Question

my question is attached.

Answer 1

where's the attachment?

Answer 2

Only question A please.

Answer 3

sin(pi/2) = 1

f(pi/2) = 1

g(1) = 1+pi/3

Answer 4

a-g(f(bi/2))\[\LARGE g(f(\frac{\Pi}{2}))\]:
First u will substitute in the equation \[F(\theta)\] instead of \[(\theta)\] put \[\frac{\Pi}{2}\]
and then substitute the result in the equation of \[g( \theta)\]

Answer 5

\(g(f(\pi/2))\). To calculate that, first we need the value of \(f(\pi/2) = \sin{\pi/2} = 1\)
So, \(g(f(\pi/2)) = g(1) = 1 + \pi/3\), replace \(f(\pi/2)\) with 1

Answer 6

by text gives the answer π+3/3

Answer 7

combine the 2 fractions into 1 fraction with denominator of 3

1 + pi/3 = 3/3 + pi/3 = (3+pi)/3