## anonymous one year ago find the exact value of cos75 degrees

1. anonymous

Use a half angle identity (150/2 = 75) or sum and difference identity (30 + 45) = 75 half angle (Θ = 150°): $\cos \frac{ \theta }{ 2 }=\sqrt{\frac{ 1+\cos \theta }{ 2 }}$ sum and difference (u = 30°, v = 45°) $\cos (u + v) = \cos u \cos v - \sin u \sin v$

2. anonymous

That is just as confusing as the notes

3. anonymous

this is the example I have in the notes

4. anonymous

ok so it looks like they used a sum and difference identity: $$cos(u+v)= \cos u \cos v− \sin u \sin v$$ The first thing you have to do is find 2 numbers on your unit circle that add up to 75. That's where the 30 and 45 come from. Then let u be 30 and let v be 45 and we're going to plug them in \(cos 75°=cos(30°+45°) With me so far?

5. anonymous

so far, yes lol

6. anonymous

I have one as a tab, not printed out

7. anonymous

that's good enough.

8. anonymous

So this is our identity cos (u + v) = cos u cos v - sin u sin v Plug in the 30 and 45 cos (30° + 45°) = cos 30° cos 45° - sin 30° sin 45° Now use the unit circle to find the 4 values in the identity: cos 30° cos 45° sin 30° sin 45°

9. anonymous

why do you subtract it by sin ?

10. anonymous

your formula is different than the one in my notes :/

11. anonymous

that's just the identity. Here's a list of the identities. Look under sum and difference http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

12. anonymous

so how do you know which one to use ?

13. anonymous

which sign to use, or which identity?

14. anonymous

identity

15. anonymous

It depends on what you're trying to find. For this one you can use multiple identities. I started off wanting to use a half angle because then all you'd have to do is find cos 150° and plug it into the formula. You could also have use the difference identity for cosine because 120 - 45 = 75. You're just looking for any combination that works. They'll all give you the same answer

16. anonymous

so pretty much you can us any, just depending if you're adding or subtracting ?

17. anonymous

yeah as long as it's the correct trig function and the numbers add/subtract right

18. anonymous

okay , I understand that part so far

19. anonymous

cos (30° + 45°) = cos 30° cos 45° - sin 30° sin 45° ok so now we plug in the numbers from the unit circle $\cos 75°=\frac{ \sqrt3 }{ 2 }\times \frac{ \sqrt2 }{ 2 }-\frac{ 1 }{ 2 }\times \frac{ \sqrt2 }{ 2 }$

20. anonymous

$\cos 75°=\frac{ \sqrt6 }{ 4 }-\frac{ \sqrt2 }{ 4 }$ $\cos 75°=\frac{ \sqrt6 -\sqrt2 }{ 4 }$

21. anonymous

where did the 1/2 go

22. anonymous

jk

23. anonymous

you're multiplying xD

24. anonymous

yeah :)

25. anonymous

I don't remember how to add and subtract radicals. I always mix it up

26. anonymous

the numbers under the radicals have to be the same to add or subtract, so this can't be simplified any further

27. anonymous

so it'd have to be $\sqrt{4} - \sqrt{4}$ to subtract ?

28. anonymous

right

29. anonymous

but wouldn't that just always equal 0 ?

30. anonymous

yes

31. anonymous

but if you had something like this, just subtract the "coefficients" $5\sqrt6-3\sqrt6=2\sqrt6$

32. anonymous

oooh okay, it's coming back lol. It's been a longer summer

33. anonymous

haha everyone's saying that

34. anonymous

I've been taking summer school for algebra 2 and trigonometry was thrown into algebra 2, which makes no sense to me

35. anonymous

yeah that doesn't make sense. This is more advanced trig so it really shouldn't be mixed in there

36. anonymous

I passed both semester of algebra 2 with a D, that being an accelerated class . I'm taking regular algebra 2 for summer school . Idk how the state thinks this is possible for an algebra 2 class lol

37. anonymous

That sounds like GA. I've tutored here since 2009 and the curriculum has changed 4 times. They keep moving stuff from one class to the next. It's ridiculous

38. anonymous

Common core is just awful . This year is the first year for common core to be fully enforced by math . Trig is introduced in accelerated algebra 2 when I was taking the class , but it wasn't so in depth

39. anonymous

@peachpi could you help me with tan105 degrees ?

40. anonymous

105 = 150 - 45

41. anonymous

$\tan(u-v)=\frac{ \tan u - \tan v }{ 1+\tan u \tan v }$ $\tan(150°-45°)=\frac{ \tan 150° - \tan 45° }{ 1+\tan 150° \tan 45° }$

42. anonymous

yeah?

43. anonymous

to find tan , you have to divide sin by cos right ?

44. anonymous

yes

45. anonymous

and then you multiply the denominator out to only have a numerator ?

46. anonymous

right

47. anonymous

okay , ima try it and see what I come up with

48. anonymous

ok

49. anonymous

you divide by the reciprocal right ?

50. anonymous

multiply *

51. anonymous

yes

52. anonymous

so for tan150 , I got $-\frac{ 2 }{ 2\sqrt{3} }$ is that right ?

53. anonymous

uh , no . where did that come from ?

54. anonymous

oh, wait never mind. I misread what you had. That is correct for tan 150. I read tan 105

55. anonymous

oh no lol. I'm doing it step by step

56. anonymous

so then tan45 is 1 ?

57. anonymous

yes

58. anonymous

and to substitute , I got $\frac{ - \frac{ 2 }{ 2\sqrt{3} } - 1 }{ 1+(-\frac{ 2 }{ 2\sqrt{3} }) \times 1}$

59. anonymous

Yes that's right

60. anonymous

how do you factor that lol

61. anonymous

start with simplifying: -2/(2√3) = -1/√3 Then multiply by √3/√3 to clear denominators |dw:1440272933438:dw|

62. anonymous

oooh okay so you have to get rid of the denominator for tan150 fraction ?

63. anonymous

for this one yeah because you can't have the radical in the denominator

64. anonymous

thank you!

65. anonymous

you're welcome

66. anonymous

to rationalize the denominator , how do you do that ? ... xD

67. anonymous

jk , I looked at your answer and I figured it out

68. anonymous

multiply by the conjugate

69. anonymous

oh ok. yeah foil will do it

70. anonymous

so is the answer $-4-\sqrt{6}$

71. anonymous

Somehow your numbers doubled. Should be $-2-\sqrt3$

72. anonymous

so when it's two radicals with the same number inside , it stays the same number ?

73. anonymous

|dw:1440273681350:dw|

74. anonymous

|dw:1440273726704:dw|

75. anonymous

oh okay , I see what I did wrong

76. anonymous

thank you again

77. anonymous

you're welcome

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