During the preparation for my calculus course next semester I bumped into the question of proving $\cos(\sin^{-1}x)=\sqrt{1-x^2}$.
My prove:
Now my idea to give a valid proof was the following, but I'm not sure if it's correct this way.
\begin{alignat}{2}
\cos(\sin^{-1}x) & = \sqrt{1-x^2} &\qquad&\\
\cos^2(\sin^{-1}x) & = 1-x^2 &&\text{squaring both sides}\\
1-\sin^2(\sin^{-1}x)& =1-x^2 &&\text{because} \cos^2(\theta)= 1 – \sin^{2}(\theta) \\
\sin^2(\sin^{-1}x) & = x^2 && \text{cancel out the 1's and divide by} -1 \\
\sin(\sin^{-1}x) & = x && \text{taking the square root on both sides}
\end{alignat}
Now $\sin(\sin^{-1}x)=x$ holds because of the cancellation laws, but for that you will need to have the interval $-1 \leq x \leq 1$, as there is nothing said about this interval I'm wondering if the prove still holds the way I did it.
Best Answer
First, $\sin^{-1}(x)$ is defined for $-1 \le x \le 1$. Consequently, the proof implicitly presumes that $-1 \le x \le 1$.
Second, you did not write $\iff$ in each step.
Third, your proof did not specify why $\cos^2(\sin^{-1}(x))=1-x^2 \iff \cos(\sin^{-1}(x))=\color{red}{+}\sqrt{1-x^2}$. You need to justify that by mentioning that $-\pi/2 \le \sin^{-1}(x) \le \pi/2$ and $\cos(\theta) \ge 0$ for $\theta \in [-\pi/2,\pi/2]$.