Question

Find expansion of tan(x+π/4) up to x^4 hence find tan(43)

Answer 1

Answer:

$\tan(x+\frac{\pi}{4})=1+2x+2x^2+2x^3++2x^4+.....$

$\implies \tan 43^\circ=0.9325$

Step-by-step explanation:

A question related to Taylor's series

The expansion of $\tan x$ is

$\tan x=1+2(x-\frac{\pi}{4})+2(x-\frac{\pi}{4})^2+...$

Therefore,

$\tan(x+\frac{\pi}{4})=1+2x+2x^2+2x^3++2x^4+.....$

Now,

$\tan 43^\circ=\tan(45^\circ-2^\circ)$

$2^\circ=\frac{\pi}{90}$ radian

Thus,

$\tan 43^\circ=\tan(\frac{\pi}{4}-\frac{\pi}{90})$

$\implies \tan 43^\circ=1+2\times (-\frac{\pi}{90})+2\times (-\frac{\pi}{90})^2+2\times (-\frac{\pi}{90})^3+2\times (-\frac{\pi}{90})^4+....$

Let $(-\frac{\pi}{90})=t$

$\tan 43^\circ=1+2t+2t^2+2t^3+...$

$\implies \tan 43^\circ=2+2t+2t^2+2t^3+...-1$

$\implies \tan 43^\circ=2(1+t+t^2+t^3+...)-1$

$\implies \tan 43^\circ=2\times \frac{1}{1-t}-1$

$\implies \tan 43^\circ=2\times \frac{1}{1+\frac{\pi}{90}}-1$

$\implies \tan 43^\circ=2\times \frac{90}{90+\pi}-1$

$\implies \tan 43^\circ=\frac{180}{90+3.14159}-1$

$\implies \tan 43^\circ=1.9325-1$

$\implies \tan 43^\circ=0.9325$

Hope this answer is helpful.