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