- anonymous

find the exact value of sec(pi/8)

- katieb

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- freckles

\[\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\]

- freckles

1/2*pi/4 means
half of pi/4
think half angle identity

- anonymous

I'm confused where the half angle identity comes in

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## More answers

- freckles

well we know pi/4 is on the unit circle
and half of pi/4 is pi/8
so
since pi/8 is in the first quadrant we know cos to be positive
\[\cos(\frac{\pi}{8}) \\ =\cos(\frac{1}{2} \cdot \frac{\pi}{4})=\sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}}\]

- anonymous

I haven't seen that equation before

- freckles

\[\cos(\frac{\theta}{2})= \pm \sqrt{\frac{1+\cos(\theta)}{2}}\]?
you haven't seen this one?

- freckles

have you seen this one:
\[\cos^2(\theta)=\frac{1}{2}(1+\cos(2 \theta))\]

- anonymous

maybe I wrote it down for the wrong thing

- anonymous

My notes are all wrong lol

- freckles

http://www.purplemath.com/modules/idents.htm
here are a lot of trig identities with the corresponding names

- freckles

maybe you can use this to fix your notes

- anonymous

\[\cos \frac{ x }{ 2 }=\sqrt{\frac{ 1+sinx }{ 2 }}\] that's what I have

- anonymous

+- in the front

- freckles

err yes that sin(x) thing is definitely suppose to be cos(x)

- anonymous

Apex is terrible

- freckles

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \]

- freckles

there is still a couple things for you to do

- freckles

find cos(pi/4) using unit circle

- freckles

and simplify

- anonymous

how'd you get cos pi/4 ?

- freckles

x is pi/4

- freckles

i replace x with pi/4

- freckles

\[\frac{1}{2} \cdot \frac{\pi}{4} =\frac{\pi}{8} \\ \text{ so } x=\frac{\pi}{4}\]

- anonymous

oooh okay

- anonymous

cos pi/4 is sqrt2/2

- anonymous

and 45 degrees

- freckles

right so you can replace cos(pi/4) is sqrt(2)/2

- freckles

well pi/4 is 45 deg

- freckles

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}}\]

- anonymous

yeah I didn't know if I needed the degrees or radians so I put both lol

- freckles

you don't need either you just need to find cos(45) or cos(pi/4)

- freckles

cos(45 deg) is equivalent to finding cos(pi/4)

- anonymous

do I need to get rid of the radical or leave that there ?

- freckles

I would get rid of the compound fraction action inside the radical and then see if there are any perfect squares afterwards

- anonymous

so would I multiply top and bottom by the opposite of the denominator ?

- freckles

\[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\]
notice the numerator inside the squareroot is a perfect square

- freckles

\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]

- anonymous

wouldn't the rad 2 or rad 2 equal 1 ?

- freckles

yes sqrt(2/2)=1
that is why I multiplied it to get rid of the compound fraction action inside the square root

- anonymous

I'm confused on the multiplying of 2 in the denominator and numerator

- freckles

let's look inside the square root
we have inside the square root:
\[\frac{2}{1+\frac{\sqrt{2}}{2}}\]
this has a mini-fraction the sqrt(2)/2
the sqrt(2)/2 has a 2 in the denominator
I multiply top and bottom inside the square root by 2
so I have inside the square root:
\[\frac{2}{1+\frac{\sqrt{2}}{2}} \cdot \frac{2}{2} \\ \frac{2(2)}{(1+\frac{\sqrt{2}}{2})(2)} \\ \frac{4}{1(2)+\frac{\sqrt{2}}{2}(2)} \\ \frac{4}{2+ \frac{2}{2} \sqrt{2}} \\ \\ \frac{4}{2+(1) \sqrt{2}} \\ \frac{4}{2+ \sqrt{2}}\]

- anonymous

why did you transfer the radical to the two ?

- freckles

what 2 ?

- freckles

are you asking me about this:
\[\frac{\sqrt{2}}{2}(2)=\frac{2}{2} \sqrt{2}\]

- anonymous

this part

##### 1 Attachment

- freckles

multiplication is commuative

- freckles

\[\frac{\sqrt{2}}{2}(2) =\frac{\sqrt{2}}{2} \cdot \frac{2}{1} = \frac{\sqrt{2} \cdot 2 }{2 \cdot 1 } \\ \text{ you change order \in multiplication } \\ =\frac{2 \sqrt{2}}{2 \cdot 1} \\ \text{ you can reseperate the fraction } \\ =\frac{2}{2} \frac{\sqrt{2}}{1} \\ =1 \frac{\sqrt{2}}{1} \\ 1 \cdot \sqrt{2} \ = \sqrt{2}\]

- anonymous

oooh okay , that makes sense

- freckles

all I did was use the fact that a*b is the same as b*a

- anonymous

okay, that makes a lot of sense

- anonymous

for my final answer , I got \[-\sqrt{2}+2\] is that right ?

- freckles

\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \\ \]
\[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\]
\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]
\[\sec(\frac{\pi}{8})=\frac{2}{\sqrt{2+\sqrt{2}}}\]

- anonymous

wouldn't the rad 2 cancel out since it's two radicals ?

- anonymous

so it'd be \[\frac{ 2 }{ \sqrt{2}+2 }\]

- freckles

no \[\sqrt{a+\sqrt{b}} \neq \sqrt{a}+b \\ \text{ take } a=4 \text{ and } b=16 \\ \text{ so } \sqrt{a}=2 \text{ and } \sqrt{b}=4 \\ \sqrt{a+\sqrt{b}}=\sqrt{4+4}=\sqrt{8} \\ \text{ while } \sqrt{a}+b=2+16=18 \\ \text{ and } 18 \text{ is definitely not equal to } \sqrt{8} \\ \text{ so } \sqrt{a+\sqrt{b}} \text{ is not } \sqrt{a}+b\]

- anonymous

so it stays like that ?

- freckles

it stays like \[\frac{2}{\sqrt{2+\sqrt{2}}}\]
yes
you could rationalize the bottom but it will still look ugly

- freckles

so I would leave it as is

- anonymous

I never learned that lol . I have to get ready for work now though , thank you for the help !

- freckles

good luck

- anonymous

If I post up a new question with this activity I have to do , do you think you could figure it out ? I tried but I couldn't get it . I got stuck

- anonymous

The activities that are worth points in this packet are super confusing

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