Find the value of this is the task:
[log base 2 (sin(11pi/6))^2]^3 + [log base 3 (tan(11pi/6))^2]^3
All I'm going to say is that this is seems very hard.
Forgive me if I am mistaken.
THANKS!
The first think you need to know is that $\displaystyle 11 \pi /6$ is a special angle, and that:
$\displaystyle
\sin(11 \pi /6)=-1/2
$ and $\displaystyle
\ \ \ \tan(11 \pi /6)=-1/\sqrt{3}
$
so:
$\displaystyle
[\log_2 ((\sin(11\pi /6))^2)]^3 + [\log_3(\tan(11\pi /6))^2)]^3=$$\displaystyle
[\log_2 (1/2^2)]^3 + [\log_3 (1/3)]^3
$
......=$\displaystyle =[-2]^3 + [-1]^3=-9$
RonL