## anonymous one year ago Verify the identity. cotangent of x to the second power divided by quantity cosecant of x plus one equals quantity one minus sine of x divided by sine of x

1. triciaal

|dw:1443371271009:dw|

2. triciaal

the original as written is not clear.

3. anonymous

1-2sin^2 x/ sin^2 x / 1+sin x/ sin x = 1- sin^2 x/ sinx * 1+ sinx

4. Nnesha

$\huge\rm \frac{ \cot^2 (x) }{ \csc(x) +1 } =\frac{ 1-\sin(x)}{ \sin(x)}$ like this ?

5. anonymous

I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx

6. anonymous

yes like that^

7. Nnesha

alright gimme a sec

8. Nnesha

are you working on left or right side ?

9. anonymous

left

10. Nnesha

alright first we can rewrite cot and csc in terms of sin and cos

11. Nnesha

csc = 1/sin right so cot = ??

12. anonymous

1/tan?

13. anonymous

cos/sin

14. Nnesha

yeah but we need to write in terms of sin cos

15. Nnesha

yes right $\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }$ now first deal with the denominator (red part)

16. Nnesha

$\frac{ 1 }{ \sin(x) }+1$ find the common denominator

17. anonymous

well 1/ sin = csc and 1+ csc = cot^2?

18. anonymous

I'm confused.

19. Nnesha

yes right you can use the identity but i found the other one easy

20. Nnesha

but if you use the identity you still have to write csc in terms of sin and cos

21. anonymous

so what do I do, I know csc is 1/sin

22. Nnesha

$\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }$ now first deal with the denominator (red part) we need to find common denominator 1 is same as 1/1 so $\frac{ 1}{ \sin(x)} +\frac{ 1 }{ 1 }$ common denomiantor is sin x right

23. Nnesha

$\large\rm \frac{ 1 }{ \color{blue}{\sin(x)} }+\frac{ 1 }{ \color{red}{1}}$ $\huge\rm \frac{ \color{red}{1}(1) +\color{blue}{sinx }(1)}{ \sin(x) }$ when we find common denominator we should multiply first fraction with the denominator of 2nd one and multiply the numerator of 2nd fraction with the denominator of first fraction

24. Nnesha

let me know if you have a question about that part ??^^^

25. anonymous

No it makes sense.

26. Nnesha

alright $\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1+sin(x) }{ \sin(x) }} }$ now change division to multiplication multiply first fraction with the reciprocal of the 2nd fraction

27. Nnesha

here is an example $\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } } =\frac{ a }{ b} \times \frac{ d }{ c }$

28. Nnesha

no multiply top fraction with the reciprocal of bottom one what's reciprocal of 1+sinx/sin ???

29. anonymous

sin/1+sinx

30. Nnesha

right so multiply $\huge\rm \frac{ \cos^2(x) }{\sin^2(x) } \times \frac{ \sin(x) }{1+\sin(x)}$

31. Nnesha

sin^2 is same as sin times sin so you cancel it out

32. Nnesha

$\huge\rm \frac{ \cos^2(x) }{\color{reD}{sin^2(x)} } \times \frac{ \sin(x) }{1+\sin(x)}$ sin^2 is same as sin times sin so you cancel it out $\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times sinx} } \times \frac{ \sin(x) }{1+\sin(x)}$

33. anonymous

so does that mean sinx cancels out?

34. Nnesha

yes right

35. Nnesha

$\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}$

36. Nnesha

now use the identity cos^2(x) = ???

37. anonymous

cos^2 = cosx * cosx?

38. Nnesha

yes but we need identity

39. Nnesha

sin^2(x) +cos^2(x) =1 solve for cos^2(x)

40. anonymous

oh wait its a path. identity.

41. Nnesha

that's the identity

42. Nnesha

ye

43. anonymous

That's it?

44. Nnesha

well ye just replace cos^2 with the identity then done

45. Nnesha

cos^2x = ???

46. anonymous

sin^2 + cos^2

47. anonymous

cos^2=sin^2?

48. Nnesha

how would you cancel out sin^2 from left side $\huge\rm sin^2x+\cos^2=1$ ?

49. Nnesha

subtract/add/divide /mutliply both sides by sin^2 x ?

50. anonymous

I don't understand, how? ^

51. Nnesha

yes i'm asking how would you cancel out sin^2x from left its just like simple equation

52. Nnesha

lets say we have x +y = 1 how would you solve for y ?

53. anonymous

you move x to the other side right?

54. Nnesha

yes right what would you get ?

55. anonymous

not enough info to solve, because x=?

56. Nnesha

well lets say 2+x =1 how would you solve for x ?

57. anonymous

-2 both sides x = -1

58. Nnesha

yes right subtract 2 both sides so for this equation subtract sin2x both sides $\huge\rm sin^2x+\cos^2=1$

59. Nnesha

|dw:1443375444006:dw| what would you get at right side ?

60. anonymous

cos^2= 1-sin^2

61. Nnesha

yes right now u can replace cos^2x with that $\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}$ $\huge\rm \frac{ 1-sin^2(x)}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}$

62. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @clairvoyant1 I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx $$\color{blue}{\text{End of Quote}}$$ now as u said you can write 1-sin^2x as (1+sinx)(1-sinx)

63. Nnesha

$\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}$

64. Nnesha

you can cancel out something right after that done!

65. Nnesha

got it ?? :=)

66. anonymous

Computer froze logged in on my kindle Lol. Not quite

67. anonymous

@Nnesha

68. Nnesha

$\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}$ what can you cancel out ?

69. anonymous

1+sin(x)

70. Nnesha

$\large\rm \frac{ (1-sin(x))\cancel{(1+sin(x))} }{\color{reD}{sinx } } \times \frac{{ 1}}{\cancel{1+\sin(x)}}$

71. Nnesha

that's it

72. Nnesha

both sides are the same right ?

73. anonymous

Ok could you do a quick recap of what we just went over please.

74. anonymous

@Nnesha

75. Nnesha

first rewrite cot and csc in terms of sin and cos then we found the common denominator change division to multiplication use the identity cos^2=1-sin^2 that's it

76. Nnesha

if you need the work you can reread the post :=) feel free to ask q

77. anonymous

Thank You so much for your patience and help :)

78. Nnesha

np :=)

79. anonymous

Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

80. anonymous

@Nnesha

81. anonymous

Can you help me with this one.

82. Nnesha

i'm not sure about that one sorry i just know the basic :(

83. Nnesha

make new post so other people will b able to help chu

84. anonymous

Find the domain of the given function. f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two. a) x > 0 b) All real numbers c) x ≥ -3, x ≠ 2 d) x ≠ -8, x ≠ -3, x ≠ 2

85. Nnesha

can you post the pic ?

86. anonymous

Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x @campbell_st

87. campbell_st

well cos(4x )= 1 - 2sin^2(2x) and cos(2x) = 1- 2sin^2(x) now make the substitution and simplify

88. anonymous

@campbell_st how would I do that?

89. campbell_st

exactly what it says... replace cos(4x) with the right hand side of the expression then the same with cos(2x) and collect like terms. what would probably be best, is to post it as a new question... as there is an lot of information that precedes the question you asked.

90. anonymous

alright I'll post it as a new question