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anonymous
 one year ago
find the exact value of sec(pi/8)
anonymous
 one year ago
find the exact value of sec(pi/8)

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.31/2*pi/4 means half of pi/4 think half angle identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused where the half angle identity comes in

freckles
 one year ago
Best ResponseYou've already chosen the best response.3well 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
 one year ago
Best ResponseYou've already chosen the best response.0I haven't seen that equation before

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\cos(\frac{\theta}{2})= \pm \sqrt{\frac{1+\cos(\theta)}{2}}\]? you haven't seen this one?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3have you seen this one: \[\cos^2(\theta)=\frac{1}{2}(1+\cos(2 \theta))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe I wrote it down for the wrong thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My notes are all wrong lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.3http://www.purplemath.com/modules/idents.htm here are a lot of trig identities with the corresponding names

freckles
 one year ago
Best ResponseYou've already chosen the best response.3maybe you can use this to fix your notes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos \frac{ x }{ 2 }=\sqrt{\frac{ 1+sinx }{ 2 }}\] that's what I have

freckles
 one year ago
Best ResponseYou've already chosen the best response.3err yes that sin(x) thing is definitely suppose to be cos(x)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3there is still a couple things for you to do

freckles
 one year ago
Best ResponseYou've already chosen the best response.3find cos(pi/4) using unit circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how'd you get cos pi/4 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3i replace x with pi/4

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{1}{2} \cdot \frac{\pi}{4} =\frac{\pi}{8} \\ \text{ so } x=\frac{\pi}{4}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3right so you can replace cos(pi/4) is sqrt(2)/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah I didn't know if I needed the degrees or radians so I put both lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you don't need either you just need to find cos(45) or cos(pi/4)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3cos(45 deg) is equivalent to finding cos(pi/4)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do I need to get rid of the radical or leave that there ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I would get rid of the compound fraction action inside the radical and then see if there are any perfect squares afterwards

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would I multiply top and bottom by the opposite of the denominator ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\] notice the numerator inside the squareroot is a perfect square

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wouldn't the rad 2 or rad 2 equal 1 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yes sqrt(2/2)=1 that is why I multiplied it to get rid of the compound fraction action inside the square root

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused on the multiplying of 2 in the denominator and numerator

freckles
 one year ago
Best ResponseYou've already chosen the best response.3let's look inside the square root we have inside the square root: \[\frac{2}{1+\frac{\sqrt{2}}{2}}\] this has a minifraction 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
 one year ago
Best ResponseYou've already chosen the best response.0why did you transfer the radical to the two ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3are you asking me about this: \[\frac{\sqrt{2}}{2}(2)=\frac{2}{2} \sqrt{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3multiplication is commuative

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{\sqrt{2}}{2}(2) =\frac{\sqrt{2}}{2} \cdot \frac{2}{1} = \frac{\sqrt{2} \cdot 2 }{2 \cdot 1 } \\ \text{ you change order \in multiplication } \\ =\frac{2 \sqrt{2}}{2 \cdot 1} \\ \text{ you can reseperate the fraction } \\ =\frac{2}{2} \frac{\sqrt{2}}{1} \\ =1 \frac{\sqrt{2}}{1} \\ 1 \cdot \sqrt{2} \ = \sqrt{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooh okay , that makes sense

freckles
 one year ago
Best ResponseYou've already chosen the best response.3all I did was use the fact that a*b is the same as b*a

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, that makes a lot of sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for my final answer , I got \[\sqrt{2}+2\] is that right ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\cos(\frac{x}{2})= \pm \sqrt{\frac{1+\cos(x)}{2}} \\ x=\frac{\pi}{4} \\ \cos(\frac{\pi/4}{2})=\pm \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \\ \cos(\frac{\pi}{8}) \text{ is positive since } \frac{\pi}{8} \text{ is \in the first quadrant } \\ \cos(\frac{\pi}{8})= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}} \\ \sec(\frac{\pi}{8})=\frac{1}{\cos(\frac{\pi}{8})} =\sqrt{\frac{2}{1+\cos(\frac{\pi}{4})}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \\ \] \[\sec(\frac{\pi}{8})=\sqrt{\frac{2}{1+\frac{\sqrt{2}}{2}}} \cdot \sqrt{\frac{2}{2}} \\ \sec(\frac{\pi}{8}) = \sqrt{ \frac{2 (2)}{(1+\frac{\sqrt{2}}{2})(2)}} \\ \sec(\frac{\pi}{8})=\sqrt{\frac{4}{2+\sqrt{2}}}\] \[\sec(\frac{\pi}{8})=\frac{\sqrt{4}}{\sqrt{2+\sqrt{2}}}\] \[\sec(\frac{\pi}{8})=\frac{2}{\sqrt{2+\sqrt{2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wouldn't the rad 2 cancel out since it's two radicals ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it'd be \[\frac{ 2 }{ \sqrt{2}+2 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3no \[\sqrt{a+\sqrt{b}} \neq \sqrt{a}+b \\ \text{ take } a=4 \text{ and } b=16 \\ \text{ so } \sqrt{a}=2 \text{ and } \sqrt{b}=4 \\ \sqrt{a+\sqrt{b}}=\sqrt{4+4}=\sqrt{8} \\ \text{ while } \sqrt{a}+b=2+16=18 \\ \text{ and } 18 \text{ is definitely not equal to } \sqrt{8} \\ \text{ so } \sqrt{a+\sqrt{b}} \text{ is not } \sqrt{a}+b\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it stays like that ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3it stays like \[\frac{2}{\sqrt{2+\sqrt{2}}}\] yes you could rationalize the bottom but it will still look ugly

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so I would leave it as is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I never learned that lol . I have to get ready for work now though , thank you for the help !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0The activities that are worth points in this packet are super confusing
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