Question

Find the approximate values of sin (29° 30'), given that 1° = 0.0175°, `sqrt(3) = 1.732`.

Answer 1

Let f'(x) = sin x

Then f'(x) = `d/dx (sin x) = cos x`

Take a = 30° = `pi/(6)` and

h = – 30'

= `-(1/2)^\circ`

= `-(1/2) xx 0.0175`

= –0.00875

Then f(a) = `f(pi/6)`

= `sin  pi/(6)`

= `(1)/(2)`

= 0.5

and

f'(a) = `f'(pi/6)`

= `cos  pi/(6)`

= `sqrt(3)/(2)`

= `(1.732)/(2)`

= 0.866

The formula for approximation is

f(a + h) ≑ f(a) + h.f''(a)

∴ sin (29° 30')

= f(29°30')

= `f(pi/6 + 0.00875)`

≑ `f(pi/6) - (0.00875).f'(pi/6)`

≑ 0.5 – 0.00875 × 0.866

≑ 0.5 – 0.0075775

= 0.4924

∴ π sin (29°30') ≑ 0.4924.