Question

Find the value of tan 33° by Lagrange’s formula if

tan 30° = 0.5774, tan 32° = 0.6249,

tan 35° = 0.7002, tan 38° = 0.7813.

Answer 1

Here the intervals are unequal.

$$\begin{matrix} x_0=30 & y_0=0.5774 \\ \\ x_1=32 & y_1=0.6249 \\ \\ x_2=35 & y_2=0.7002 \\ \\ x_3=38 & y_3=0.7813 \\ \end{matrix}$$

By Lagrange’s interpolation formula we have

$$y=f(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3$$

Put $x=33$

$$f(33)=\dfrac{(33-32)(33-35)(33-38)}{(30-32)(30-35)(30-38)}\times 0.5774$$

$$+\dfrac{(33-30)(33-35)(33-38)}{(32-30)(32-35)(32-38)}\times 0.6249$$

$$+\dfrac{(33-30)(33-32)(33-38)}{(35-30)(35-32)(35-38)}\times 0.7002$$

$$+\dfrac{(33-30)(33-32)(33-35)}{(38-30)(38-32)(38-35)}\times 0.7813$$

$$=0.6494$$

$\tan33\degree=0.6494$