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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 $\theta=1.1$.

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    $\begingroup$ !!! Can you just one second verify your trigonometru identities? $\endgroup$ – Martigan Nov 7 '14 at 14:43
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    $\begingroup$ I'm guessing it's a bug. $\endgroup$ – Akiva Weinberger Nov 7 '14 at 14:51
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$\cos \theta=\sqrt{1-\sin^2 \theta} \iff \cos \theta \in [0,1] \iff \theta \in [-\frac{\pi}{2},\frac{\pi}{2}]+2n\pi$ for some integer $n$.

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  • $\begingroup$ 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? $\endgroup$ – Jack Nov 7 '14 at 14:50
  • $\begingroup$ @Jack It answers the core or your question. $\endgroup$ – Jean-Claude Arbaut Nov 7 '14 at 15:17
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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.

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  • $\begingroup$ Thank you. You answered my question. I thought WA was concluding 1.1 was the solution. As you mention, it is one of many. Appreciate it everyone! $\endgroup$ – Jack Nov 7 '14 at 14:53
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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).

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