Question

sin^-1(3/5)-cos^-1(63/65)​

Answer 1

Answer:

let sin^-1(3/5) = A

sinA = 3/5

sin^2A = 9/25

1 - cos^2A = 9/25

cos^2A = 1 - 9/25

cos^2A = 16/25

cosA = 4/5

let cos^-1(63/65) = B

cosB = 63/65

cos^2B = 3969/4225

1 - sin^2B = 3969/4225

sin^2B = 1 - 3969/4225 = 256/4225

sinB = 16/65

now,

cos(A - B) = cosAcosB + sinAsinB

= (4/5)(63/65) + (3/5)(16/65)

= 252/325 + 48/325

= 300/325

= 12/13

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

sin(A-B) = sinAcosB - cosAsinB

= (3/5)(63/65) - (4/5)(16/65)

= 189/325 - 64/325

= 125/325

= 5/13

A - B = sin^-1(5/13)