# sin(z) = -1 - i with solutions (need explanation)

#### Lalaluye

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 = 2-sqrt(5)
x= 2pi - sin(sqrt(5)-2)^-1 + 2kpi

Can someone explain?

#### romsek

Math Team
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 $a=b=y$

$\sinh(2y) = 2\sinh(y)\cosh(y)$

$\dfrac{\sinh(2y)}{2} = \sinh(y)\cosh(y)$

$\dfrac{\sinh^2(2y)}{4} = \sinh^2(y)\cosh^2(y)$

has a good explanation of how

$\cosh^{-1}(y) = \log(y\pm\sqrt{y^2-1}),~y>1$

• 1 person

#### Lalaluye

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 $a=b=y$

$\sinh(2y) = 2\sinh(y)\cosh(y)$

$\dfrac{\sinh(2y)}{2} = \sinh(y)\cosh(y)$

$\dfrac{\sinh^2(2y)}{4} = \sinh^2(y)\cosh^2(y)$

has a good explanation of how

$\cosh^{-1}(y) = \log(y\pm\sqrt{y^2-1}),~y>1$

Thanks! that helped a lot!

Do you know why he set sinx = 2 - sqrt(5) though?

#### Lalaluye

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 $a=b=y$

$\sinh(2y) = 2\sinh(y)\cosh(y)$

$\dfrac{\sinh(2y)}{2} = \sinh(y)\cosh(y)$

$\dfrac{\sinh^2(2y)}{4} = \sinh^2(y)\cosh^2(y)$

has a good explanation of how

$\cosh^{-1}(y) = \log(y\pm\sqrt{y^2-1}),~y>1$

Thanks! that helped a lot!

Do you know why he set sinx = 2 - sqrt(5) though?

#### Prove It

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*}.

• 1 person