\sin \theta \cos^3 \theta- 4 \sin^3 \theta \cos \theta$$

$$\tan 4\theta= \frac{4t-4t^3}{1-6t^2+t^4}$$

where $t=\tan \theta$

QUESION

By considering the argument of $(2+i)$, explain why $t=\frac{1}{2}$ is a root for the following equation:

$$\tan 4\theta= \frac{4t-4t^3}{1-6t^2+t^4}=\frac{-24}{7}$$

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The first part:

$(2+i)^4 = (2+i)^2 (2+i)^2 = (3+4i)^2 = -7+24i$

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Please help me in the last part, I don't quite know how to do this part, but here are my thoughts:

The argument of $(2+i)$ is :

$$\theta = \arctan \frac{1}{2}\implies \tan \theta= \frac{1}{2}$$

However I don't know how this relates to the latter equation. I've also noticed something cheesy here that :

$$\tan 4\theta= \frac{4t-4t^3}{1-6t^2+t^4}=\frac{-27}{7}= \tan \gamma$$

Where $\gamma$ is the argument of $(-7+24i)$ or $(2+i)^4$

Answer 1

The argument of $(2+i)$ is $\arctan(1/2)$.

When you raise this number to the 4th power, you have $(-7+24i)$, and this number's argument is $\arctan(-24/7)+\pi$.

Since raising a complex number to the 4th power multiplies its argument by $4$, we have that $$4\arctan(1/2)=\arctan(-24/7)+\pi\text{.}$$ Apply $\tan$ to both sides, and $$\tan(4\arctan(1/2))=-{\frac{24}{7}}\text{.}$$

Therefore $t=1/2$ is a solution to $\tan(4\arctan(t))=-{\frac{24}{7}}\text{.}$

Answer 2

Let $2+i=re^{i\theta}$ so that $(2+i)^4=r^4e^{4i\theta}$

We have $$2+i=r\cos \theta +ir\sin \theta$$ and $$(2+i)^4=r^4\cos 4\theta+ir^4\sin 4\theta$$

From these we can compute $\tan \theta$ and $\tan 4\theta$.

To complete the question you will need to compute $\tan 4\theta$ in two ways - first by using $\tan \theta =\frac 12$ and the expression for $\tan 4\theta$ in terms of $\tan \theta$, and secondly by computing $(2+i)^4$ directly.