Question

Find the value of cos (sin⁻¹$\frac{3}{5}$ + sin⁻¹$\frac{5}{13}$).

Answer 1

Solution :

i ) Let cos^-1 ( 3/5 ) = A

=> cosA = 3/5 ---( 1 )

ii ) sin² A

= 1 - cos²A

= 1 - ( 3/5 )²

= 1 - 9/25

= ( 25 - 9 )/25

= 16/25

sin A = 4/5 ----( 2 )

iii ) cos^-1 ( 12/13 ) = B

=> cos B = 12/13 ---- ( 3 )

iv ) sin² B

= 1 - cos² B

= 1 - ( 12/13 )²

= 1 - 144/169

= ( 169 - 144 )/169

= 25/169

sin B = 5/13 --- ( 4 )

Now ,

v ) cos { cos^-1(3/5) + sin^-1(5/13) }

= cos( A + B )

= cosAcosB - sinAsinB

= (4/5)(12/13) - (3/5)(5/13)

[ from(1),(2),(3) and (4) ]

= 48/65 - 15/65

= ( 48 - 15 )/65

= 33/65

••••

Answer 2

Answer:

33/65

Step-by-step explanation:

Formula used:

cos(A+B) = cosA cosB - sinA sinB

cos (sin⁻¹(3/5))+ sin⁻¹(5/13)).

sin⁻¹(3/5) = A implies sinA = 3/5

sin⁻¹(5/13) =B implies sinB = 5/13

sinA = 3/5, cos²A = 1 - sin²A= 16/25

cosA = 4/5

sinB = 5/13, cos²B = 1 - sin²B= 144/169

cosB = 12/13

Now,

cos (sin⁻¹(3/5))+ sin⁻¹(5/13))

= cos(A+B)

= cosA cosB - sinA sinB

= (4/5)(12/13) - (3/5)(5/13)

= (48/65) - (15/65)

= 33/65

I hope this answer helps you