All Questions
12
questions
3
votes
1
answer
62
views
$\displaystyle P(n):|\sum_{k=1}^{n} \sin(k)\sin(k^2)| \leq 1, \forall n\geq 1$
Prove that $\displaystyle P(n):|\sum_{k=1}^{n} \sin(k) \sin(k^2)| \leq 1, \forall n\geq 1$ is true
What I've tried:
For $n=1 \implies \left|\sin(1)\sin(1^2)\right| \leq1$ is true.
Suppose that $P(n)$...
- 996
1
vote
0
answers
27
views
Prove (via induction) that $\forall n\ge1$, $2\sum_{j=1}^n \sin x\cos^{2j-1}x=\sin(2nx)$
So, the start of the proof is fine. Base case, easy enough. But when I start using the inductive hypothesis to try and prove P(k+1), I run into the following wall:
$\begin{align*}
2\sum_{j=1}^{k+1} \...
- 11
0
votes
2
answers
200
views
How to prove $\sum^n_{k=0}\binom{n}k\cos\big((n-2k)\theta\big)=2^n\cos^n\theta$?
Given that $n\in\mathbb{Z}$, for any $\theta\in\mathbb{R}$, prove that
$$\sum^n_{k=0}\binom{n}k\cos\big((n-2k)\theta\big)=2^n\cos^n\theta\,.$$
I tried to finish the proof by Mathematical Induction. ...
- 323
3
votes
4
answers
70
views
how do I show this :$\sum_{k=0}^{2n }\binom{2n}{k} \sin ((n-k)x)=0$ , for every real $x$ and for every integer $n$?
My attempt fails to show this formula $\sum_{k=0}^{2n }\binom{2n}{k} \sin ((n-k)x)=0$ which I have accrossed in my textbook, using induction proof, but I think by induction seems very hard, I want to ...
user517526
3
votes
2
answers
248
views
Trigonometric identity of finite terms
Prove that:
$$\dfrac{1}{\cos x+\cos {3x}} + \dfrac{1}{\cos x+ \cos {5x}}+\dots+\dfrac{1}{\cos x+ \cos {(2n+1)x}} \\= \frac{1}{2}\csc x \,[ \tan{(n+1)x}-\tan{x}]$$
I tried to prove this using the ...
- 3,789
1
vote
2
answers
18k
views
Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$ [duplicate]
Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$
My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}$$$$=\sum_{k=1}^...
- 123
6
votes
1
answer
4k
views
Formula for cos(k*x)
I need to prove that:
\begin{align}
c_k =&\; \cos(k\!\cdot\!x)\\
c_k :=&\; c_{k-1} +d_{k-1}\\
d_k :=&\; 2d_0\!\cdot\!c_k +d_{k−1}\\
d_0 :=&\; −2\!\cdot\!\sin^2{(x/2)}\\
\end{align}
I'...
- 2,341
1
vote
4
answers
2k
views
Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
$\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $...
- 3,479
1
vote
1
answer
128
views
Prove by induction: $\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$
$\sum\limits_{k=1}^{n}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{2^n}\cot\frac{x}{2^n}-\cot x,x\neq k\pi,k\in \mathbb{Z}$
Base Case: For $n=1$, $\frac{1}{2}\tan\frac{x}{2}=\frac{1}{2}\cot\frac{x}{2}-\...
- 3,479
0
votes
2
answers
3k
views
Using induction to prove a formula for $\sin x+\sin 3x+\dots+\sin (2n-1)x$
I'm working from the text "Intro To Real Analysis" by William Trench. Here is what I have thus far.
I will prove using Mathematical Induction that $\sin x+\sin 3x+...+\sin (2n-1)x=\frac{1-\cos 2nx}{...
- 722
6
votes
1
answer
220
views
Prove |cos(x−1)|+|cos(x)|+|cos(x+1)|≥3/2
I'm working on an induction proof, but I keep coming up against a brick wall.
While working through the induction proof process I keep ending up with $$|\cos(m)|\ge\frac12$$ ,but clearly this isn't ...
- 1,250
1
vote
1
answer
415
views
Expressing $\int \tan^n x\,dx$ with a sum
I was playing around with integrals of $\tan x$, because I knew that both $\int\tan x\,dx$ and $\int\tan^2x\,dx$ were solvable. I then came across the fact that
$$\begin{align}
\int \tan^n x\,dx &...
- 1,535