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I am currently studying Integration (a very basic introduction) and I have a question regarding the summation of trigonometric functions. Given $f(x) = \sin x$, determine the area under the curve between a and b. By definition of a definite integral (using sigma notation rather than antiderivatives),

$$ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^\infty f(x_i)\,\Delta x $$

Before I proceed in determining the integral, is there a way that I can determine the value of

$$ \sum_{i=1}^n\sin i $$

To put this in context, I am wondering if there is a formula that one can use for the summation of trig functions. There are formulas for summation of polynomials. i.e

$$ \sum_{i=1}^n i = \frac{n(n+1)}{2} $$

I would like to expand this method to all trig functions! Any help would be much appreciated.

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  • $\begingroup$ Wolfram Alpha seems to be able to find it: wolframalpha.com/input/?i=sum%28sin%28i%29%2C%7Bi%2C1%2Cn%7D%5D but I'm not entirely sure how it's derived. From a quick look: mathworld.wolfram.com/Sine.html if you look at equation number 12 then that if you let $x=1$ it gives the Wolfram Alpha result too. $\endgroup$ – Jay Nov 29 '14 at 0:18
  • $\begingroup$ hey thanks Jay! I will check it out! $\endgroup$ – Shane M Nov 29 '14 at 0:20
  • $\begingroup$ No worries; from the looks of it, it seems like a neat derivation! $\endgroup$ – Jay Nov 29 '14 at 0:24
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What you really want for the Riemann sum of $\int_a^b \sin x \, dx$ is to take $\sin x$ at $n$ uniform steps within the interval $[a,b].$ So you want $x_i$ to be something like $a + i\Delta x,$ or even better, $$x_i = a + i\Delta x - \tfrac12\Delta x \quad \text{where} \quad \Delta x = \frac{b-a}{n},$$ so that $x_1 = a + \tfrac12\Delta x$ and $x_n = b - \tfrac12\Delta x.$ If you define $x_i$ and $\Delta x$ in that way, the summation you're looking for is $$ \sum_{i=1}^n \sin\left(x_i\right) \Delta x. $$

Here's a handy trigonometric identity you can use for this problem: $$\sin A \sin B = \tfrac12 \cos(A−B) − \tfrac12 \cos(A+B).$$ We can apply it as follows: let $A = x_i$ and let $B = \frac12 \Delta x.$ Then $$\sin (x_i) \sin \left(\tfrac12 \Delta x\right) = \tfrac12 \cos\left(x_i - \tfrac12 \Delta x\right) − \tfrac12 \cos\left(x_i + \tfrac12 \Delta x\right). \tag1$$ Since you want to compute a summation over $\sin (x_i) \Delta x$ rather than $\sin (x_i) \sin \left(\tfrac12 \Delta x\right),$ let's multiply both sides of equation$\ (1)$ by $\dfrac{\Delta x}{\sin \left(\tfrac12 \Delta x\right)}$ to obtain $$\sin (x_i) \Delta x = \frac{\Delta x}{2\sin\left(\tfrac12\Delta x\right)} \cos\left(x_i - \tfrac12\Delta x\right) − \frac{\Delta x}{2\sin \left(\tfrac12 \Delta x\right)} \cos\left(x_i + \tfrac12 \Delta x\right). $$ The big fraction on the right side of this equation will occur at least once in every equation we write after this; to reduce clutter, let $k = \dfrac{\Delta x}{2\sin\left(\tfrac12\Delta x\right)}$ so that we can write $$\sin (x_i) \Delta x = k \cos\left(x_i - \tfrac12\Delta x\right) − k \cos\left(x_i + \tfrac12 \Delta x\right).$$ Now let's take a look at the next term in the summation, $\sin (x_{i+1}) \Delta x.$ Since $x_{i+1} = x_i + \Delta x,$ $$\begin{eqnarray} \sin (x_{i+1}) \Delta x &=& k \cos\left(x_{i+1} - \tfrac12 \Delta x\right) − k \cos\left(x_{i+1} + \tfrac12 \Delta x\right)\\ &=& k \cos\left(x_i + \tfrac12 \Delta x\right) − k \cos\left(x_{i+1} + \tfrac12 \Delta x\right). \end{eqnarray}$$ Now notice what happens if we add $\sin (x_{i+1})$ to $\sin (x_i).$ The two terms $k \cos\left(x_i + \tfrac12 \Delta x\right)$ cancel, and we're left with $$\sin (x_i) \Delta x + \sin (x_{i+1}) \Delta x = k \cos\left(x_i - \tfrac12 \Delta x\right) − k \cos\left(x_{i+1} + \tfrac12 \Delta x\right). $$ This is what we call a "telescoping sum," and it simplifies the summation wonderfully: if we write each term $\sin (x_i) \Delta x$ as a difference of two cosines (times a constant), as in the equations above, a pair of cosines cancel each other each time we add another term to the sum, and we're left with just a difference of two cosines (times a constant) at the end.

More formally, $$\begin{eqnarray} \sum_{i=1}^n \sin\left(x_i\right) \Delta x &=& \sum_{i=1}^n \left( k \cos\left(x_i - \tfrac12 \Delta x\right) − k \cos\left(x_i + \tfrac12 \Delta x\right) \right) \\ &=& k \left( \sum_{i=1}^n \cos\left(x_i - \tfrac12 \Delta x\right) − \sum_{i=1}^n \cos\left(x_i + \tfrac12 \Delta x\right) \right). \end{eqnarray}$$ The last $n - 1$ terms of $\sum_{i=1}^n \cos\left(x_i - \tfrac12 \Delta x\right)$ are the same as the first $n - 1$ terms of $\sum_{i=1}^n \cos\left(x_i + \tfrac12 \Delta x\right),$ so these cancel, leaving $$\begin{eqnarray} \sum_{i=1}^n \sin\left(x_i\right) \Delta x &=& k \left( \cos\left(x_1 - \tfrac12 \Delta x\right) − \cos\left(x_n + \tfrac12 \Delta x\right) \right) \\ &=& \frac{\Delta x}{2 \sin \left(\tfrac12 \Delta x\right)} (\cos a − \cos b). \tag2 \end{eqnarray}$$ That's the sum. To derive a definite integral from this, take the limit of the right-hand side of equation$\ (2)$ as $\Delta x \to 0.$ You'll find it's the same answer as you get from the antiderivative.

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Consider $$S=\sum_{k=1}^n\sin (k)$$ $$C=\sum_{k=1}^n\cos (k)$$ $$C+i S=\sum_{i=1}^n e^{ik}=\frac{e^i \left(1-e^{i n}\right)}{1-e^i}$$ So, $$C=\Re\left(\frac{e^i \left(1-e^{i n}\right)}{1-e^i}\right)=\frac{1}{2} \left(\cos (n)+\cot \left(\frac{1}{2}\right) \sin (n)-1\right)$$ $$S=\Im\left(\frac{e^i \left(1-e^{i n}\right)}{1-e^i}\right)=\frac{1}{2} \left(\sin (n)-\cot \left(\frac{1}{2}\right) \cos (n)+\cot \left(\frac{1}{2}\right)\right)$$

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