My attempt:  First I converted this expression to sum notation:  $\sin(1^\circ) + \sin(3^\circ) + \sin(5^\circ) + ... + \sin(175^\circ) + \sin(177^\circ) + \sin(179^\circ)$ = $\sum_{n=1}^{90}\sin(2n-1)^\circ$  Next, I attempted to use Euler's formula for the sum, since I needed this huge expression to be simplified in exponential form:  $\sum_{n=1}^{90}\sin(2n-1)^\circ$ = $\operatorname{Im}(\sum_{n=1}^{90}cis(2n-1)^\circ)$  $\operatorname{Im}(\sum_{n=1}^{90}cis(2n-1)^\circ)$ = $\operatorname{Im}(\sum_{n=1}^{90}e^{i(2n-1)^\circ})$  $\operatorname{Im}(\sum_{n=1}^{90}e^{i(2n-1)^\circ})$ = $\operatorname{Im}(e^{i} + e^{3i} + e^{5i} + ... + e^{175i} + e^{177i} + e^{179i})$  Next, I used the sum of the finite geometric series formula on this expression:  $\operatorname{Im}(e^{i} + e^{3i} + e^{5i} + ... + e^{175i} + e^{177i} + e^{179i})$ = $\operatorname{Im}(\dfrac{e^i(1-e^{180i})}{1-e^{2i}})$  $\operatorname{Im}(\dfrac{e^i(1-e^{180i})}{1-e^{2i}})$ = $\operatorname{Im}(\dfrac{2e^i}{1-e^{2i}})$  Now I'm stuck in here;
1806
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:  $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$  There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$   How do we prove the above two identities?
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.
47576
Just as the title asks: I'm looking for a proof of Euler's identity without first proving the general case of $e^{ix} = \cos x + i\sin x$.
77367
I'm trying to calculate the trigonometric sum : $$\sum\limits_{k=1}^{n}\cos(k x)$$  This is what I've tried so far : $$\renewcommand\Re{\operatorname{Re}} \begin{align*} \sum\limits_{k=1}^{n}\cos(k x) &amp;= \Re\left(\sum\limits_{k=1}^{n}e^{i k x}\right)\\ &amp;= \Re\left(e^{i x}\frac{1 - e^{inx}}{1 - e^{ix}}\right) \end{align*}$$  How can I go on ?
17841
I tried and got this   $$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ where $m$ is an integer. $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$  Is it correct?
98662
At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result)
41912
In     However I am unable to work this out myself. When I multiply things out I get $\frac{1}{nS_{xx}}(\sum_{j=1}^n x_j^2 -2n\bar{x}x_i+nx_i^2)$. For things to be true, the terms inside the parenthesis can be rearranged to be $S_{xx}+n(x_i-\bar{x})^2$. I tried rearranging the terms so $\sum_{j=1}^n x_j^2 -n\bar{x}x_i+nx_i^2-n\bar{x}x_i$, but I can't seem to get to the answer. There must be some form of $S_{xx}$ that I am not aware of that is buried in their somewhere. Any help would be appreciated.
33804
$$\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$$  could you help me with how exactly I work this out?
97512
To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:  $e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$  $\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$  $\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$  $\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$  In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$  However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$.  What did I do wrong? I can't seem to find what or where.
2308
Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
115222
For example $151x - 294\equiv 44\pmod 7 $. How would I go about solving that?  The answer says to simplify it into the $ax\equiv b\pmod c $ form first, but I have no clue on how to get rid of the $294$... help?
92471
Is there anybody who can help me show the following?      $$ \sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0  \qquad\hbox{and}\qquad  \sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0 $$   I can only prove this claim for the sine expression when $k=0$, by grouping the output values such that they all cancel out (for both $n$ even and odd). I have thought about using complex numbers in exponential form to prove it, but I can't do it ... :P   I would really appreciate an algerbraic proof.
97081
This is where I'm at:  I know $$ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} , \hspace{2mm} \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}, $$ where $$ \tan(z) = \frac{\sin(z)}{\cos(z)}. $$ Applying the above, with a little manipulation, gives me: $$ \tan(z) = \frac{i\left(e^{-iz} - e^{iz}\right)}{e^{iz} + e^{-iz}}.$$  My thoughts are that I could use $e^{z} = e^{x+iy} = e^x\left(\cos(y) + i\sin(y) \right)$ to express both the numerator and denominator in trig form. Then I could times both by the denominator's complex conjugate as to get a real denominator, which would then, in turn, allow me to express the function in its real and imaginary parts.   However, as I'm sure you'd agree, this is messy, and it's hard to shake the idea that I'm missing something far more elegant.
18581
I am a little weak in trigonometry. I have two questions:   Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $ Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $   I am asking these questions because: 1. I am weak and unable to solve these. 2. I want to know the difference in these questions.
80108
I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it. $$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$  Thanks.
21643
Euler’s formula states that $e^{i  x} = \cos(x) + i \sin(x)$.  I can see from the MacLaurin Expansion that this is indeed true; however, I don’t intuitively understand how raising $e$ to the power of $ix$ produces rotation.  Can anyone give me an intuitive understanding?
79654
Question: Find the sum of the series:   $$\cos^3 \alpha +\cos^3 {3\alpha} + \cos^3 {5\alpha}+....+\cos^3 {(2n-1)\alpha}$$   The book from which this question was taken says that the answer is $\frac{3\sin{n\alpha}\cos{n\alpha}}{4\sin\alpha}+\frac{\sin{3n\alpha}\cos{3n\alpha}}{4\sin{3\alpha}}$.  My attempt to solve this question: $$\text{Let S be the trigonometric series,}$$ $$\cos {3\theta} = 4\cos^3\theta-3\cos\theta \implies 4\cos^3 \theta=\cos{3\theta}+3\cos\theta$$ Applying formula on $\cos^3\theta$,.. $$4S = 3\cos\alpha + \cos3\alpha +3\cos3\alpha + \cos9\alpha+...+3\cos{(2n-1)\alpha}+\cos{(6n-3)\alpha}$$ $$4S= 3(\cos \alpha + \cos 3\alpha+\cos5\alpha+...)+(\cos3\alpha + \cos9\alpha+..)$$ Applying the summation of cosine formula, $$4S= 3\frac{\sin{n\alpha}}{\sin\alpha}\cdot\cos{(\alpha+(n-1)\alpha)}+?$$  So my problem is I don't know how to apply the formula for the second series (denoted by '?') for a fixed number of terms $n$ or should I treat it as a general and separate series.  To clarify my doubt again, what I am asking is that would the $(2n-1)$ affect the formula for the second series (denoted by '?').  My working here looks most probably correct but if there is any mistake please correct it.