Is 2tan2b the Simplified Form of tan(45° + b) - tan(45° - b)?

  • Thread starter Thread starter philipp2020
  • Start date Start date
AI Thread Summary
The discussion focuses on the simplification of the expression tan(45° + b) - tan(45° - b). The initial attempt led to confusion regarding the correct numerator, which was clarified to be just 4 tan b, rather than including an additional 1. The correct denominator was acknowledged as valid, and it was noted that this simplification leads to the final result of 2 * tan 2b. The conversation emphasizes the importance of accurately applying tangent sum and difference identities for proper simplification. The conclusion confirms that the simplified form is indeed 2tan2b.
philipp2020
Messages
34
Reaction score
0
hi

i have a question on a simplification

tan (45° + b) - tan (45° -b)

Then I put on both sides the Theorems for Addition. But I am not sure if my result is right. What is most possible simplification here?

is it 1 + 4 tan b / 1 - tan^2 b ?

Thanks very much for an answer.

Greetings

Philipp
 
Last edited:
Physics news on Phys.org
Double check the tangent sum and difference identities. Your denominator is good, but I can't figure out what mistake you could have made to get the numerator that you did.

If you get the numerator correct, there's another identity that allows you to simplify the fraction.
 
ah so the numerator is only 4 tan b and not plus 1 yes...

oh ok... i see

then at the end it will be just 2 * tan 2b
 
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