Question

3. If SinA= - 3/5,sinB= 12/13, A lies in 3rd quadrant, B lies in 2nd quadrant, find cos(A+B) and sin(A-B)

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Answer 1

Answer:

Sin(A-B) = 63/65

Cos(A+B) = 56/65

Step-by-step explanation:

SinA = - 3/5

Now,

$cos ^{2} a \: + sin^{2} a \: = 1 \\ cos ^{2} a = 1 - sin ^{2} a \\ cos ^{2} a = 1 - 9 \div25 \\ cos \: a = \sqrt{25 - 9 \div 25} \\ cos \: a = \sqrt{16 \div 25 } \\ cos \: a = 4 \div 5$

But A is in 3rd quadrant. So, cosA = - 4/5

SinB = 12/13

Now,

$cos ^{2} b + sin ^{2} b = 1 \\ cos ^{2} b = 1 - sin ^{2} b \\ cos ^{2} b = 1 - 144 \div 169 \\ cos \: b = \sqrt{169 - 144 \div 169} \\ cos \: b = \sqrt{25 \div 169} \\ cos \: b \: = 5 \div 13$

But B is in 2nd quadrant, so cosB = - 5/13

1. sin(A-B) = sinA × cosB - cosA ×sinB

[(-3/5) × (-5/13)] - [(-4/5) ×(12/13)]

= 3/13 + 48/65

= 15+48/65 = 63/65

2. cos(A+B) = cosA×cosB - sinA× sinB

= [(-4/5)× (-5/13)] - [ (-3/5) × (12/13)]

= 4/13 + 36/65

= 20+ 36/65 = 56/65