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

1. freckles

$\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}$

2. freckles

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

3. anonymous

I'm confused where the half angle identity comes in

4. 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 $\cos(\frac{\pi}{8}) \\ =\cos(\frac{1}{2} \cdot \frac{\pi}{4})=\sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}}$

5. anonymous

I haven't seen that equation before

6. freckles

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

7. freckles

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

8. anonymous

maybe I wrote it down for the wrong thing

9. anonymous

My notes are all wrong lol

10. freckles

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

11. freckles

maybe you can use this to fix your notes

12. anonymous

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

13. anonymous

+- in the front

14. freckles

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

15. anonymous

Apex is terrible

16. freckles

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

17. freckles

there is still a couple things for you to do

18. freckles

find cos(pi/4) using unit circle

19. freckles

and simplify

20. anonymous

how'd you get cos pi/4 ?

21. freckles

x is pi/4

22. freckles

i replace x with pi/4

23. freckles

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

24. anonymous

oooh okay

25. anonymous

cos pi/4 is sqrt2/2

26. anonymous

and 45 degrees

27. freckles

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

28. freckles

well pi/4 is 45 deg

29. freckles

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

30. anonymous

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

31. freckles

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

32. freckles

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

33. anonymous

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

34. freckles

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

35. anonymous

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

36. 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

37. freckles

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

38. anonymous

39. freckles

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

40. anonymous

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

41. 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: $\frac{2}{1+\frac{\sqrt{2}}{2}} \cdot \frac{2}{2} \\ \frac{2(2)}{(1+\frac{\sqrt{2}}{2})(2)} \\ \frac{4}{1(2)+\frac{\sqrt{2}}{2}(2)} \\ \frac{4}{2+ \frac{2}{2} \sqrt{2}} \\ \\ \frac{4}{2+(1) \sqrt{2}} \\ \frac{4}{2+ \sqrt{2}}$

42. anonymous

why did you transfer the radical to the two ?

43. freckles

what 2 ?

44. freckles

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

45. anonymous

this part

46. freckles

multiplication is commuative

47. freckles

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

48. anonymous

oooh okay , that makes sense

49. freckles

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

50. anonymous

okay, that makes a lot of sense

51. anonymous

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

52. freckles

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

53. anonymous

54. anonymous

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

55. freckles

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

56. anonymous

so it stays like that ?

57. freckles

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

58. freckles

so I would leave it as is

59. anonymous

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

60. freckles

good luck

61. 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

62. anonymous

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