My Eighth Problem

Similar to Daniel Ellesar 's solution but maybe simpler. I just used the identity of $\cos{2\theta} = 1 - 2 \sin^2{\theta}$ .

$2\cos{(i\ln{4})}+2\sin^2{(i\ln{2})}+i\sin{(i\ln{4})}$

$=2\cos{(i\ln{4})}+(1-\cos{(i\ln{4})})+i\sin{(i\ln{4})}$

$=1+\cos{(i\ln{4})}+i\sin{(i\ln{4})}$

$= 1 + e^{i(i\ln{4})} = 1 + e^{-\ln{4}} = 1 + \frac {1}{4} = \boxed{1.25}$

The problem should be more than 10 points so that more will try it.

First use the logarithmic identity that $log4=2log2$ .

Then use the double angle formulae.

The problem becomes $2cos^{2}(ilog2)-2sin^{2}(ilog2)+2sin^{2}(ilog2)+2isin(ilog2)cos(ilog2)$

Simplify and factorise by taking out $cos(ilog2)$

$={cos(ilog2)}\times{2(cos(ilog2)+isin(ilog2))}$

Now use the trigonometric identities $cos(x)=\frac{e^{2ix}+1}{2e^{ix}}$

and $cos(x)+isin(x)=e^{ix}$ .

The problem becomes ${\frac{e^{2i^{2}log2}+1}{2e^{i^{2}log2}}}\times{2e^{i^{2}log2}}$

Since $i^{2}=-1$

The problem becomes $e^{-2log2}+1$

$=e^{log(\frac{1}{4})}+1$

$=\frac{1}{4}+1=1.25$