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Can someone please explain where the $1$ goes in this expression?

Find the measure of angle in radians by solving: $$\sin^2 \theta +\cos \theta=1$$

I got $$\sin^2 \theta = 1- \cos \theta$$

Substituting, I get:

$$(1-\cos \theta) +\cos \theta -1= 0$$

I know the next step is $$\cos (1-\cos) = 0$$

But where does the $1$ go? Does it cancel out and if so why?

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    Please edit your question [using MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-02-07
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    $1=\sin^2\theta+\cos^2\theta$ !!! – 2017-02-07

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Given $$\sin^2 \theta + \cos \theta = 1,$$ and using the circular identity $$\sin^2 \theta + \cos^2 \theta = 1,$$ it follows that $$\cos^2 \theta = \cos \theta,$$ or $$(\cos \theta - 1) \cos \theta = 0.$$ Hence $$\cos \theta \in \{0, 1\},$$ which implies $$\theta \in \{\tfrac{(2k+1)}{2}\pi, 2k\pi\}, \quad k \in \mathbb Z.$$

You can check this: if $\theta$ is an integer multiple of $2\pi$, then $\sin \theta = 0$ and $\cos \theta = 1$, so $\sin^2 \theta + \cos \theta = 1$. If $\theta$ is an odd multiple of $\pi/2$, then $\sin \theta = \pm 1$, so $\sin^2 \theta = 1$, and $\cos \theta = 0$, which also checks out.

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    Nothing wrong, but I would have said $\sin^ t+\cos t=1\iff (-1+\cos t)\cos t \iff ...$ (Et Cetera) to show that you don't need to check it (Except to catch possible errors) because of all the "iff"s..................+1 – 2017-02-07
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There's quite a bit going on here. First, this is not an "identity", it's a conditional equation and you're asked to solve it. Second, replace $\sin^2 \theta$ by $1-\cos^2 \theta$ (you forgot to square the $\cos \theta.$) Third, the last time you write $\cos$, you leave it dis-embodied. You need to take the $\cos$ of something. (In this case, $\theta.$)

Then you should have something like:

$$1-\cos^2 \theta = 1-\cos \theta.$$

The left side factors by difference of squares. Then either $1-\cos \theta =0$ (and you solve that equation.) Or you can divide by $1-\cos \theta$ and you can solve the resulting equation.

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I do not get your question, however I'll be elaborate in this explanation so that you can clear your doubt. \begin{split} \sin^2 \theta + \cos \theta = 1 & \implies \cos \theta = 1 - \sin^2 \theta \\ & \implies \cos \theta = \cos^2 \theta \qquad(\text{since $\sin^2 \theta + \cos^2 \theta = 1$}) \\ & \implies 0 = \cos^2 \theta - \cos \theta \\ & \implies \cos \theta(1-\cos \theta) = 0 \\ & \implies \cos \theta = 0 \text{ or }\cos \theta = 1 \\ & \implies \theta = 0^\circ \text{ or } 90^\circ \end{split}

You can convert to radians, this will give $0$ and $\frac \pi 2$ radians respectively.

The $1$ does not vanish : it is instead replaced with $\sin^2 \theta + \cos^2 \theta$ to produce calcellations.