Saved Bookmarks
| 1. |
tanh\(\left(\frac{\pi i}{4}\right)\) – coth\(\left(\frac{\pi i}{4}\right)\) is equal to |
|
Answer» tanh\(\left(\frac{\pi i}{4}\right)\) – coth\(\left(\frac{\pi i}{4}\right)\) \(=\cfrac{\sinh\left(\frac{\pi i}{4}\right)}{\cosh\left(\frac{\pi i}{4}\right)}\) \(-\cfrac{\cosh\left(\frac{\pi i}{4}\right)}{\sinh\left(\frac{\pi i}{h}\right)}\) \(=\cfrac{\sinh^2\left(\frac{\pi i}{4}\right)-\cosh^2\left(\frac{\pi i}{4}\right)}{\sinh\left(\frac{\pi i}{4}\right)\cosh\left(\frac{\pi i}{4}\right)}\) \(=\cfrac{-1}{\frac{1}{2}\times 2\sinh\left(\frac{\pi i}{h}\right)\cosh\left(\frac{\pi i}{h}\right)}\) (\(\because\) 1 + sinh2x = cosh2x ⇒ sinh2x – cosh2x = –1) \(=\cfrac{-2}{\sinh\left(\frac{\pi i}{2}\right)}\) (\(\because\) 2sinh(x) cosh(x) = sinh(2x)) \(=\frac{-2}{i}\) \(\big(\because\) i sinh(x) = sinix ⇒ sinh\(\left(\frac{\pi}{2}i\right)\) \(=\cfrac{\sin\left(i^2\frac{\pi}{2}\right)}{i}\) \(=\cfrac{-\sin \frac{\pi}{2}}{i^2}i = i\big)\) \(=\frac{-2}{i^2}i\) \(=2i\) \(\big(\because\) i2 = –1) |
|