Question

Use the half angle formulas to come up with an exact expression for each function value below. You DO NOT have to simplify your answer.

a. cos(π/8)
b. tan(π/8)
c. tan(π/16)

Answer 1

I've finished the first two! ** π/8 is half of π/4 ** \[a) = \sqrt{\frac{ 2+\sqrt{2} }{ 4 }}\]\[b) = \sqrt{2} - 1\]

Answer 2

Haha, nice. Alright, let's see about #3. Gotta use it twice D:

Answer 3

Use what twice? π/8?

Answer 4

Well I'm just starting from pi/4 for tan is all xD Because yeah, I didn't come up with what you did for tan(pi/8)

Answer 5

Should I write out what I did for B. tan(π/8) ?

Answer 6

If you want to. but I did come up with quite a different answer.

Answer 7

wait....I guess yours is correct just....simplified way more than mine I suppose xD

Answer 8

Oh right! It did say you do not have to simplify it... Whoops!

Answer 9

I'm not even sure how you simplified it that far, haha.

Answer 10

lets see...

Answer 11

Nvm, I see it.

Answer 12

\[\tan\frac{ π }{ 8 } = \tan\frac{( \frac{ π }{ 4 }) }{ 2 } \]\[\tan(\frac{ π }{ 8 }) = \frac{ 1 - \cos \frac{ π }{ 4 } }{ \sin \frac{ π }{ 4 } }\]

Answer 13

oh okay! dont mind me then :)

Answer 14

I was about to write it all out... lolz

Answer 15

I just had to check it, lol. Hmm...you started off in adirection I wouldn't know, haha. (1-cosx)/sinx?

Answer 16

It just worked! My next step was:\[\frac{ 1-\frac{ \sqrt{2} }{ 2 } }{ \sqrt{2} }\]

Answer 17

Looks like you missed the 2 on the bottom xD Lol, yeah, not sure what ya did o.o I just memorize the formula after all this silly time of abusing the hell out of it:
\[\pm \sqrt{\frac{ 1-cosx }{ 1+cosx }}\]

Answer 18

Do you mean the other √2/2 ? You're right, I did!

Answer 19

And I use that formula, just not for this problem in particular...

Answer 20

Yeah, lol. But nah, I just memorize a lot of these. So I just jotted that down and pluggedin numbers.

Answer 21

Okay, well, just do what you did but with your sqrt(2) - 1, lol.

Answer 22

Wait, where?

Answer 23

Lol, nvm. If you don't need to simplify, then I guess you could just plug in pi/8 into all your formula numbers. Unless we need to get the valueof cos and sinpi/8, too. And I see where you got your formula from, lol. I found it in my textbook. I just remember it the long way I guess xD

Answer 24

All it really says is "to come up with an expression for each function".

Answer 25

Bleh, too many stacked square roots x_x

Answer 26

I just don't know how to approach this.. lolz

Answer 27

You need to get the value of cos(pi/8), lol. That way you can plug it into your formula xD

Answer 28

The answer I got is correct but stupid xD

Answer 29

Look at this brutality:

\[\sqrt{\frac{ 2-\sqrt{\sqrt{2}+2} }{ 2+\sqrt{\sqrt{2}+2} }}\]

Answer 30

oh damn. I don't like this! UGH

Answer 31

Without just plugging in cos(pi/8) in, you kinda have to find the value of that and plugitin, which resultsin a ton of roots xD

Answer 32

I don't even feel like attempting to get an answer like that

Answer 33

Is it required? xd

Answer 34

Kinda? I need to be able to understand/remember/solve these types of problems when school starts

Answer 35

True. Well, I just did the half angle formula for cos, using pi/4 as my x and that got me the value of cospi/8. Then I just plugged that directly into the formula for half angle of tangent to finally get tan(pi/16)

Answer 36

Well thats just perfect! Let me try, and thanks again :)

Answer 37

Mhm. Because yeah, if you use x = pi/8 in order to find tan of pi/16, then this would be the plugging in:
\[\sqrt{\frac{ 1-\cos \frac{ \pi }{ 8 } }{ 1+\cos \frac{ \pi }{ 8 } }}\]

Since this would be the required formula anyway, there is no practical way to get that without actually finding the value of cos(pi/8) and plugging that directly in. So using thehalf angle formula for cos, using x = pi/4, I have:
\[\cos \frac{ \pi }{ 8 } = \sqrt{\frac{ 1+\frac{ \sqrt{2} }{ 2 } }{ 2 }}\]
So yeah, lots of subs with square roots, but then I could use that in the same spot I have cos(pi/8) in the tan half angle formula and go from there to get more roots than a tree.

Answer 38

How come you used the less simplified equation for cos(π/8)? Does mine not work in the problem?

Answer 39

Because I didn't realize you already had it :D

Answer 40

Oh okay! But would your be easier?

Answer 41

But even with yours you could still do one itty bitty thing :P
\[\frac{ \sqrt{2+\sqrt{2}} }{ 2 }\]

And probably not xD I just did it with unsimplified versions. I would use this. Id have simplified mine to this eventually anyway, I just didnt do it immediately.

Answer 42

I've gotten my final answer! \[\sqrt{\frac{2-\sqrt{2+\sqrt{2}} }{ 2+\sqrt{2+\sqrt{2}} } }\]

Answer 43

haha, nice xD

Answer 44

THANK YOU, THANK YOU, THANK YOU!

Answer 45

Yep, np ^_^