Trigonometry Trigonometry Math Forum

 October 16th, 2019, 01:35 PM #1 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 64 Math Focus: Area of Circle Inequality $$\left| \sin \left( \sum_{k=1}^n x_k \right) \right| \le \sum_{k=1}^n \sin (x_k)$$ $\; \; \; \; \; (0 \le x_k \le \pi ; \; \; k=1,2,...,n)$ October 16th, 2019, 02:26 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 True. (Trig. identities) October 16th, 2019, 03:32 PM   #3
Senior Member

Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 64

Math Focus: Area of Circle
Quote:
 Originally Posted by DarnItJimImAnEngineer True. (Trig. identities)
I know it is true. Prove it... October 16th, 2019, 03:45 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 376 Thanks: 202 sin(a+b) = sin(a)cos(b)+cos(a)sin(b) The cosines will have a magnitude less than or equal to 1. Whether they add or subtract, the sum will have a magnitude less than the sum of the sines. Last edited by DarnItJimImAnEngineer; October 16th, 2019 at 04:34 PM. October 17th, 2019, 01:16 AM #5 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math I will post the solution if I prove it. Can you give at least a hint? Last edited by skipjack; October 18th, 2019 at 10:01 PM. October 17th, 2019, 04:32 AM #6 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math The bound of LHS occurs for $\displaystyle x_k = \pi /2n$. $\displaystyle 1\leq n\sqrt[n]{\sin(x_1 ) \sin(x_2) \cdot ...\cdot \sin(x_n )}=n\sqrt[n]{\sin^{n}(\pi /2n ) }=n\sin(\pi /2n ).$ $\displaystyle n\sin(\pi /2n ) \geq 1$ proves the inequality. Last edited by skipjack; October 18th, 2019 at 09:59 PM. October 17th, 2019, 11:21 AM   #7
Senior Member

Joined: Jun 2019
From: USA

Posts: 376
Thanks: 202

Quote:
 Originally Posted by idontknow The bound of LHS occurs for $\displaystyle x_k = \pi /2n$. $\displaystyle 1\leq n\sqrt[n]{\sin(x_1 ) \sin(x_2) \cdot ...\cdot \sin(x_n )}=n\sqrt[n]{\sin^{n}(\pi /2n ) }=n\sin(\pi /2n ).$ $\displaystyle n\sin(\pi /2n ) \geq 1$ proves the inequality.
What has that got to do with the price of tea in China?
Also, set n=1, and it is clearly false.

I just demonstrated that $\sin(x_1 + x_2) = c_1 \sin(x_1) + c_2 \sin(x_2)$, where $c_1, c_2 \in [-1, 1]$. Is that not sufficient proof that $|\sin(x_1 + x_2)| \leq \sin(x_1) + \sin(x_2)$ if the sines are positive? Then you can extend this to $|\sin((x_1 + x_2) + x_3)| \leq (\sin(x_1) + \sin(x_2)) + \sin(x_3)$, and keep adding terms up to $x_k$.

Last edited by skipjack; October 18th, 2019 at 09:58 PM. October 17th, 2019, 11:50 AM   #8
Senior Member

Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 64

Math Focus: Area of Circle
Quote:
 Originally Posted by DarnItJimImAnEngineer What has that got to do with the price of tea in China? Also, set n=1, and it is clearly false. I just demonstrated that $\sin(x_1 + x_2) = c_1 \sin(x_1) + c_2 \sin(x_2)$, where $c_1, c_2 \in [-1, 1]$. Is that not sufficient proof that $|\sin(x_1 + x_2)| \leq \sin(x_1) + \sin(x_2)$ if the sines are positive? Then you can extend this to $|\sin((x_1 + x_2) + x_3)| \leq (\sin(x_1) + \sin(x_2)) + \sin(x_3)$, and keep adding terms up to $x_k$.
How much does tea cost in China? It is sufficient.

Last edited by skipjack; October 18th, 2019 at 09:56 PM. October 18th, 2019, 05:08 AM #9 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math $\displaystyle n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N}$, remains to prove the inequality. October 19th, 2019, 02:56 AM   #10
Senior Member

Joined: Dec 2015
From: Earth

Posts: 823
Thanks: 113

Math Focus: Elementary Math
Quote:
 Originally Posted by idontknow $\displaystyle f(n)=n\cdot \sin(\pi /2n ) \geq 1 \; , n\in \mathbb{N}$, remains to prove the inequality.
Since $\displaystyle \frac{f(1+n) }{f(n)} >1$ then $\displaystyle f(n)\uparrow$ and min{$\displaystyle n\cdot \sin(\pi /2n )$}$\displaystyle =1\geq 1$. Tags inequality Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post StillAlive Calculus 5 September 3rd, 2016 12:45 AM matemagia Algebra 2 November 15th, 2014 01:53 PM jiasyuen Algebra 10 September 28th, 2014 07:21 AM abhishake Algebra 3 June 5th, 2011 05:45 PM supermikong Algebra 8 May 2nd, 2011 03:24 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      