How can I find the value of sin(sin^-1(4/5)+tan^-1(5/12)) without using calculator?.Use angle sum or different identities.

Question

Please help me in this..Tomorrow is my exam.

Answer 1

Angle addition formula

Let X= arcsin(4/5) y=arcan(5/12)

Then sin x=4/5 --> opposite=4 and hypotenuse=5;

Then adjacent=3 by pythagorean.

So then cos X=3/5

Llikewise, tan y= 5/12 --> opposite=5 and adjacent =12. So then hypotenuse=13 by pythagorean.

So then sin y = 5/13 and cos y=12/13

Sin(x+y)= sin x cos y + cos x sin y

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

= (48 + 15)/65

63/65

Answer 2

sin(A+B) = sinAcosB +sinBcosA

if A=sin^-1(4/5) then sinA = 4/5, cosA =3/5 (it's a 3-4-5 right triangle, 3^2 +4^2 = 5^2)

B= tan^-1(5/12) (this is a 5-12-13 right triangle, 5^2 + 12^2 = 13^2)

tanB= 5/12, sinB=5/13, cosB = 12/13

sin(A+B) = (4/5)(12/13) + (3/5)(5/13) = (48+15)/65= 63/65 = about 0.97

check the answer with a calculator with inverse trig functions

the angle whose sine is 4/5 = about 53.13

the angle whose tangent is 5/12 = 22.62

sum the angles = 75.75

sin(75.75) = about 0,97

Answer 3

sin(sin^-1(4/5)+tan^-1(5/12))

Let A = sin^-1(4/5) and B = tan^-1(5/12).

Then sin A = 4/5, cosA = √1 - 16/23 = 3/5;

tanB = 5/12; cosB = 1/√(1 + 25/144) = 12/13; sinB = 5/12·12/13 = 5/13

sin(sin^-1(4/5)+tan^-1(5/12)) = sin(A + B) = sinAcosB + cosAsinB = 4/5·12/13 + 3/5·5/13 =

48/65 + 15/65 = 63/65

Answer 4

To use the sin addition formula, we need to know the sin and cos of the angles which have 1) sin of 4/5, and 2) tan of 5/12

For θ₁, If the sin is 4/5, the cos is 3/5 (think 3-4-5 right triangle). For θ₂, If the tan is 5/12, the sin is 5/13 and the cos is 12/13 (think 5-12-13 right triangle).

Thus, sin(θ₁ + θ₂) = sinθ₁cosθ₂ + cosθ₁sinθ₂ = 4/5 * 12/13 + 3/5 * 5/13 = 63/65