Question inside! :)

Question

Question inside! :)

anonymous · Answer

|dw:1367029300752:dw| what do u think @terenzreignz ? :)

terenzreignz · Answer

It won't be a function if there was more than one :)
Good job.

terenzreignz · Answer

Wait... this isn't a function, is it :D

anonymous · Answer

lol :P

wait so is the answer still 1 then? :)

terenzreignz · Answer

Nope.

anonymous · Answer

aww darn :( then how can i find the right one @terenzreignz ? :D

anonymous · Answer

wait so if it isn't a function, then what is it? :/

anonymous · Answer

there are infinite solutions

$$\cos (\pi/2)=0\
\cos(3\pi/2)=0\
\vdots\
\cos\left[(2n+1){\pi\over2}\right]=0
$$

terenzreignz · Answer

Because these functions (sines and cosines) aren't one-to-one...
For instance, $\large \frac{\pi}{2}$  a solution, but also  $\large \frac{3\pi}{2}$

and many others ;)

anonymous · Answer

|dw:1367029515976:dw|
every time you trace the angle to these veritical positions, cosine ratio becomes "0"