Question

express in radical form : (sin pi over 4) (tan pi over 6) - (tan pi over 3) (cos pi over 2) plz help

Answer 1

dang what happened to my answer?

Answer 2

ok second term is 0 since
\[cos(\frac{\pi}{2})=0\]so we can ignore it

Answer 3

first term is
\[sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\]

Answer 4

and
\[tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}\]

Answer 5

multiplying gives
\[\frac{\sqrt{2}}{2\times \sqrt{3}}\]

Answer 6

to write in simplest radical form multiply top and bottom by
\[\sqrt{3}\] to get
\[\frac{\sqrt{6}}{6}\]

Answer 7

ok thats it?