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