Finding sin(A+B) given cos A=1/3 and sin B=1/4

[majinknight]

Ok So i finished my questions and was wondering if someone could check my work to see if i did it right.
Question: If CosA= 1/3, with 0<A<pie/2, and SinB=1/4, with pie/2<B<pie, Calculate sin(A+B)

My Solution:

Sin(A+B)= SinACosB + CosASinB
=SinACosB + (1/3)(1/4)
SinACosB + 1/12

Ok Now i drew Triangle A and labeled the bottom 1 and hypotaneous 3. Then used the theorum to get the last side is square root 8.
I made Triangle B and labelled hypotaneous 4, adjacent side being 1 and used theorum to get the opposite side to be square root of 15.

=(square root8/3)(square root15/4) +1/12
=(square root120/12) + 1/12
to get my final answer of squareroot 120 +1 /12.

Is that correct?

 

[HallsofIvy]

Ouch! Your trigonometry is excellent- your notation is horrible!

You are exactly right (and for the right reasons!) that
sin(A+B)= =(&radic;(8)/3)(&radic;(15)/4) +1/12

but then you seem to have lost control of your parentheses.

sin(A+ B)= &radic(120)/12+ 1/12
= (&radic(120)+ 1)/12

which is not the same as &radic(120)+ 1/12 which is how I would have interpreted "squareroot 120 +1 /12".

It's not your use of "squareroot" instead of &radic; I am complaining about- it's your use of parentheses.

(squareroot(120)+ 1)/12 would be a correct answer.

 

[himanshu121]

shouldn't cosB be negative for $$\frac{\pi}{2}<B< \pi$$