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February 5th, 2017, 11:50 AM  #1 
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  sin(z) = 1  i with solutions (need explanation)
So I look at the prof's solutions  see attached. I dont understand how he got from (coshy)^2 + (sinhy)^2 = (sinhy)^2(coshy)^2 to cosh2y = 1/4(sinh2y)^2 Then cosh2y = sqrt(5) +2 y = (1/2)ln(sqrt(5)+2 +/ sqrt(8+4sqrt(5)) Then sinx = 2sqrt(5) x= 2pi  sin(sqrt(5)2)^1 + 2kpi Can someone explain? 
February 5th, 2017, 12:08 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,605 Thanks: 817 
basic identity $\cosh(a+b)=\cosh(a)\cosh(b) + \sinh(a)\sinh(b)$ let $a=b=y$ $\cosh(2y) = \cosh^2(y) + \sinh^2(y)$ basic identify $\sinh(a+b) = \sinh(a)\cosh(b)+\cosh(a)\sinh(b)$ again let $a=b=y$ $\sinh(2y) = 2\sinh(y)\cosh(y)$ $\dfrac{\sinh(2y)}{2} = \sinh(y)\cosh(y)$ $\dfrac{\sinh^2(2y)}{4} = \sinh^2(y)\cosh^2(y)$ This link has a good explanation of how $\cosh^{1}(y) = \log(y\pm\sqrt{y^21}),~y>1$ 
February 5th, 2017, 12:30 PM  #3  
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Quote:
Thanks! that helped a lot! Do you know why he set sinx = 2  sqrt(5) though?  
February 5th, 2017, 12:40 PM  #4  
Newbie Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0  Quote:
Thanks! that helped a lot! Do you know why he set sinx = 2  sqrt(5) though?  
February 5th, 2017, 05:54 PM  #5 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
It's because the solution to the quadratic equation $\displaystyle \begin{align*} x^2  4\,x  1 = 0 \end{align*}$ is $\displaystyle \begin{align*} x = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\,\sqrt{5}}{2} = 2 \pm \sqrt{5} \end{align*}$.


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