trigonometric equations calculations and their implementation in practise.

Question

$$\sin (11x)\sin (x) = \cos (10x) $$

How can one solve this ? The formulas for summation or doubling dont seem to be of much help here.

score 3 · Answer 1 · answered Oct 27 '13 at 19:20

3

Hint:

$$\sin(10x+x) = \sin(10x)\cos(x) + \cos(10x)\sin(x)$$

answered Oct 27 '13 at 19:20

Nathaniel Bubis

lab bhattacharjee · Accepted Answer · 2013-10-28T03:43:45.083

3

As $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B),$

$\displaystyle\implies 2\sin11x\sin x=\cos(10x)-\cos12x$

We have $\displaystyle2\sin11x\sin x=2\cos10x$

$\displaystyle\implies 2\cos10x=\cos(10x)-\cos12x$

$\displaystyle\implies\cos10x=-\cos12x=\cos(\pi-12x)$ as $\cos(\pi-y)=-\cos y$

$\displaystyle\implies 10x=2n\pi\pm(\pi-12x)$ where $n$ is any integer

edited Oct 28 '13 at 03:43

answered Oct 28 '13 at 03:11

lab bhattacharjee

comparing two cosines? – Bak1139 Oct 28 '13 at 07:13
how did you get to the last stage? – Bak1139 Oct 28 '13 at 07:20
@Bak1139, we know $\cos A=\cos B\implies A=2n\pi\pm B$ – lab bhattacharjee Oct 28 '13 at 07:28
@Bak1139, See also, http://mathworld.wolfram.com/WernerFormulas.html and http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – lab bhattacharjee Jan 24 '14 at 18:36
man, you're a necromancer!:) thanks... – Bak1139 Jan 24 '14 at 18:37
@Bak1139, not sure how I've come across this , but immediately I felt to add those links. Anyway, my pleasure – lab bhattacharjee Jan 24 '14 at 18:41

2 Answers2