How to Simplify This Trigonometric Equation Using Substitutions?

AI Thread Summary
The discussion focuses on simplifying a trigonometric equation using substitutions. The equation presented involves terms like sin and cos functions, ultimately equating to tan(α/2). Participants suggest substituting specific values, such as α = 60°, to verify the equation's correctness. There is also a request for guidance on using substitutions, specifically with s = sin(α) and c = cos(α), and a prompt to recall formulas for cos(2α) in terms of cos(α). The conversation emphasizes the need for clarity in problem statements and the application of trigonometric identities.
Fred1230
Messages
2
Reaction score
1
Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions
 
Physics news on Phys.org
s=\sin\alpha and c=\cos\alpha
 
Fred1230 said:
Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions
Substitute ##\alpha=60^{\circ}## in your expression and check if you come out with ##\tan30^{\circ}##. If not it's back to the drawing board!
 
Fred1230 said:
Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions
Do you know a formula for ##\cos(2\alpha)## in terms of ##\cos(\alpha)##?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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