Question

sinA + sinB = -21/65 and cosA+cosB =-27/65 then value of cos[(A-B)] is?

Answer 1

cos A cos B + sin A sin B 

(4/5) (5/13) + (3/5)(12/13) 

20/65 + 63/65 

83/65

Answer 2

5 answers · Mathematics 

 Best Answer

Do you know this identity: cos(a - b) = cos(a).cos(b) + sin(a).sin(b) 

You need to calculate cos(a) and sin(b) 


To calculate cos(a) 

cos²(a) + sin²(a) = 1 

cos²(a) = 1 - sin²(a) → given sin(a) = 3/5 

cos²(a) = 1 - (3/5)² 

cos²(a) = (25/25) - (9/25) 

cos²(a) = 16/25 

cos²(a) = (± 4/5)² 

→ cos(a) = ± 4/5 


To calculate cos(a) 

cos²(b) + sin²(b) = 1 

sin²(b) = 1 - cos²(b) → given cos(b) 

sin²(b) = 1 - (5/13)² 

sin²(b) = (169/169) - (25/169) 

sin²(b) = 144/169 

sin²(b) = (± 12/13)² 

→ sin(b) = ± 12/13 



cos(a - b) = cos(a).cos(b) + sin(a).sin(b) 



First case: 

given sin(a) = 3/5 → cos(a) = 4/5 

given cos(b) = 5/13 → sin(b) = 12/13 

= cos(a).cos(b) + sin(a).sin(b) 

= (4/5).(5/13) + (3/5).(12/13) 

= (20/65) + (36/65) 

= 56/65 



Second case: 

given sin(a) = 3/5 → cos(a) = 4/5 

given cos(b) = 5/13 → sin(b) = - 12/13 

= cos(a).cos(b) + sin(a).sin(b) 

= (4/5).(5/13) + (3/5).(- 12/13) 

= (20/65) - (36/65) 

= - 16/65 



Third case: 

given sin(a) = 3/5 → cos(a) = - 4/5 

given cos(b) = 5/13 → sin(b) = 12/13 

= cos(a).cos(b) + sin(a).sin(b) 

= (- 4/5).(5/13) + (3/5).(12/13) 

= - (20/65) + (36/65) 

= 16/65 



Fourth case: 

given sin(a) = 3/5 → cos(a) = - 4/5 

given cos(b) = 5/13 → sin(b) = - 12/13 

= cos(a).cos(b) + sin(a).sin(b) 

= (- 4/5).(5/13) + (3/5).(- 12/13) 

= - (20/65) - (36/65) 

= - 56/65 




Conclusion: 

cos(a - b) = ± 16/35 

or 

cos(a - b) = ± 56/65