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1. anonymous

2. anonymous

@freckles

3. freckles

which one do you looking at?

4. freckles

all of them?

5. anonymous

yes all

6. freckles

ok well lets look at: $\cos(2x)=\frac{1}{\sqrt{2}} \\ \text{ or also known as } \\ \cos(2x)=\frac{\sqrt{2}}{2}$ Before we solve this one, do you know how to solve: $\cos(\theta)=\frac{\sqrt{2}}{2}$ also are we solving in a specific interval ?

7. anonymous

the interval is between 0 and 360

8. anonymous

45 and 315

9. freckles

ok if we have $0<x<360 \\ \text{ and we \let } theta=2x \\ \text{ then } x=\frac{\theta}{2} \\ \text{ and so we have } \\ 0<\frac{\theta}{2}<360 \\ \text{ multiplying both sides by 2 we have } \\ 0< \theta<720 \\ \text{ so you said } \cos(\theta)=\frac{\sqrt{2}}{2} \text{ has solutions } \\ \theta=45^o,315^o, 45^o+360^o,315+360^o \\ \text{ after simplifying } \\ \theta=45^o,315^o,405^o,675$

10. freckles

now recall $\theta=2x \\ \text{ so we have } \\ 2x=45^o ,315^o,405^o,675^o$

11. freckles

just divide both sides by 2

12. anonymous

so 22.2?

13. anonymous

157.5?

14. freckles

ok and two more solutions

15. freckles

$x=\frac{45^o}{2},\frac{315}{2}^o,\frac{405^o}{2},\frac{675^o}{2}$

16. freckles

we had to solve cos(theta)=sqrt(2)/2 in (0,720) since x was between 0 and 360

17. freckles

theta was 2 times more than x

18. anonymous

ahh gotcha.. and thats the answer?

19. freckles

yes those 4 solutions right there at mentioned are the answers

20. anonymous

is that considered double identity angles?

21. freckles

we didn't use the double angle identity but we could have... $\cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ so we have } 2\cos^2(x)-1=\frac{\sqrt{2}}{2} \\ 2 \cos^2(x)=\frac{\sqrt{2}}{2}+1 \\ 2\cos^2(x)=\frac{2+\sqrt{2}}{2} \\ \cos^2(x)=\frac{2+\sqrt{2}}{4}$ but yucky this looks a whole bunch uglier

22. anonymous

ooo yeah definitely better the way you showed me.. thank you.. next one?

23. anonymous

btw i like how you actually explain whats going on its greatly appreciated

24. freckles

take the square root of both sides you will end solving two equations

25. freckles

$\sin^2(x)=\frac{1}{2} \\ \text{ implies you have } \\ \sin(x)=\frac{1}{\sqrt{2}} \text{ or } \sin(x)=\frac{-1}{\sqrt{2}}$ find the solutions to both equations and the solution will be the union of the sets of solutions you found

26. anonymous

45?

27. anonymous

and -45?

28. freckles

earlier you said you wanted the solutions between 0 and 360 $\sin(x)=\frac{\sqrt{2}}{2} \text{ has solutions } x=45^o \text{ or } x=? \\ \sin(x)=-\frac{\sqrt{2}}{2} \text{ has solutions } x=-45^o+360^o \text{ or } x=?$ you still have 2 more solutions (1 for each equation)

29. anonymous

315

30. freckles

well you already have the solution

31. freckles

that was the -45+360 one

32. anonymous

i found 45, 315, 135 225

33. freckles

awesome sauce

34. freckles

number 59 is very similar to 57

35. freckles

number 60 is a quadratic try solve 2u^2-1-u=0 ( notice I just replaced cos(X) with u)

36. freckles

solve for u you can use the quadratic formula if you want

37. freckles

then replace u with cos(x) and then solve for x

38. freckles

i will be back in like 30 minutes (sorry)

39. anonymous

OMGGGG I GET IT! THANK YOU SO MUCH!!!!!!!! (hope the pie tastes bomb!)

40. freckles

l messed up on the crust but the filling should taste great.

41. anonymous

haha aww its okay.. tbh ive never had pecan pie.. but i have one last question

42. freckles

sure what is it

43. anonymous

just curious how old are you bc youre INCREDIBLE at math

44. freckles

There is some math superior to me (a lot of math actually) but I'm 30.

45. anonymous

46. anonymous

ahh gotcha... that makes me feel ALOT better haha.. im only a senior in high school and if you were my age and that mathematically inclined id be awfully embarrasssed lol

47. freckles

I would write purely in terms of sin(x) which will result in a quadratic equation again: $\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=(1-\sin^2(x))-\sin^2(x) \\ \cos(2x)=1-2 \sin^2(x)$ you would be embarrassed?

48. freckles

oops I mean in terms of cos(x)

49. freckles

one sec

50. anonymous

yes i would bc compared to you im math illiterate...

51. freckles

$\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ and } \sin^2(x)=1-\cos^2(x)$ I forgot about that one term being cos(x) :p

52. freckles

$\sin^2(x)+\cos(2x)-\cos(x)=0 \\ \text{ is equivalent to } \\ 1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ I replaced } \sin^2(x) \text{ with } 1-\cos^2(x) \\ \text{ and I replaced } \cos(2x) \text{ with } 2\cos^2(x)-1$

53. freckles

$1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ combine like terms on the \left } \\ \cos^2(x)-\cos(x)+1-1=0 \\ \cos^2(x)-\cos(x)=0$ and actually this is tons easier to solve than using the quadratic formula this one is easily factorable on the the left hend side

54. freckles

$a^2-a=a(a-1)$ $\cos^2(x)-\cos(x)=\cos(x)(\cos(x)-1)$ you will have two equations to solve $\cos(x)(\cos(x)-1)=0 \\ \text{ gives you } \cos(x)=0 \text{ or } \cos(x)-1=0$

55. anonymous

soooo 0 90 180

56. anonymous

not 180 i meant 270

57. freckles

cos(x)=0 when x=90 or x=270 good job there the other equation can be written as cos(x)=1

58. anonymous

360

59. anonymous

for when cos(x)=1

60. freckles

one question is the interval in which we solve the equation (0,360) or [0,360) or (0,360] or [0,360]

61. anonymous

[0,360)

62. anonymous

so it would be 0 not 360

63. freckles

ok so [0,360) means we don't look at 360 just everything up to it and we do include 0 and things after you know until we get to 360 (which we do not include)

64. freckles

so instead of saying the solution to cos(x)=1 is 360 we will say 0

65. freckles

however if we were solving cos(x)=1 on [0,360] then we would say x=0 or x=360 since we can include both endpoints

66. anonymous

yesss gotcha! seriously thank you soooo much :)

67. freckles

np

68. anonymous

have a great night!

69. freckles

you too!