Given that sin A=3/5 and cos B=5/13, find the value of cos ( A− B )?

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

given sin(A) = 3/5

given sin = opposite/hypotenuse and cos = adjacent/hypotenuse

using Pythagoras' Theorem a = ±√(5² − 3²) = ±4

=> cos(A) = ±4/5

given cos(B) = 5/13

=> o ±√(13² − 5²) = ±12

=> sin(B) = ±12/13

given cos(A − B) = cos(A)cos(B) + sin(A)sin(B)

=> cos(A − B) = (±4/5 *5/13) + (3/5*±12/13)

=> cos(A − B) = (±4/13) + (±36/65)

=> cos(A − B) = (±20/65) + (±36/65)

=> cos(A − B) = ±56/65, ±16/65

basically

A = sin⁻¹(3/5) if it also equals cos⁻¹(4/5) then A ≈ 36.8699° (4 dp)

A = sin⁻¹(3/5) if it also equals cos⁻¹(-4/5) then A ≈ 143.1301° (4 dp)

B = cos⁻¹(5/13) if it also equals sin⁻¹(12/13) then B ≈ 67.3801° (4 dp)

B = cos⁻¹(5/13) if it also equals sin⁻¹(-12/13) then B ≈ 292.6199° (4 dp)

as such cos(A − B) can be cos(36.8699 − 67.3801) = cos(-30.5102) = 56/65

or cos(A − B) can be cos(143.1301 − 67.3801) = cos(75.7500) = 16/65

or cos(A − B) can be cos(36.8699 − 292.6199) = cos(-255.7500) = -16/65

or cos(A − B) can be cos(143.1301 − 292.6199) = cos(-149.4898) = -56/65

Edit:

There is No possible way given the information proved for cos(A − B) to equal -27/65

sin(A − B) = sin(A)cos(B) − cos(A)sin(B) which = 15/65 ± 48/65 = -33/65 or 63/65

sin(A + B) = sin(A)cos(B) + cos(A)sin(B) which = 15/65 ± 48/65 = -33/65 or 63/65

cos(A + B) = cos(A)cos(B) − sin(A)sin(B) which again equal ±56/65, ±16/65

cos A cos B + sin A sin B

(4/5) (5/13) + (3/5)(12/13)

20/65 + 63/65

83/65

Find Sin

well,

as

cos(a-b) = cosa cosb + sina sinb

the information is not enough : there is still ambiguity on signs of

cosa and sinb ...

hope it' ll help !!