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This question comes from Single Variable Calculus 7th Edition by Stewart, Section 11.5, Number 15. $$\sum_{n=0}^{\inf}{{\sin(n+{{1}\over{2}}})\pi\over{1+\sqrt{n}}}$$

The problem I have is that when working over this solution the proffesor made the substitution $$\sin(n+{1\over2})\pi = \sin(n\pi+{\pi\over2})$$

This is a true statement, but only for select values of $n$. We assumed that the question was missing a set of parenthesis around the $((n + {1\over2})\pi)$, but when we asked her about how she did it, she said that you could simply multiply it in. If that is true, how can you multiply it in when it only works for certain values of $n$?

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  • $\begingroup$ $\sin \theta=\sin(\theta)$ similarly $\sin(n+\frac{1}2)\pi=\sin((n+\frac{1}2)\pi)$ $\endgroup$ – Michael Medvinsky Oct 29 '15 at 18:11
  • $\begingroup$ So the lack of parenthesis doesn't mean that there are no parenthesis? It's just a notation? How do you recognize this? Some professors of mine would take off major points for lack of parenthesis. So it's weird not seeing them written in. $\endgroup$ – Fmonkey2001 Oct 29 '15 at 18:13
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    $\begingroup$ One often omits the parentheses around the argument of $\sin,\,\cos$ and the like. Out of laziness and because "everybody knows where the argument ends" [which, see above, isn't true]. $\endgroup$ – Daniel Fischer Oct 29 '15 at 18:14
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    $\begingroup$ you can omit the parentheses around the argument of trigonometric functions as long as it is clear what the argument is. Sometime something that is clean to one looks ambiguous to another. Special case is teachers and students, but you will learn it fast. $\endgroup$ – Michael Medvinsky Oct 29 '15 at 18:21
  • $\begingroup$ I clearly see that. Thanks for your clarification! $\endgroup$ – Fmonkey2001 Oct 29 '15 at 18:22
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Generally, when you want to multiply a function you put the multipliers in front of it( example $4\sin x$ ). Everything after that is considered part of the argument.

For example $4\sin(x+1)\pi x^2 + 2$

(+$2$ is the part where the argument end, so you're calculating $\sin[(x+1)\cdot \pi\cdot x^2]$

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