anonymous
  • anonymous
find the exact value of sec(pi/8)
Mathematics
katieb
  • katieb
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freckles
  • freckles
\[\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\]
freckles
  • freckles
1/2*pi/4 means half of pi/4 think half angle identity
anonymous
  • anonymous
I'm confused where the half angle identity comes in

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freckles
  • 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
  • anonymous
I haven't seen that equation before
freckles
  • freckles
\[\cos(\frac{\theta}{2})= \pm \sqrt{\frac{1+\cos(\theta)}{2}}\]? you haven't seen this one?
freckles
  • freckles
have you seen this one: \[\cos^2(\theta)=\frac{1}{2}(1+\cos(2 \theta))\]
anonymous
  • anonymous
maybe I wrote it down for the wrong thing
anonymous
  • anonymous
My notes are all wrong lol
freckles
  • freckles
http://www.purplemath.com/modules/idents.htm here are a lot of trig identities with the corresponding names
freckles
  • freckles
maybe you can use this to fix your notes
anonymous
  • anonymous
\[\cos \frac{ x }{ 2 }=\sqrt{\frac{ 1+sinx }{ 2 }}\] that's what I have
anonymous
  • anonymous
+- in the front
freckles
  • freckles
err yes that sin(x) thing is definitely suppose to be cos(x)
anonymous
  • anonymous
Apex is terrible
freckles
  • 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
  • freckles
there is still a couple things for you to do
freckles
  • freckles
find cos(pi/4) using unit circle
freckles
  • freckles
and simplify
anonymous
  • anonymous
how'd you get cos pi/4 ?
freckles
  • freckles
x is pi/4
freckles
  • freckles
i replace x with pi/4
freckles
  • freckles
\[\frac{1}{2} \cdot \frac{\pi}{4} =\frac{\pi}{8} \\ \text{ so } x=\frac{\pi}{4}\]
anonymous
  • anonymous
oooh okay
anonymous
  • anonymous
cos pi/4 is sqrt2/2
anonymous
  • anonymous
and 45 degrees
freckles
  • freckles
right so you can replace cos(pi/4) is sqrt(2)/2
freckles
  • freckles
well pi/4 is 45 deg
freckles
  • 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
  • anonymous
yeah I didn't know if I needed the degrees or radians so I put both lol
freckles
  • freckles
you don't need either you just need to find cos(45) or cos(pi/4)
freckles
  • freckles
cos(45 deg) is equivalent to finding cos(pi/4)
anonymous
  • anonymous
do I need to get rid of the radical or leave that there ?
freckles
  • freckles
I would get rid of the compound fraction action inside the radical and then see if there are any perfect squares afterwards
anonymous
  • anonymous
so would I multiply top and bottom by the opposite of the denominator ?
freckles
  • 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
  • freckles
\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]
anonymous
  • anonymous
wouldn't the rad 2 or rad 2 equal 1 ?
freckles
  • 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
  • anonymous
I'm confused on the multiplying of 2 in the denominator and numerator
freckles
  • 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
  • anonymous
why did you transfer the radical to the two ?
freckles
  • freckles
what 2 ?
freckles
  • freckles
are you asking me about this: \[\frac{\sqrt{2}}{2}(2)=\frac{2}{2} \sqrt{2}\]
anonymous
  • anonymous
freckles
  • freckles
multiplication is commuative
freckles
  • 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
  • anonymous
oooh okay , that makes sense
freckles
  • freckles
all I did was use the fact that a*b is the same as b*a
anonymous
  • anonymous
okay, that makes a lot of sense
anonymous
  • anonymous
for my final answer , I got \[-\sqrt{2}+2\] is that right ?
freckles
  • 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
  • anonymous
wouldn't the rad 2 cancel out since it's two radicals ?
anonymous
  • anonymous
so it'd be \[\frac{ 2 }{ \sqrt{2}+2 }\]
freckles
  • 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
  • anonymous
so it stays like that ?
freckles
  • freckles
it stays like \[\frac{2}{\sqrt{2+\sqrt{2}}}\] yes you could rationalize the bottom but it will still look ugly
freckles
  • freckles
so I would leave it as is
anonymous
  • anonymous
I never learned that lol . I have to get ready for work now though , thank you for the help !
freckles
  • freckles
good luck
anonymous
  • 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
  • anonymous
The activities that are worth points in this packet are super confusing

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