Question

Find the approximate value of sin (30° 30′). Give that 1° = 0.0175^c and cos 30° = 0.866

Answer 1

Let f(x) = sin x  ...(I)

Differentiate w. r. t. x.

f'(x) = cos x

Now, 30° 30' = 30° + 30' = `30^circ + (1/2)^circ`

= `π/6 + (0.0175)/2`

30° 30' = `π/6 + 0.00875`  ...(II)

Let a = `π/6`, h = 0.00875

For x = a = `π/6`, from (I) we get

f(a) = `f(π/6) = sin(π/6) = 1/2` = 0.5  ...(III)

For x = a = `π/6`, from (II) we get

f'(a) = `f^'(π/6) = cos(π/6)` = 0.866  ...(IV)

We have, f(a + h) = f(a) + hf'(a)

`f(π/6 + 0.00875) = f(π/6) + (0.00875).f^'(π/6)`

f(30° 30′) = 0.5 + (0.00875) × (0.866)  ...[From (III) and (IV)]

= 0.5 + 0.0075775

∴ f(30° 30′) = sin (30° 30′) = 0.5076