Having abit of problems with trig homework.
For acute angles P and Q, sin P = 12/13 and sin Q = 3/5 . Show that the exact value of sin (P + Q) is 63/65
So far I have done the following steps
sin(P+Q) = sin(p)cos(Q)+cos(P)Sin(Q)
then subsituted the values into the equation
sin(12/13+3/5)sin(12/13)cos(3/5)+cos(12/13)sin(3/5)
Doing the calculations doesn't give me the answers im looking for.
Since it's homework , if i've made mistakes can you please explain in detail so I have a better understanding ( I'd rather not just wing my homework then fail the exam :P)
The trig identity you've written down is correct, but you can't substitute anything on the left hand side , sin(P+Q). That's the value that you are trying to calculate.
What you have to do is calculate the value of the two cosines on the right hand side.
To find the value of cosP, sketch a right angled triangle, call one of the angles P and fill in the lengths of the opposite side and the hypotenuse, (from the given value of sinP). Then use the pythagoras theorem to calculate the length of the third side and so read off the value of cosP.
Repeat the routine to find cosQ.
Then substitute for each of the terms on the right hand side.
