To find the Taylor series expansion of $\tan\left(x + \frac{\pi}{4}\right)$ up to terms in $x^4$, we proceed as follows:

1. Expression Setup:

We know the trigonometric identity: $\tan\left(x + \frac{\pi}{4}\right) = \frac{\tan x + 1}{1 - \tan x}$ Let $t = \tan x$, so we expand $\tan\left(x + \frac{\pi}{4}\right)$ as: $\tan\left(x + \frac{\pi}{4}\right) = \frac{t + 1}{1 - t}.$

2. Taylor Expansion of $\tan x$:

The Taylor series of $\tan x$ around $x = 0$ is: $\tan x = x + \frac{x^3}{3} + \mathcal{O}(x^5).$ Substituting this into the expression for $\tan\left(x + \frac{\pi}{4}\right)$, we expand step by step.

3. Substitute $t$ into the Formula:

Replace $t \approx x + \frac{x^3}{3}$ in: $\frac{t + 1}{1 - t}.$

Numerator:

$t + 1 \approx \left(x + \frac{x^3}{3}\right) + 1 = 1 + x + \frac{x^3}{3}.$

Denominator:

$1 - t \approx 1 - \left(x + \frac{x^3}{3}\right) = 1 - x - \frac{x^3}{3}.$

Thus: $\tan\left(x + \frac{\pi}{4}\right) \approx \frac{1 + x + \frac{x^3}{3}}{1 - x - \frac{x^3}{3}}.$

4. Simplification Using Series Division:

To simplify $\frac{1 + u}{1 - v}$ for small $u$ and $v$, we use: $\frac{1 + u}{1 - v} \approx 1 + u + v + u v \quad (\text{neglect higher-order terms}).$

Here:

• $u = x + \frac{x^3}{3}$,
• $v = x + \frac{x^3}{3}$.

Substitute: $u + v = \left(x + \frac{x^3}{3}\right) + \left(x + \frac{x^3}{3}\right) = 2x + \frac{2x^3}{3}.$ $u v = \left(x + \frac{x^3}{3}\right)\left(x + \frac{x^3}{3}\right) = x^2 + \frac{2x^4}{3} \quad (\text{keeping terms up to } x^4).$

Thus: $\tan\left(x + \frac{\pi}{4}\right) \approx 1 + \left(2x + \frac{2x^3}{3}\right) + x^2.$

5. Combine and Finalize:

Combine all terms up to $x^4$: $\tan\left(x + \frac{\pi}{4}\right) \approx 1 + 2x + x^2 + \frac{2x^3}{3}.$

6. Approximation for $\tan 43^\circ$:

Since $43^\circ \approx \frac{\pi}{4} + x$ where $x = 1^\circ$ in radians ($x = \frac{\pi}{180}$), substitute $x \approx 0.01745$ into the series: $\tan\left(\frac{\pi}{4} + x\right) \approx 1 + 2(0.01745) + (0.01745)^2.$ Calculating this gives: $\tan 43^\circ \approx 1 + 0.0349 + 0.000304 \approx 1.0352.$

Final Results:

1. Taylor Series: $\tan\left(x + \frac{\pi}{4}\right) \approx 1 + 2x + x^2 + \frac{2x^3}{3}.$
2. Approximation for $\tan 43^\circ$: $\tan 43^\circ \approx 1.0352.$

Let me know if you need further explanations or simplifications!

Related Questions:

1. How is the Taylor series for $\tan x$ derived step by step?
2. Can you expand other trigonometric functions like $\sin(x + \frac{\pi}{4})$ using Taylor series?
3. How accurate is the approximation of $\tan 43^\circ$ when using higher-order terms?
4. What happens when we include $x^5$-terms in the series expansion?
5. How can the series be applied for small angles in engineering or physics?

Tip:

When working with small angles, using approximations like $\sin x \approx x$ and $\tan x \approx x$ simplifies calculations significantly!