SOLUTION: cos(A) = 9/41, with A in QI and sin(B) = -4/5 in QIII Compute the following: sin (A+B) cos (A-B) sin 2 A cos (B/2)

Algebra ->  Trigonometry-basics -> SOLUTION: cos(A) = 9/41, with A in QI and sin(B) = -4/5 in QIII Compute the following: sin (A+B) cos (A-B) sin 2 A cos (B/2)      Log On


   

Question 1030532: cos(A) = 9/41, with A in QI and sin(B) = -4/5 in QIII
Compute the following:
sin (A+B)
cos (A-B)
sin 2 A
cos (B/2)
Answer by ikleyn(52833) About Me  (Show Source):
You can put this solution on YOUR website!
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cos(A) = 9/41, with A in QI and highlight%28cross%28sin+B=4%2F5%29%29 sin(B) = -4/5 in QIII
Compute the following:
a) sin (A+B)
b) cos (A-B)
c) sin (2A)
d) cos(B/2)
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Notice
   Hello, in your original post the data was written as "sin(B) = 4/5 in QIII" which is IMPOSSIBLE and doesn't make sense.

   I fixed it in a way "sin(B) = -4/5 in QIII."

   What is written below relates to the fixed version.
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We are going to use these formulas 
   
   sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)  and  cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B),

therefore, we need to know (to find)  sin(A)  and  cos(B),  in addition to the given  cos(A)  and  sin(B).


OK. We have  sin(A) = sqrt%281-cos%5E2%28A%29%29 = sqrt%281+-+%289%2F41%29%5E2%29 = sqrt%281+-+81%2F1681%29 = sqrt%28%281681-81%29%2F1681%29 = sqrt%281600%2F1681%29 = 40%2F41. 

The sign of the square root was chosen "+" for sin(A) since angle A is in Q1.

Next, we have  cos(B) = -sqrt%281-sin%5E2%28B%29%29 = -sqrt%281+-+%28-4%2F5%29%5E2%29 = -sqrt%281+-+16%2F25%29 = -sqrt%28%2825-16%29%2F25%29 = -sqrt%289%2F16%29 = -3%2F5. 

The sign of the square root was chosen "-" for cos(B) since angle B is in Q3.


Now, 

a)  sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B) = %2840%2F41%29%2A%28-3%2F5%29+%2B+%289%2F41%29%2A%28-4%2F5%29 = -120%2F205+%2B+%28-36%2F205%29 = %28-120-36%29%2F205 = -156%2F205.

b)  cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B) = %289%2F41%29%2A%28-3%2F5%29+%2B+%2840%2F41%29%2A%28-4%2F5%29 = -27%2F205+%2B+%28-164%2F205%29 = %28-27+-164%29%2F205 = -191%2F205.

c)  sin(2A) = 2*sin(A)*cos(A) = 2%2A%2840%2F41%29%2A%289%2F41%29 = 720%2F41%5E2.

d)  cos(B/2) = -sqrt%28%281%2Bcos%28B%29%29%2F2%29 = -sqrt%28%281+%2B+%28-3%2F5%29%5E2%29%2F2%29 = -sqrt%28%281+%2B+%289%2F25%29%29%2F2%29 = -sqrt%28%2825%2B9%29%2F%282%2A25%29%29 = -sqrt%2834%2F50%29 = -sqrt%2817%2F25%29 = -sqrt%2817%29%2F5.

    The sign of the square root was chosen "-" for cos(B/2) since angle B/2 lies in Q2.