Is there anything wrong with this algebraically?

  • Thread starter Thread starter tahayassen
  • Start date Start date
AI Thread Summary
The discussion focuses on simplifying a given algebraic expression involving trigonometric functions. It emphasizes that taking the reciprocal of both terms in a denominator does not change the value of the expression, highlighting the importance of not misapplying reciprocal properties. The simplification process is demonstrated, leading to the expression Cos(x)Sin(x) = Sin(2x)/2. Participants clarify that the simplification cannot be reduced further beyond this point. Overall, the conversation centers on proper algebraic manipulation and simplification techniques.
tahayassen
Messages
269
Reaction score
1

Homework Statement



http://img220.imageshack.us/img220/7554/daumequation13237539948.png

Homework Equations



N/A

The Attempt at a Solution



The denominator is two terms. If I take the reciprocal of both terms, does that change the value?
 
Last edited by a moderator:
Physics news on Phys.org
You would need to multiply by
\frac{1}{1 + \frac{\sin^2 x}{\cos^2 x}}
Just remember that \displaystyle \frac{1}{A + B} \ne \frac{1}{A} + \frac{1}{B}
 
Ah, thanks for clearing that up.
 
You could simplify it to

\frac{\frac{Sin(x)}{Cos(x)}}{1+\frac{Sin^2(x)}{Cos^2(x)}} = Cos(x)Sin(x)
by multiplying by \frac{Cos^2(x)}{Cos^2(x)}
And then that could become Cos(x)Sin(x)=\frac{Sin(2x)}{2}
That's about as simple as you'll be able to get it though
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top