Find the real and imaginary parts of $$\sin\left(\frac{\pi}{2}+i\ln2\right)$$
Using the double angle formula I have gotten $$\sin\left(\frac{\pi}{2}\right)\cos(i\ln2)+\cos\left(\frac{\pi}{2}\right)\sin(i\ln2)$$
Which is then $$\cos(i\ln2)$$
But I do not know where to go after this?
[Math] Find the real and imaginary parts of $\sin(\frac{\pi}{2}+i\ln2)$
complex numbers
Best Answer
$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2} \Rightarrow \cos(i\ln 2)=\frac{e^{-\ln 2}+e^{\ln 2}}{2}=\frac{\frac{1}{2}+2}{2}=\boxed{\dfrac{5}{4}}$$
So the imaginary part is zero and real part is $5/4$.