+118658 tan(2π/λ-λ/8)= tan π/4 How could this be correct; please explain.
There is more than one answer to this because tan pi/4 = tan 5pi/4 = tan 9pi/4 etc
The most basic answer is found by solving
\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}\\ 8\lambda (\frac{2\pi}{\lambda}-\frac{\lambda}{8})=8\lambda(\frac{\pi}{4})\\ 16\pi-\lambda^2=2\pi \lambda\\ 0=\lambda^2+2\pi \lambda-16\pi\\ \lambda=\frac{-2\pi\pm \sqrt{4\pi^2 +64\pi}}{2}\\ \lambda=\frac{-2\pi\pm 2\sqrt{\pi^2 +16\pi}}{2}\\ \lambda=-\pi\pm \sqrt{\pi^2 +16\pi}\\\)
To get the general solution is the same method just the algebra is a bit worse
\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}+\pi n \qquad n\in Z\\\)
andthen continue the same as the first one. :)
+118658 tan(2π/λ-λ/8)= tan π/4 How could this be correct; please explain.
There is more than one answer to this because tan pi/4 = tan 5pi/4 = tan 9pi/4 etc
The most basic answer is found by solving
\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}\\ 8\lambda (\frac{2\pi}{\lambda}-\frac{\lambda}{8})=8\lambda(\frac{\pi}{4})\\ 16\pi-\lambda^2=2\pi \lambda\\ 0=\lambda^2+2\pi \lambda-16\pi\\ \lambda=\frac{-2\pi\pm \sqrt{4\pi^2 +64\pi}}{2}\\ \lambda=\frac{-2\pi\pm 2\sqrt{\pi^2 +16\pi}}{2}\\ \lambda=-\pi\pm \sqrt{\pi^2 +16\pi}\\\)
To get the general solution is the same method just the algebra is a bit worse
\(\frac{2\pi}{\lambda}-\frac{\lambda}{8}=\frac{\pi}{4}+\pi n \qquad n\in Z\\\)
andthen continue the same as the first one. :)
