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Given that sin A=3/5 and cos B=5/13, find the value of cos ( A− B )?
Please show steps.
That's the question as is and the answer can be worked out.
The answer should be -27/65 = - 0.415 (3dp). I just need to know how to get to that answer please.
5 Answers
- la consoleLv 71 decade agoFavourite 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
...Show more - RogueLv 71 decade ago
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
...Show more - ComoLv 71 decade ago
cos A cos B + sin A sin B
(4/5) (5/13) + (3/5)(12/13)
20/65 + 63/65
83/65
...Show more - ?Lv 71 decade ago
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 !!
...Show more