Question

Trig.

Answer 1

I'm not sure how to plug in the Half Angle formula..

Answer 2

@Astrophysics ? :)

Answer 3

\[\tan(\pi/12)=\frac{ 1-\cos(\pi/12) }{ \sin(\pi/12) }\] ?

Answer 4

Isn't the double angle formula =
tan(a/2) = (1-cosa)/(sina) though?

Answer 5

Yes that should work, \[\tan(1/2a) = \frac{ sina }{ 1+cosa }\] we let \[a = \pi/12\]

Answer 6

Just to note that \[\frac{ 1-cosa }{ sina } = \frac{ \sin a }{ 1+\cos a }\] it's the same

Answer 7

Right. Why is tan (1/2a) instead of tan(a/2)?

Answer 8

1/2a and a/2 are the same thing

Answer 9

\[\frac{ a }{ 2 } = \frac{ 1 }{ 2 } a\]

Answer 10

So we need to now figure out \[\cos(\pi/12)~~~and~~~\sin(\pi/12)\]

Answer 11

oh xD so now it would be
\[\tan \frac{ \pi }{ 6 }=\frac{ \sin \frac{ \pi }{ 6 } }{1+\cos \frac{ \pi }{ 6 } }\]

Answer 12

Just give me one second let me do it on paper quickly

Answer 13

ok :)

Answer 14

Ok nice!

Answer 15

So we have \[\tan (\pi/6) = \frac{ 1-\cos(\pi/6) }{ \sin(\pi/6) }\] right?

Answer 16

as that is our half angle formula

Answer 17

Oh, that does work? Yay! hah :P

Answer 18

Yup! It should, ok so we can use a reference triangle for pi/6 do you know how to do it?

Answer 19

Yep, so cos(pi/6) = \[\sqrt{3}/2\]
and sin(pi/6) = 1/2

Answer 20

this is one of the three reference triangles you should've learned about, it's very helpful to remember them, we actually call them "special triangles"