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Question
Find the approximate values of : tan (45° 40'), given that 1° = 0.0175°.
Solution
Let f(x) = tan x
Then f'(x) = `d/dx(tanx) = sec^2x`
Take a = 45°
= `pi/(4)`
and
h = 40'
= `(40/60 xx 0.0175)^c`
= 0.01167c
Then f(a) = `f(pi/4)`
= `tan pi/(4)`
= 1
and
f'(a) = `f'(pi/4)`
= `sec^2 pi/(4)`
= `(sqrt(2))^2`
= 2
The formula for approximation is
f(a + h) ≑ f(a) + h.f'(a)
∴ tan(45° 40')
= f(45° 40')
= `f(pi/4 + 0.01167)`
≑ `f(pi/4) + (0.01167).f'(pi/4)`
≑ 1 + 0.01167 x 2
= 1 + 0.02334
= 1.02334
∴ tan (45° 40') ≑ 1.02334.
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