This ain't an identity, right?

Geometry Level 3

$1+\sin^2 ax =\cos x$

If the above equation has infinitely many solutions, which of the following is the most general statment that is still true about $a$ ?

An irrational number A rational number Equal to

\sqrt 3

Equal to 1

1 solution

Akhil Bansal
Apr 13, 2016

$\large 1 + \sin^2 ax = \cos x$

The above equation is true only if $\sin^2 ax = 0$ and $\cos x = 1$ .
$\large \cos x = 1 \Rightarrow x = 2n\pi \pm \frac{\pi}{2} \ , \ (n \in Z) \ ..(1)$ $\large \sin^2 ax = 0 \Rightarrow ax = n\pi \ , \ (n \in Z) \ .. (2)$
Substituting value of $x$ from eqn(1) in eqn(2) $\large a\left( 2n\pi \pm \dfrac{\pi}{2}\right) = n\pi$ $\large a = \dfrac{n}{2n \pm \frac{1}{2}} \ , \ (n \in Z)$ Hence, a is a rational number .

Moderator note:

Several issues

You should explain why "The above equation is true only if $\sin^2 ax = 0$ and $\cos x = 1$ . "
The solution for $\cos x = 1$ is incorrect.
Avoid letting your notation do double duty, since you are likely to forget when that happens. There are many more rational numbers than just $\frac{ 2n } { 4n \pm 1 }$ which satisfy the condition.

Nice approach, but I have a couple of suggestions. First, I believe that

$\cos(x) = 1 \Longrightarrow x = 2n\pi, n \in \mathbb{Z}$ .

Next, you should probably write equation $(2)$ as $ax = m\pi, m \in \mathbb{Z}$ , since the integral multiple in this equation need not necessarily be the same as the integral multiple $n$ in equation $(1)$ . As a result we have that

$a*2n\pi = m\pi \Longrightarrow a = \dfrac{m}{2n}, m,n \in \mathbb{Z}, n \ne 0$ , i.e., $a$ must be a rational number.

Brian Charlesworth - 5 years, 2 months ago

I agree with you!

Nihar Mahajan - 5 years, 2 months ago

Thanks! :)

Akhil Bansal - 5 years, 2 months ago

This ain't an identity, right?

1 solution

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