Solution Set 1: Math
Solutions: first set, mathematics
No. 1
It simplifies to $a^2 = b^2$. Assume non-zero values for both $a$ and $b$.
$a^4 = b^4$ can be further simplified. Partial points if you left it like this.
Partial points if you left it as $a=b$. Not the right answer but still something.
No. 2
It doesn’t always mean that $a=b$. We can have, as a counter-example $a=1$ and $b=-1$. They meet the requirements but don’t equal.
No. 3
Yes, if $a^4=b^4$, we always have $a^2=b^2$ for real numbers $a,b$. This because the powers are both even so any negatives will go away in the simplification too.
If $a^6=b^6$, it doesn’t mean that $a=b$, always. Try $a=2$ and $b=-2$ as counter examples.
No. 4
$A$ and $a$ are directly related. This means that reducing one also reduces the other. Since they correspond to the angle and opposing side of any triangle, the set of values, ignoring the other sides and angles, is $a,A \in (0,\infty)$. You can say they may approach $0$ as an answer.
No. 5
Start with
\[\sin^2(a) + \cos^2(a) = 1 \\ \cos(a) = \sin(\pi/2 - a)\]For radius $1$. Now, since $1$ is the radius, $1 \in (0,\infty)$ which makes
\[R^2 = 1 \in (0,\infty)\]All radii are positive, so $r = 1$. That’s it!
Hope it went well!