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November 23, 2014

November 23, 2014

Posted by **Chelsea** on Monday, November 15, 2010 at 10:15pm.

Tan 44 degrees

Please help.

- Calc. -
**MathMate**, Monday, November 15, 2010 at 10:42pmf(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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