Posted by Chelsea on Monday, November 15, 2010 at 10:15pm.
f(x)=tan(x)
f'(x)=sec²(x)
Using linear approximation:
f(xo+δ)=f(xo)+δf'(xo) (approx.)
Put xo=π/4 (in radians, = 45°)
δ=π/180 (1°)
f(44°)
=f(xo-δ)
=tan(π/4)-δ*sec²(π/4) (approx.)
=1-π/180/cos²(π/4)
=1-π/180/(1/2)
=1-π/90
=0.96509...
accurate value of tan(44°)=0.96569...
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