SOLUTION: let sinA= 3/5 with A in Q2 and sinB= -5/13 with b in quadrant 3. find sin(A+B), cos(a+b), and tan(a+b). where does a and b terminate

Algebra ->  Trigonometry-basics -> SOLUTION: let sinA= 3/5 with A in Q2 and sinB= -5/13 with b in quadrant 3. find sin(A+B), cos(a+b), and tan(a+b). where does a and b terminate      Log On


   

Question 1056959: let sinA= 3/5 with A in Q2 and sinB= -5/13 with b in quadrant 3. find sin(A+B), cos(a+b), and tan(a+b). where does a and b terminate
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
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let sinA= 3/5 with A in Q2 and sinB= -5/13 with B in quadrant 3. find sin(A+B), cos(a+b), and tan(a+b). where does a and b terminate
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Use the formulas 

sin%28alpha%2Bbeta%29 = sin%28alpha%29%2Acos%28beta%29+%2B+cos%28alpha%29%2Asin%28beta%29    (1)

   and

cos%28alpha%2Bbeta%29 = cos%28alpha%29%2Acos%28beta%29+-+sin%28alpha%29%2Asin%28beta%29.   (2)


Regarding these formulas, see the lesson Addition and subtraction formulas in this site.


In addition to the given  sin(A) = 3%2F5  and  sin(B) = -5%2F13, you need to know  cos(A) and cos(B).


1.  cos(A) = sqrt%281-sin%5E2%28A%29%29 = -sqrt%281+-+%283%2F5%29%5E2%29 = -sqrt%281+-+9%2F25%29 = -sqrt%28%2825-9%29%2F25%29 = -sqrt%2816%2F25%29 = -4%2F5.

   The sign "-" was chosen for the square root because cosine is negative in Q2.


2.  cos(B) = = -sqrt%281-sin%5E2%28B%29%29 = -sqrt%281+-+%28-5%2F13%29%5E2%29 = -sqrt%281+-+25%2F169%29 = -sqrt%28%28169-25%29%2F169%29 = -sqrt%28144%2F169%29 =-12%2F13 = -12%2F13.

   The sign "-" was chosen for the square root because cosine is negative in Q3.


Now all you need to do is to substitute everything into the formulas (1) and (2) and make the calculations.


sin%28A%2BB%29 = %283%2F5%29%2A%28-12%2F13%29+%2B+%28-4%2F5%29%2A%285%2F13%29 = -36%2F65+-+20%2F65 = %28-36-20%29%2F65 = -56%2F65,   and

cos%28A%2BB%29 = %28-4%2F5%29%2A%28-12%2F13%29+-+%283%2F5%29%2A%28-5%2F13%29 = 48%2F65+%2B+15%2F65 = %2848%2B15%29%2F65 = 63%2F65.

Finally, tan(A+B) = sin%28A%2BB%29%2Fcos%28A%2BB%29 = %28%28-56%2F65%29%29%2F%28%2863%2F65%29%29 = -56%2F63.

The last question "where does a and b terminate" is just answered in your condition.