Vtyjkyuk

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:

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

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

$$fractan(x)+tan(y)1-tan(x)tan(y) - tan(x)$$

$$fractan(x)+tan(y)1-tan(x)tan(y) + frac-tan(x)+tan^2(x)tan(y)1-tan(x)tan(y)$$

$$fractan(y) + tan^2(x)tan(y)1-tan(x)tan(y)$$

$$fractan(y) [1+tan^2(x)]1-tan(x)tan(y)$$

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

$$fractan(y) sec^2(x)1-tan(x)tan(y) = boxedfracsec^2(x)tan(y)1-tan(x)tan(y)$$

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 $frac f(x+h)-f(x)h$

$frac[(x+h)^2 - 4(x+h) + 7] - (x^2 - 4x + 7)h = frach(2x+h-4)h = 2x+h-4$

...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.

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.

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 $ xin mathbbR $, 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, $tan(x+y)=fractan(x)+tan(y)1-tan(x)tan(y)$. 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, $f(x)=acdot x$ 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.

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.