Theta and alpha are each acute angles in standard position, sin theta = 3/5 and cos alpha = 12/13

a) Determine cos (alpha+theta)
b) Determine sin (theta + alpha)

Here's a list of all the basic trig identities you should know to do further trig identities:

cosx=\frac{1}{secx} and secx=\frac{1}{cosx}

cotx=\frac{1}{tanx} and tanx=\frac{1}{cotx}

cosecx=\frac{1}{sinx} and sinx=\frac{1}{cosecx}

Then, the square identities;

cos^2x + sin^2x=1

1 + cot^2x=cosec^2x

tan^2x=1+sec^2x

Addition and subtraction of angles;

cos(A+B)= cosAcosB - sinAsinB and
cos(A-B)= cosAcosB + sinAsinB

sin(A+B)= sinAcosB + cosAsinB and
sin(A-B)= sinAcosB - cosAsinB

tan(A+B)=\frac{tanA+tanB}{1-tanAtanB} and tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}

Double Angles;

sin2A=2sinAcosA

cos2A = cos^2A - sin^2A

OOOOO=1 - 2sin^2A

OOOOO=2cos^2A-1

tan2A = \frac{2tanA}{1-tan^2A}

Now use them to solve your questions. :)

For cos (alpha + theta)
would the answer be cos (12/13 + 4/5)?

No. Start by converting cos(\alpha+\theta) to cos{\alpha}cos{\theta} - sin{\alpha}sin{\theta}

Now have a drawing, you know some of the Pythagorean triples I hope:

Then, substitute your values of sin theta, sin alpha, cos theta and cos alpha. Do the same procedure for the second one. :)

EDIT: The second triangle has a mistake.

OK this is what I got
sin alpha cos theta- sinalpha sintheta
= (12/13) (4/5) - (11/13) (3/5)
= 3/13

and for the second one

sin theta cos alpha - costheta sin alpha
= (3/5)(12/13) - (4 /5 ) (11/13)
= -11/65

Sorry, yesterday, it was very late and I was very tired. The second triangle has sides 12, 13, 5 instead of 12, 13, 11.

Ok you did the question well, except that you haven't notice my mistake over there. You understood the concept though, good! :) now, just re-work up the actual answers.

sin\alpha = \frac{5}{13}