Please show $|\sin(n+1)x| = |\sin(nx+x)|$ Pretty straightforward. I'm looking at a textbook for analysis, in chapter 1, I am not familiar with how
$|\sin(n+1)x| = |\sin(nx+x)|$
Could someone show me the proof or method here? I'm sure it's quite simple but I don't see it.
 A: I presume that on the left-hand side, we actually have $|\sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $\sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$. 
It's as simple as that.
A: Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $\sin$, $\exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$\sin x+\sin y =2\sin\tfrac12(x+y)\sin\tfrac12(x-y),$$$$\ln xy=\ln x+\ln y,\qquad\quad$$$$\exp(x+y)=\exp x+\exp y.\qquad\qquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$\sin nx+x,$$which is the sum of $\sin nx$ and $x$.
