## anonymous 5 years ago Prove the identity. csc(X)-sin(x)/csc(x)+sin(x)=cos^2x/1+sin^2x

1. Yuki

sorry I tried, but it doesn't seem straight forward.

2. anonymous

its okay, at least you put effort into it

3. anonymous

This isn't as hard as it looks :)

4. anonymous

$\csc x =\frac{1}{\sin x}$by definition, so$\frac{\csc x - \sin x}{\csc x + \sin x}=\frac{\frac{1}{\sin x}-\sin x}{\frac{1}{\sin x}+\sin x}=\frac{\frac{1-\sin^2x}{\sin x}}{\frac{1+\sin^2 x}{\sin x}}=\frac{1-\sin^2 x}{\sin x}\frac{\sin x}{1+\sin^2x}$$=\frac{1-\sin^2 x}{1+\sin^2x}=\frac{\cos^2x}{1+\sin^2x}$

5. anonymous

Do you need it to be explained more?

6. anonymous

yes if you don't care to

7. anonymous

Is there a specific part?

8. anonymous

9. anonymous

You can refer to the part by the number on the attachment, if that's easier.

10. anonymous

brb

11. anonymous

yes that made it alot easier to see thank you! do u care to help me on another one?

12. anonymous

Oh, so you're okay with the last one?

13. anonymous

Yes i knew how to get it started i just didnt know how to simplify that complex fraction, but i understand now.

14. anonymous

15. anonymous

solve this trig equation on the interval 0<x<2pi, express answer for angles as exact value in radians. 2cos^2x+3cosx+1=0

16. anonymous

Alright...this is a quadratic equation in cos(x). If you 'see' cos(x) as something like, 'u', then you have the equation$2u^2+3u+1=0$

17. amistre64

2 or 3 more and we can start a club ;)

18. anonymous

It can be factored into $(2u+1)(u+1)=0$

19. anonymous

Yeah, but in 30s you'll be on infinity hero, so... :)

20. amistre64

lol.... by then ill just need another account :)

21. anonymous

No, by then it's like TRON and you're sucked into the website.

22. anonymous

Sorry mj

23. anonymous

is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!

24. anonymous

Yes it is.

25. anonymous

That whole thing is equal to zero, though.

26. anonymous

Which means either 2cos^2x+1 = 0 OR cos(x) + 1 = 0 OR they are BOTH zero. The solution set is all x's that satisfy those conditions (just like a 'normal' quadratic).

27. anonymous

how do i find the angles? thats where i have a problem

28. anonymous

Yep...I'll do them on paper first.

29. anonymous

Okay thanks!

30. anonymous

Okay, if you look at the first factor, you should see something fishy...$2\cos^2 x + 1 =0 \rightarrow \cos^2x=-\frac{1}{2}$We have something squared on the left-hand side, and a negative on the right-hand side. If you're only using real numbers (which is what's going on here), this can't happen, so there are no x-values that satisfy this particular equation.

31. anonymous

The square of any real number is always 0 or positive.

32. anonymous

For the second factor, you have$\cos x = -1$On the interval $0 \le x \lt 2\pi$this happens for $x=\pi$only.

33. anonymous

If you plu x=pi into your original equation, you will get 0.

34. anonymous

*plug

35. anonymous

Thanks! if i had (sin(x)-4)(sin(x)+1) how would i find the angles on the same interval?

36. anonymous

Is that equal to zero?

37. anonymous

yes sorry i forgot

38. anonymous

Okay...it's the same deal...always... Either one or both of the factors have to be zero, so you set each one to zero and solve for any possible x-values. Here, in your first factor, you have sin(x)-4 = 0 --> sin(x)=4 This cannot happen (sin(x) oscillates between -1 and 1...there are no values of x here that will punch out a 4, so you won't be getting any solutions from this factor. Note, if it had been something like 4sin(x)-4 = 0 --> sin(x) =4/4 = 1 and you would have solutions). Sot the second factor, you have sin(x)+1=0 which means sin(x) = -1 When does sin(x) = -1 on this interval?

39. anonymous

*'Sot' = 'For'.

40. anonymous

is it -pi/2 +2kpi =3pi/2

41. anonymous

Yes it is.

42. anonymous

3pi/2

43. anonymous

Be careful when using formulas like that; you need to ensure you give the angle that lies in the interval they want (even though -pi/2 is the same as 3pi/2 when considered in the context of plugging into sine, cosine, etc.).

44. anonymous

so sinx=3pi/2

45. anonymous

No, $\sin \frac{3\pi}{2}=-1$

46. anonymous

You're trying to find the x's that, when plugged into sine, give -1.

47. anonymous

oh so x=3pi/2 i made a mistake

48. anonymous

Yes.

49. anonymous

you have been a lifesaver, that helped me out alot, especially with the identities.

50. anonymous

You're welcome, mj :)

51. anonymous

So that's it? You're done?

52. anonymous

Uhm this one seems simple but i forgot the identity to use 5sin^2x+2cos^2x=5-3cos^2x

53. anonymous

Okay...this might be more than you have to do, but technically, this is how you would prove it 'properly'.

54. anonymous

Assume that the statement is not true. Then,$5\sin^2 x + 2\cos^2 x \ne 5-3 \cos^2 x$But then adding 3cos^2x to both sides gives,$5\sin^2 x + 5 \cos^2 x \ne 5$Now, the left-hand side is just$5\sin^2 x + 5 \cos^2 x=5(\sin^2 x + \cos^2 x) = 5$which would mean,$5=5\sin^2 x + 5 \cos^2 x \ne 5$That is,$5 \ne 5$which is false.

55. anonymous

Our assumption that the original statement was false led to a contradiction, so we must take the original statement as true.

56. anonymous

The statement is either true or false. If it's not false, it must me true.

57. anonymous

If you're not happy with that way, I have another.

58. anonymous

i dont think thats what im lookin for my teacher makes us make one side the same as the other side :/

59. anonymous

Yeah...I'll do it another way.

60. anonymous

Take the right-hand side first. You have$5-3\cos^2 x$which equals,$5(1)-3\cos^2 x$$=5(\sin^2x + \cos^2x)-3\cos^2x$$=5\sin^2x + 5\cos^2x-3\cos^2 x$$=5\sin^2 x+2\cos^2x$

61. anonymous

ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this

62. anonymous

No more questions then?

63. anonymous

tan(x/2)sinx=1-cosx LHS tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx did i do that right?

64. anonymous

I'm not sure what you did. This is what I would do:

65. anonymous

$\tan \frac{x}{2}\sin x = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}.2\sin \frac{x}{2}\cos \frac{x}{2}=2\sin^2 \frac{x}{2}=1-\cos x$

66. anonymous

Since by the double angle formula for cosine,$\cos x = \cos ^2 \frac{x}{2}-\sin^2 \frac{x}{2}=1-2\sin^2 \frac{x}{2} \rightarrow 2\sin^2 \frac{x}{2}=1-\cos x$

67. anonymous

cos(alpha+beta)/sin(alpha)cos(beta)=cot(alpha)-tan(beta) cos(alpha)cos(beta)-sin(alpha)sin(beta)/.5(sin(alpha+beta)+sin(alpha-beta) how would i finish that

68. anonymous

$\frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}=\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta}$$=\frac{\cos \alpha \cos \beta}{\sin \alpha \cos \beta }-\frac{\sin \alpha \sin \beta }{\sin \alpha \cos \beta}$$=\frac{\cos \alpha}{\sin \alpha}-\frac{\sin \beta }{\cos \beta}$$=\cot \alpha - \tan \beta$

69. anonymous

That was a much easier way of doing it, identities must take alot of practice to get use to.

70. anonymous

Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!

71. anonymous

I can tell the preparation and follow through is the hardest part! cos(x/2) cos(x/2)=square root of 1+4/5/2 =square root of 9/5/2 =square root of 9/10 is that right if cosx=4/5 with 3pi/2<x<2pi ?

72. anonymous

Um, it's hard for me to tell since I'm not sure of your order of operations.

73. anonymous

What's under the square root?

74. anonymous

$\frac{\sqrt{1+4/5}}{2}$?

75. anonymous

i used the half angle formula for cos YES that is correct

76. anonymous

Yes, and it's positive root if you're in that interval.

77. anonymous

yes that is what i used how bout tan(2x) on same interval if i typed out what i have it would be hard for you to understand

78. anonymous

Do you know how to use the equation editor? The button below will take you there.

79. anonymous

oh i didnt know that thank you!

80. anonymous

$\tan2(\theta)=1-\cos(2\theta)\div2\div1+\cos(2\theta)\div2$

81. anonymous

lol...it's not clear...is it...$\tan 2 \theta =\frac{\frac{1-\cos 2 \theta}{2}}{\frac{1+ \cos 2 \theta}{2}}$?

82. anonymous

yes thats what i was trying to do i got -3/13 for the answer but im not sure about it

83. anonymous

Are you trying to find tan(2 theta) and the right-hand side is what you came up with?

84. anonymous

yes im trin to find tan(2theta) on the interval 3pi/2<x<2pi i used the double angle formula for tan

85. anonymous

For you to get a number, tan(2theta) would have to have been set equal to something in the first place. Is that what you have?

86. anonymous

Because tan(2theta) takes an infinity of values on that interval...unless tan(2theta)=(something) to begin with.

87. anonymous

yes thats what i have it equal to

88. anonymous

ok...let me see.

89. anonymous

okay!

90. anonymous

I'll tell you what I'm doing (because I'm being distracted)...since cos2theta is all the rage in this equation, try to get everything in terms of cos(2theta) and determine a polynomial in cos(2theta),$P(\cos 2 \theta)=0$Then solve for the roots of the polynomial.

91. anonymous

okay i will do that! can you tell me how to simplify real quick (-1/2)(4/5)-square root of 3/2(3/5)

92. anonymous

I need you to use the equation editor ::

93. anonymous

$-1/2(4/5)-(\sqrt{3/2})(3/5)$

94. anonymous

$-\frac{1}{2}\frac{4}{5}-\frac{3}{5}\sqrt{\frac{3}{2}}$do you mean this?

95. anonymous

yess that is right

96. anonymous

97. anonymous

Thank you for everything! I appreciate your time!

98. anonymous

welcome

99. anonymous

did you finish the last one?

100. anonymous

nope :/

101. anonymous

Is this due online or something?

102. anonymous

no i have to turn in next week

103. anonymous

Okay. I have to go sort some things out (like my life). I can look at the last problem and post back here. You should get an e-mail from the site saying it's been updated. I'll probably be back during the day to do it...probably :)

104. anonymous

Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir

105. anonymous

You too ;)

106. anonymous

Here you are...remember to check it.

107. anonymous

108. anonymous

109. anonymous

110. anonymous

111. anonymous

112. anonymous

So, in summary, you have a quartic polynomial in tan(theta), and so, four possible roots. Two of these roots will be complex conjugates of each other, so are not included. The roots are:$\tan \theta = 0$$\tan \theta \approx -1.52138$which means that your angles are (for this interval)$\theta = 2\pi$(from the first root)$\theta \approx 5.294^c$(from the second root). The 'c' superscript means, 'radians'.

113. anonymous

The plot is for f(t)=t^3-t+2