Answer to Question #225116 in Trigonometry for Amit

Question

Problem C

Determine the numerical value of the following expression without the use of a calculator: 

problem is found on the website below

https://docs.google.com/viewerng/viewer?url=https://iymc.info/docs/IYMC_Qualification_Round_2021.pdf

Answer 1

$\sqrt{\sin\left(\frac{\pi+\log_2(\sqrt{2^\pi\cdot2^\pi})}{2^5-2^4-2^3}\right)\cdot\sqrt[3]{\frac{2^{4-1}}{2^{3/2}}}+\log_2(\log_3(9^{15})+\pi^{1+(-1)^{17}}+1)+\frac{(-1)^5+(-1)^{27}}{(-1)^{766}}}=\\ {}=\sqrt{\sin\left(\frac{\pi+\log_2(\sqrt{2^{2\pi}})}{32-16-8}\right)\cdot\sqrt[3]{\frac{2^{3}}{2^{3/2}}}+\log_2(\log_3(3^{30})+\pi^{1-1}+1)+\frac{-1-1}{1}}=\\ {}=\sqrt{\sin\left(\frac{\pi+\log_2(2^{\pi})}{8}\right)\cdot\sqrt[3]{2^{3/2}}+\log_2(30+1+1)-2}=\\ {}=\sqrt{\sin\left(\frac{\pi+\pi}{8}\right)\cdot\sqrt{2}+\log_2(32)-2}=\sqrt{\sin\left(\frac{\pi}{4}\right)\cdot\sqrt{2}+\log_2(2^5)-2}=\\ {}=\sqrt{\frac{\sqrt2}2\cdot\sqrt{2}+5-2}=\sqrt{1+3}=\sqrt4=2.$