Question

Use the half-angle formula to evaluate tan(17pi/12)

Answer 1

well
\[\tan(\frac{17\pi}{12}) \\ y=\tan(x) \text{ has period } \pi \\ \tan(\frac{17\pi}{12}-\pi)=\tan(\frac{17\pi}{12}-\frac{12\pi}{12})=\tan(\frac{5\pi}{12}) \\ \text{ now } 2 \cdot \frac{5 \pi}{12}=\frac{5\pi}{6} \\ \text{ so we have that we want to use the half-angle formula on } \\ \tan( \frac{1}{2} \cdot \frac{5 \pi}{6})\]

Answer 2

do you know the half angle identity for tan?

Answer 3

no, only for sin

Answer 4

so you can't use the half angle identity for tan?

Answer 5

you can only use the one for sin?

Answer 6

the weird thing is you have tan here not sin

Answer 7

or do you mean you can use it but you just don't know it?

Answer 8

http://www.purplemath.com/modules/idents.htm
do you see half-angle identities on this page?

Answer 9

notice the three different ones they wrote for tan(x/2)

Answer 10

use one of them

Answer 11

im confused because I don't really know how to do this lol

Answer 12

well first let's just copy of of the half-angle identities over from that page
\[\tan(\frac{1}{2}x)=\frac{\sin(x)}{1+\cos(x)} \\ \text{ and we want to evaluate } \\ \tan(\frac{1}{2} \frac{5\pi}{6})=?\]
do you see what to replace x with ?

Answer 13

compare tan(1/2*5pi/6) to tan(1/2*x)

Answer 14

x has to be 5pi/6 right?

Answer 15

so replace the x's in the identity with 5pi/6

Answer 16

ok one sec

Answer 17

tan(1/2*5pi/6)= sin(5pi/6) / 1+cos(5pi/6) ??

Answer 18

ok and ( ) around the bottom

Answer 19

so sin(5pi/6)=?
and cos(5pi/6)=?

Answer 20

1/2 and -sqrt 3/2 ??

Answer 21

\[\tan(\frac{1}{2} \cdot \frac{5\pi}{6})=\frac{\frac{1}{2}}{1+\frac{-\sqrt{3}}{2}}\]
multiply top and bottom by 2

Answer 22

\[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}}\]

Answer 23

can you simplify from there?

Answer 24

you know 2/2=1

Answer 25

Yes one second

Answer 26

2+ sqrt3

Answer 27

@freckles is that my answer?

Answer 28

let me check one sec

Answer 29

\[\frac{\frac{2}{2}}{2+\frac{2(-\sqrt{3})}{2}} \\ \frac{1}{2-\sqrt{3} } \\ \frac{1}{2 -\sqrt{3}} \cdot \frac{2 +\sqrt{3}}{2+\sqrt{3}} \\ \frac{2+\sqrt{3}}{4-3} \\ \frac{2 +\sqrt{3}}{1} \\ 2+\sqrt{3}\]

Answer 30

yep looks fabulous

Answer 31

thanks so much!!!

Answer 32

np