Can't you just substitute and as given?
This is as exact as you're going to get...
I am told that and , where x is in the second quandrant and y is in the third quandrant.
I need to find the exact value of:
I know I need to use the difference formula some how and I realise that sin is negative in the 3rd and 4th quadrant; however, I didn't know how to apply the negative sign for . Do I apply it within the sin function or does it make the entire sin function negative? The confusion was because there are two values in different quadrants.
Also, how would I go about converting x and y into values that I can derive exact values from? Do I need to rearrange them before I apply the sum/difference formula or after?
How did knowing the quadrants play any role at all then? I just figured that the statement was made for a reason. Also, I just assumed exact meant using the values for 30, 45, 60 as it also said without using a calculator.
PS. should be right?
I suspect you have written the problem wrong- that the fractions are NOT x and y but are either sine or cosine of x and y. Otherwise, it makes no sense to say that a single number is in any specific quadrant.
It should be:
Durrrrr, my bad.
So would I draw two triangles using cosx and tany and then use the difference formula and substitute in the values from the traingles?
And would I apply the necessary sign to cos(x) and tan(y) now? ie -17/25 becomes -- = +17/25 and 15/8 stays +?
Working on it now will see what I get...
Because and is in the 2nd quadrant, where sin(x) is positive, that would mean the 7 is negative and the 25 is positive right? So when substituting values into the sum formula equation I should be using a -7 shouldn't I?
Sorry if I am talking nonsense; can't seem to get it out of my head.
So for I would have:
Just out of curiosity, in where you have made the two edges (15 & 8) negative and the hypotenuse (17) positive which makes tan positive and both sin & cos negative, which is correct for . Does it matter that you do it this way? for example, if you were to make the hypotenuse negative and the two edges positive, you would still have tan +, sin -, cos -.