value of $\tan(A)$

Question

What is value of $tanA$ if $2\tan(2A)+4\tan(4A)+\frac{8}{\tan(8A)}=0$ writing everything in $\tan(A)$ and solving for $t$ is next to impossible without maths engines. So i am seeking for a shorter way. Thanks

Answer 1

First this equation is defined if and only if $2A\not\equiv \dfrac\pi2\mod\pi$, $\;4A\not\equiv \dfrac\pi2\mod\pi$ and $8A\not\equiv 0\mod\dfrac\pi2$, i.e. if and only if $\begin{cases}A\not\equiv \dfrac\pi4,\dfrac \pi8\mod\dfrac\pi2,\\A\not\equiv 0\mod\dfrac\pi{16}. \end{cases}$

The first condition is redundant w.r.t. the second, so it all comes down to the simple $$A\not\equiv 0\mod\dfrac\pi{16}.$$

Now use the duplication formulae for the tangent: $$4\tan 4A+\frac8{\tan 8A}=4\tan 4A+ \frac{8}{\cfrac{2\tan4A}{1-\tan^24A}}=4\Bigl(\tan4A+\frac{1-\tan^24A}{\tan 4A}\Bigr)=\frac4{\tan 4A}$$ Similarly $$2\tan 2A+4\tan 4A+\frac8{\tan 8A}=2\tan 2A+\frac4{\tan 4A}=\frac2{\tan2A},$$ which can't be $0$.

However you might consider it means $\tan 2A=\pm\infty$, which means $2A\equiv \dfrac\pi2\mod\pi$, i.e. $A\equiv\dfrac\pi4\mod\dfrac\pi2$.

Answer 2

Let $x=4A$. Then, the equation $2\tan(2A)+4\tan(4A)+8\cot(8A)=0$ is equivalent to

$$\tan(x/2)+2\tan(x)+4\cot(2x)=0 \tag 1$$

Using the double angle formula for the tangent function, we write

$$\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)} \tag 2$$

Using $(2)$ we find that

$$2\tan(x)+4\cot(2x)=2\cot(x) \tag 3$$

This reduces $(1)$ to the equation

$$\tan(x/2)+2\cot(x)=0 \tag 4$$

But comparing $(4)$ to $(3)$, we see that

$$\tan(x/2)+2\cot(x)=\cot(x/2)$$

Therefore, we have reduced the problem to finding the roots of

$$\cot(x/2)=0$$

These roots are at

$$x=(2n+1)\pi$$

for all integer values of $n$.

Finally, we find that since $x=4A$,

$$\bbox[5px,border:2px solid #C0A000]{A=(2n+1)\pi/4}$$

However, note that $\tan(2A)$ is not defined. And $\cot(8A)$ is likewise not defined. We have found spurious solutions that one might interpret as a solution in the sense of a limit or in the sense of the extended reals.

Answer 3

$\tan(\phi)+{2 \over \tan(2\phi)} = {\sin(\phi) \over \cos(\phi)} + {\cos(2\phi) \over \sin(\phi) \cos(\phi)} = {\sin^2(\phi) + \cos^2(\phi) - \sin^2(\phi) \over \sin(\phi) \cos(\phi)} = {\cos(\phi) \over \sin(\phi)} = {1 \over \tan(\phi)}$

$\tan(a)+2\tan(2a)+{4 \over \tan(4a)} = tan(a) + {2 \over \tan(2a)} = {1 \over \tan(a)} = 0$

Answer 4

HINT:

$$\cot A-\tan A=\cdots=2\cot2A$$

$$\implies2\tan2A+4\tan4A+8\cot8A$$

$$=2\cot2A-2(\cot2A-\tan2A)+4\tan4A+8\cot8A$$

$$=2\cot2A-2(2\cot4A)+4\tan4A+8\cot8A$$

$$=2\cot2A-8(\cot8A-\cot8A)$$

$$=2\cot2A$$

$$\implies\cot2A=0\iff\cos2A=0$$

$\implies2A$ must be odd multiple of $\dfrac\pi2$