What is the mathematical property that dictates that $f(x+y) = \tan(x+y)$ given that $f(x) = \tan(x)$?

Question

We were asked in our Calculus class to prove that,

$f(x+y) - f(x) = \frac {\sec^2(x) \tan(y)} {1 - \tan(x) \tan(y)}$ given that $f(x) = \tan(x)$

I have gotten so far as:

$$f(x+y) - f(x)$$

$$\tan(x+y) - \tan(x)$$

$$\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} - \tan(x)$$

$$\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)} + \frac{-\tan(x)+\tan^2(x)\tan(y)}{1-\tan(x)\tan(y)}$$

$$\frac{\tan(y) + \tan^2(x)\tan(y)}{1-\tan(x)\tan(y)}$$

$$\frac{\tan(y) [1+\tan^2(x)]}{1-\tan(x)\tan(y)}$$

Substituting the pythagorean identity, $$1+\tan^2(x) = \sec^2(x)$$

$$\frac{\tan(y) \sec^2(x)}{1-\tan(x)\tan(y)} = \boxed{\frac{\sec^2(x)\tan(y)}{1-\tan(x)\tan(y)}}$$

I don't quite understand how $f(x+y)$ became $\tan(x+y)$. I've had a few search results stating that $f(x+y) = f(x)+f(y)$ but it does not quite fit the bill.

I got the idea for my solution above because of a textbook example I've read, where:

Given $f(x)=x^2-4x+7$, find $\frac {f(x+h)-f(x)}{h}$

$\frac{[(x+h)^2 - 4(x+h) + 7] - (x^2 - 4x + 7)}{h} = \frac{h(2x+h-4)}{h} = 2x+h-4$

...but the book did not describe what property was used in order to 'insert' the value of $f(x)$ into $f(x+h)$, and by extension the $f(x)$ into the $f(x+y)$ of my problem. They feel... similar.

Is there a name for this mathematical property? Thank you very much.

Answer 1

It's simply substitution of the argument to $f$.

On the other hand: $f(x+y)=f(x)+f(y)$ that you claim to have found several sources for, is not generally true.

Answer 2

For your first query: "how $f(x+y)$ became $tan(x+y)$?" - Inside the bracket, we write the parameters of a function. As, $f(x)=tan(x)$ for all $ x\in \mathbb{R} $, hence, $f(z)=tan(z)$ also with $z=x+y$, as reals are closed under addition(means for any two real $x$ and $y$ , $x+y$ is also a real number). Writing $x+y$ in $f()$, we mean how the value of $f(z),z=x+y$. As, $x$ and $y$ are the partitions of $z$, using property of the function we are finding $f(z)$ in terms of $x$ and $y$, here, $tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$. There is no name for this property (as far I know).

For the last part, suppose you want to find $f(z)$ for $z=x+h$, so, you know that $f(z)=z^2-4z+7$, now substitute $z$ by $x+h$(as, $z=x+h$) and you will get $f(z)=(x+h)^2-4(x+h)+7$ which is nothing but $f(x+h)$ !. Think $x+h$ as a single number, then it will be clear to you.

The functions that satisfies $f(x+y)=f(x)+f(y)$ are known as linear functions. All function does not satisfy this property. For example, $f(x)=a\cdot x$ with constant $a$, find out $f(x+y)$ and $f(x)+f(y)$ separately, see that this function satisfies the property. On the other hand, suppose another function $g(x)=x^2$, observe that $g(x+y)=(x+y)^2=x^2+y^2+2xy$ is not equal to $g(x)+g(y)=x^2+y^2$ for all $x$ and $y$ in it's domain.

I think this explains , please feel free to ask for any doubt.

Answer 3

You might be unnecessarily hung up on the $x$ in $f(x)=\tan(x)$. Remember that your function $f$ is simply a mapping between some input to some output. The definition of $f$ just so happens to use $x$ to stand for any real number in this context.

You can consider what happens to another real number $x'=x+y$. That results in $f(x')=\tan(x')$, which is $f(x+y)=\tan(x+y)$ by substituion property of quality.