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March 11th, 2010, 10:59 PM  #1 
Newbie Joined: Mar 2010 Posts: 1 Thanks: 0  Brain fried from trig angles question
Hi everyone, I am currently stuck on this extension question from my textbook: If sin(x) = 1/4 where sin(x) is 0 < x < pi/2 and sec(y) = 6/5 with 0 < y < pi/2, find the exact value of sin(x + y). What I have done is: Use Trig identity: sin(x+y) = sin(x)cos(y) + sin(y)cos(x) sec(y) = 6/5 => 1/cos(y) = 6/5 => cos(y) = 5/6 Therefore... sin(x+y) = (1/4 * 5/6) + cos(x)sin(y) And I kind of get stuck since I'm not sure how to find cos(x) or sin(y). I'm thinking of using the identities, sin^2(x) + cos^2(x) = 1 and sin^2(y) + cos^2(y) = 1, to find what cos(x) and sin(y) are respectively but I am not confident that I am going in the right direction. Thank you in advance. 
March 12th, 2010, 06:04 AM  #2 
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Brain fried from trig angles question
If I had a right triangle and knew that the sine of an angle in it was 1/4, would that not give me an idea of the possible [proportionate] lengths of two sides? Would that not then enable me to figure the 3rd side using Pythagoras, then all the other ratios?

March 13th, 2010, 08:22 AM  #3 
Member Joined: Dec 2009 Posts: 66 Thanks: 0  Re: Brain fried from trig angles question
Like David stated, draw a right triangle out and fill in the known proportions. From there it will be a lot more easier to understand your dilemma

March 22nd, 2010, 08:15 AM  #4 
Newbie Joined: Mar 2010 Posts: 10 Thanks: 0  Re: Brain fried from trig angles question
I think you're heading in the right direction. cos(x) = +squareroot( 1  sin^2(x) ) = +squareroot( 15/16 ) sin(y) = +squareroot( 1  cos^2(y) ) = +squareroot( 11/36 ) 

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