#arcsin(4/5)# is some #alpha# between #-pi/2# and #pi/2# with #sin alpha = 4/5#

#arctan(12/5)# is some #beta# between #-pi/2# and #pi/2# with #tan beta = 12/5#

We are asked to find #sin(alpha + beta)#. (Do you have some suspicions about why I rewrote the problem this way?)

We know that

#sin(alpha + beta) = sin alpha cos beta+cos alpha sinbeta# .

We know #sin alpha = 4/5# which is positive and with the restriction we already mentioned, we conclude that #0 <= alpha <= pi/2#. We want #cos alpha#.

At this point in our study of trigonometry we have already developed at least one method to find #cos alpha# (We may have three or four or more methods, but we only need one right now.)

Use your chosen method to get #cos alpha = 3/5#

So we have:

#sin alpha = 4/5" "# and #" "cos alpha = 3/5#

Similarly, given #tan beta = 12/5# and #0 <= beta <= pi/2#, find #sin beta# and #cos beta#

#sin beta = 12/13" "# and #" "cos beta = 5/13#

#sin(alpha + beta) = sin alpha cos beta+cos alpha sinbeta#

# = 4/5 5/13 + 3/5 12/13#

# = 56/65#.

If you want to write this without the names #alpha# and #beta# 9Or something like that, you could write:

#sin(arcsin(4/5)+arctan(12/5))#

# = sin(arcsin(4/5))cos(arctan(12/5))+ cos(arcsin(4/5))sin(arctan(12/5))#

But I, personally do not think that is more clear. (I do think it is good for students to see it this way, however.)