## anonymous one year ago Need help with some trig problems

1. anonymous

2. anonymous

@jim_thompson5910

3. jim_thompson5910

Use the identity cos(A - B) = cos(A)*cos(B) + sin(A)*sin(B)

4. jim_thompson5910

you're given tan(B). Use that to find the value of sin(B)

5. anonymous

wait we are solving question number 1 right

6. jim_thompson5910

yes

7. anonymous

ok how do I do that

8. jim_thompson5910

draw a right triangle with the angle beta |dw:1438649540480:dw|

9. jim_thompson5910

if tan(B) = -1/2, then what does that mean for the right triangle above?

10. anonymous

ok one sec

11. anonymous

sin is $\sqrt{5}/5$

12. jim_thompson5910

correct |dw:1438649735884:dw|

13. jim_thompson5910

what's cos(B) equal to?

14. anonymous

$\frac{ 2\sqrt{5} }{ 5 }$

15. jim_thompson5910

yes

16. jim_thompson5910

actually, B is in Q2, so cosine is negative

17. jim_thompson5910

now you do the same for the angle alpha |dw:1438649901995:dw|

18. jim_thompson5910

|dw:1438649921905:dw|

19. anonymous

oh yeah

20. jim_thompson5910

sin(alpha) will be negative since alpha is in Q4

21. anonymous

$\sqrt{23}/5$

22. jim_thompson5910

that will be negative, but otherwise, good

23. anonymous

ok

24. jim_thompson5910

$\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$ $\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)$ $\Large \cos(\alpha - \beta) =??$

25. anonymous

so $\frac{ \sqrt{2} }{ 5 }\times \frac{ -2\sqrt{5} }{ 5 }+-\frac{ \sqrt{23} }{ 5 }\times- \frac{\sqrt{5} }{ 5 }$

26. jim_thompson5910

sin(B) is positive because B is in Q2

27. anonymous

I cant get an exact answer on my calculator

28. jim_thompson5910

$\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$ $\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)$ $\Large \cos(\alpha - \beta) = \frac{-2\sqrt{2*5}}{5*5} + \frac{-\sqrt{23*5}}{5*5}$ $\Large \cos(\alpha - \beta) = -\frac{2\sqrt{10}}{25} - \frac{\sqrt{115}}{25}$

29. jim_thompson5910

Optionally, you can factor out the GCF $$\Large \frac{\sqrt{5}}{25}$$ to get $\Large \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$ $\Large \cos(\alpha - \beta) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{-2\sqrt{5}}{5}\right) + \left(\frac{-\sqrt{23}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)$ $\Large \cos(\alpha - \beta) = \frac{\sqrt{5}}{25}\left(-2\sqrt{2}-\sqrt{23}\right)$

30. anonymous

$\frac{ -2\sqrt{10}-\sqrt{115} }{ 25 }$

31. anonymous

is that the answer

32. jim_thompson5910

yeah that's another way to write the answer

33. anonymous

ok how do we do the second one

34. jim_thompson5910

review this list of identities http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf look for the "tan(2theta)" identity on the second page

35. anonymous

ok i see it

36. jim_thompson5910

so you need to find tan(alpha) and plug it into that identity

37. anonymous

ok

38. anonymous

tan is $\frac{ - \sqrt{46}}{ 2}$

39. jim_thompson5910

good

40. jim_thompson5910

alpha is in Q4 where tan is negative

41. anonymous

ok so can you show me how you plug it in

42. jim_thompson5910

$\Large \tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ $\Large \tan(2\alpha) = \frac{2\left(\frac{-\sqrt{46}}{2}\right)}{1-\left(\frac{-\sqrt{46}}{2}\right)^2}$ $\Large \tan(2\alpha) = \frac{-\sqrt{46}}{1-\frac{46}{4}}$ I'll let you simplify

43. anonymous

ok i got$\frac{ 2\sqrt{46} }{ 21 }$

44. jim_thompson5910

me too

45. anonymous

great does the attachement you sent me have the equation for the third one as well

46. jim_thompson5910

yes, look under "Sum and Difference Formulas" on page 2

47. anonymous

sin(a)cos(b)+cos(a)sin(b)

48. jim_thompson5910

yes

49. anonymous

ok again I am having problems plugging the equations in here from my side, can you show me what you got when you plugged each of the values in

50. jim_thompson5910

The values of sin(alpha), sin(beta), cos(alpha), cos(beta) were determined previously. Plug them in to get $\Large \sin(\beta + \alpha) = \sin(\beta)\cos(\alpha) + \cos(\beta)\sin(\alpha)$ $\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)$ $\Large \sin(\beta + \alpha) =??$

51. anonymous

I am having a hard time typing it in

52. anonymous

one second

53. anonymous

$-\frac{ -2\sqrt{23}+\sqrt{2}}{ 5\sqrt{5} }$

54. jim_thompson5910

$\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{-2\sqrt{5*23}}{5*5}$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{-2\sqrt{115}}{25}$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{10}-2\sqrt{115}}{25}$

55. anonymous

ok I see what I did wrong, I messed up on the second step

56. jim_thompson5910

oops I should have a positive sign

57. anonymous

where in front of the 2

58. jim_thompson5910

$\Large \sin(\beta + \alpha) = \left(\frac{\sqrt{5}}{5}\right)\left(\frac{\sqrt{2}}{5}\right) + \left(\frac{-2\sqrt{5}}{5}\right)\left(\frac{-\sqrt{23}}{5}\right)$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{5*2}}{5*5} + \frac{2\sqrt{5*23}}{5*5}$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{10}}{25} + \frac{2\sqrt{115}}{25}$ $\Large \sin(\beta + \alpha) = \frac{\sqrt{10}+2\sqrt{115}}{25}$

59. jim_thompson5910

yeah

60. anonymous

ok

61. anonymous

@jim_thompson5910 i have to go to the bathroom I will be back in about 6 minutes

62. jim_thompson5910

ok

63. anonymous

In the meanwhile could you input the information for #5 so I can see how you did it

64. jim_thompson5910

you mean #4 ?

65. anonymous

@jim_thompson5910 sorry that took longer than expected

66. anonymous

ok so I tried finding the equation for question #4 and #5 in the attachment you sent, but I could not find them

67. jim_thompson5910

alpha is in Q4 270 < alpha < 360 270/2 < alpha/2 < 360/2 135 < alpha/2 < 180 so alpha/2 is in Q2 which means sin(alpha/2) will be positive ------------------------------------------------------- Look on that identity sheet on page 2. Look under the "Half Angle Formulas" section You will use the identity $\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos(\alpha)}{2}}$ the right side is positive because sin(alpha/2) is positive

68. anonymous

ok i see it now

69. anonymous

i just plug in and solve, right?

70. jim_thompson5910

yes, $\Large \cos(\alpha) = \frac{\sqrt{2}}{5}$

71. anonymous

ok so my answer is

72. jim_thompson5910

I'm getting $\Large \sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{5-\sqrt{2}}{10}}$ which is the same thing more or less

73. anonymous

ok sweet, alright last problem

74. anonymous

its the same thing, but instead of subtracting root, we add this time

75. anonymous

oh no, we need a different value

76. jim_thompson5910

you'll use a different identity

77. jim_thompson5910

and you're using beta instead of alpha

78. anonymous

yeah ok so I found cosine in terms of beta, which is $-\frac{ 2\sqrt{5} }{ 5 }$

79. jim_thompson5910

yes

80. anonymous

Is this my equation

81. jim_thompson5910

now simplify that

82. anonymous

83. jim_thompson5910

I would rationalize the denominator

84. jim_thompson5910

multiply top and bottom of that inner fraction by sqrt(5)

85. jim_thompson5910

doing that gives $\Large \cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{5-2\sqrt{5}}{10}}$

86. anonymous

Is that as simplified as it gets?

87. jim_thompson5910

yeah

88. anonymous

wait I think one more problem got cut out while I was uploading my attachment

89. anonymous

Here one sec this is simpler, but it got me confused

90. jim_thompson5910

make a new post. This one is getting very laggy

91. anonymous

ok