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UPDATE **** AGGGH, I am embarrassed, but I made an error in deriving the equation in this question. Please disregard this question, and I will start a new one if I get stuck on the corrected version. It does appear that @avz2611's hint may still come into play though, so at least this post was not for nothing. :) (I'm not sure how to "close" a question.)

I have a problem that after many manipulations was able to reduce it down to the following equation with only one unknown, $b$:

$$\sin^2(a+b) + 2 \sin(a) \cos(t) \sin(a+b) - \sin^2(b) = 0.$$

So $a$ and $t$ are known constants. The unknown, $b$, is what I am trying to solve for if it's possible. Is it possible to solve for $b$?

alfreema
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2 Answers2

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$HINT$: $sin(x+y)sin(x-y)$=$sin^2x-sin^2y$ try to find the expression and simplify

avz2611
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  • Following the hint here the equation becomes $0 = \sin(a+2b) + 2 \cos(t) , \sin(a+b)$. The solutions of which are not easy to find and with the use of Wolfram Alfa are : http://www.wolframalpha.com/input/?i=\sin%28a%2B2x%29+%2B+2+p+\sin%28a%2Bx%29+%3D+0 – Leucippus Oct 03 '14 at 18:16
  • So I just got out of an afternoon meeting, and saw this very interesting hint with a property I was unaware. Is the output from wolframalpha.com a solution? The URL is mangled. – alfreema Oct 03 '14 at 21:41
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have you try to develop $sin(a+b)$, then say $X=sin(b)$ and try to solve ?

Aurelien
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  • I'm not sure if this answers your question, but I tried converting sin(a+b) to sin(a)cos(b)+sin(b)cos(a) and then expanding everything out. I got a nice long equation, with many of the terms containing either sin(b) or cos(b) and not being able to isolate the b terms. I'd be happy to show the expansion if it helps? – alfreema Oct 03 '14 at 17:15
  • Hum yes, you are right, you'll have some trouble with cos(b). – Aurelien Oct 03 '14 at 17:38
  • it is difficult to solve this equation for $b$ – Dr. Sonnhard Graubner Oct 03 '14 at 17:53
  • OK. Maybe I need to draft up the original problem that got me to that equation and see if there is an entirely different way to attack it. I have taken many approaches and keep running into dead-ends, however the above equation is the closest I have gotten by far. – alfreema Oct 03 '14 at 21:43