# a challeging problem

• Mar 17th 2013, 03:16 PM
engineeroneday
a challeging problem
the instructions state: find the exact value of the expression given that sec x = 3/2, csc y = 3, and x and y are in the quadrant I. sin (x+y) I have no clues as to how to solve this problem from my class notes. Please help if you can. thanks alot
• Mar 17th 2013, 03:27 PM
emakarov
Re: a challeging problem
sin(x + y) can be expressed through sine and cosine of x and y. The problem statement gives you the values of cos(x) and sin(y), from where you can find sin(x) and cos(y) taking into account that they are positive.
• Mar 17th 2013, 03:37 PM
engineeroneday
Re: a challeging problem
thanks for your help. question. is it a valid statement to say that if the sec of x is 3/2 then the cos of x is 2/3? and if the csc of y is 3 then the sin of y is 1/3?
• Mar 17th 2013, 03:41 PM
emakarov
Re: a challeging problem
Quote:

Originally Posted by engineeroneday
is it a valid statement to say that if the sec of x is 3/2 then the cos of x is 2/3? and if the csc of y is 3 then the sin of y is 1/3?

Yes because by definition sec(x) = 1/cos(x) and csc(x) = 1/sin(x).
• Mar 17th 2013, 04:11 PM
engineeroneday
Re: a challeging problem
thanks again. I believe I can solve the problem now.