Question

If sin4B + sin2B =1 then prove that tan4B - tan2B =1.

Answer 1

Answer on Question #79558 – Math – Trigonometry

Question

If $\sin 4B + \sin 2B = 1$ then prove that $\tan 4B - \tan 2B = 1$.

Solution

The question is incorrect. Let’s prove it. Assume that there exists $B$ such that $\sin 4B + \sin 2B = 1$ and $\tan 4B - \tan 2B = 1$.

Let $\sin 2B = x$ and $\cos 2B = y$. Then we have: $\sin 4B = 2\sin 2B\cos 2B = 2xy$;

$\tan 2B = x / y; \tan 4B = \frac{2 \tan 2B}{1 - \tan^2 2B} = \frac{2xy}{y^2 - x^2}.$

Then, we have: $2xy + x = 1$ (**) and $\frac{2xy}{y^2 - x^2} - \frac{x}{y} = 1$, with $x^2 + y^2 = 1$.

The second condition is equivalent to $2xy^2 = (x + y)(y^2 - x^2) = (x + y)^2(y - x)$. (*)

Note that: $(x + y)^2 = x^2 + 2xy + y^2 = 1 + 2xy = 1 + (1 - x) = 2 - x$.

Also, $2xy^2 = (1 - x)y = y - xy$.

Then, from the $(*)$, we get: $y - xy = (2 - x)(y - x) = 2y - xy - 2x + x^2$,

Which is equivalent to: $y = 2x - x^2$.

On the other hand, (from **) we get that $y = \frac{1 - x}{2x}$.

Thus, $2x - x^2 = \frac{1 - x}{2x}$, so $1 - x = 4x^2 - 2x^3$, and then $2x^3 - 4x^2 - x + 1 = 0$. (°)

Since $x^2 + y^2 = 1$, we get: $x^2 + \left(\frac{1 - x}{2x}\right)^2 = 1$, which means that $4x^4 + 1 - 2x + x^2 = 4x^2$, so $4x^4 - 3x^2 - 2x + 1 = 0$,

which is equivalent to $(x - 1)(4x^3 + 4x^2 + x - 1) = 0$. Since $x = 1$ cannot be the root (because otherwise $y = 0$ and $\tan 2B$ does not exist), we have $4x^3 + 4x^2 + x - 1 = 0$.

Sum it up with $(°): 6x^3 = 0$, which means that $x = 0$, $y = \pm 1$. But I this case, the equation (**) is wrong.

Thus, the task is incorrect.

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