# Is the numerical solution for $\cos(x)=\sqrt{1-\sin^2(x)}, x=1.1$?

I typed $\cos\theta=\sqrt{1-\sin^2\theta}$ into WolframAlpha and it gave me the numerical solution $\theta=1.1$. However, it did not provide a step-by-step solution like it normally does.

Is this correct? I know that $\cos\theta=\sqrt{1-\sin^2\theta}$ is not an identity since the left-hand side can be negative and the right-hand side is always positive, so I assume a numerical solution can possibly exist, but I am very interested in figuring out how one arrives at the solution $\theta=1.1$.

• !!! Can you just one second verify your trigonometru identities? – Martigan Nov 7 '14 at 14:43
• I'm guessing it's a bug. – Akiva Weinberger Nov 7 '14 at 14:51

$\cos \theta=\sqrt{1-\sin^2 \theta} \iff \cos \theta \in [0,1] \iff \theta \in [-\frac{\pi}{2},\frac{\pi}{2}]+2n\pi$ for some integer $n$.
• Thanks for the answers on the identity issue, but the core of my question concerns the numerical solution $\theta=1.1$. Is that a correct numerical solution? – Jack Nov 7 '14 at 14:50
This is an identity that is true as you pointed out, anywhere $\cos \theta$ is positive. This includes any number in $[-\pi/2,\pi/2]$, or in any interval of the form $[-\pi/2 + 2n\pi,\pi/2 + 2n\pi], n\in\mathbb{Z}$.
Thus, $\theta = 1.1$ is a solution, among infinitely many other solutions, as WA points out later on in the page. Why WA decides specifically on $1.1$ is a bit of a mystery, but I suspect this is due to some floating point quirk.
As you noted, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ is an identity, but only for when the LHS is positive. Therefore, that equation will be true for all $\theta\in[0,\pi/2]$ (with the second interval $[\pi/2,\pi]$ cut out, because none of those $\theta$ values will work).