Question

If A and B are acute angles, sinA = 3/5, cosB = 12/13,
find cos (A+B) 
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Answer 1

SOLUTION:-

cos (A + B)  =  cosAcosB - sinAsinB 

From the formula of cos (A + B), the value of sinA and cosA are given in the question.

We need the value of sinB and cosA.

We know that,  

=>sin ² B  =  1 - cos ² B

=>sin ² B  =  1 - (12/13)²

=>sin ² B  =  1 - 144/169

=>sin ² B  =  (169 - 144)/169

=>sin ² B  =  25/169

=>sin ² B  =  (5/13)² --------> sinB  =  5/13

=>cos ² A  =  1 - sin ² A

=>cos ² A  =  1 - (3/5)²

=>cos ² A  =  1 - 9/25

=>cos ² A  =  (25 - 9)/25

=>cos ² A  =  16 / 25

=>cos ² A  =  (4/5)² --------> cosA  =  4 / 5 

Putting the values of cosA, cosB, sinA and sinB in the formula of cos (A+B), we get, 

cos (A + B)  =  (4/5) x (12/13)  -  (3/5) x (5/13)

cos (A + B)  =  (4/5) x (12/13)  -  (3/5) x (5/13)

cos (A + B)  =  (48 / 65)  -  (15 / 65)

cos (A + B)  =  (48 - 15) / 65

cos (A + B)  =  33 / 65

Hence, the value of cos(A+B) is 33/65.