## anonymous 3 years ago simplify tan x cos x

1. anonymous

Remember tangent is just sin(x)/cos(x)

2. anonymous

so its cos x (sin x / cos x) ?

3. anonymous

Correct, now what happens when you go ahead and multiply the cosine?

4. anonymous

sin x

5. anonymous

That's it. The cosines cancel.

6. anonymous

okay cool what bout sec y sin (pi/2-y)? IDK how to do this at all

7. anonymous

$\sec(y)\sin(\frac{\pi}{2-y})$ ??

8. anonymous

but isnt it like pi/2 - y not pi/ 2-y

9. hartnn

sin(pi/2-y)=cos y sec y= 1/cos y

10. anonymous

Ok, so rewrite secant as 1/cos

11. anonymous

Then you can use a fancy pants trig identity.

12. anonymous

cos y 1/cos y

13. anonymous

|dw:1347861835303:dw| Right?

14. anonymous

i think so i was kinda looking at what hartnns post too so couldn't you do it like|dw:1347861948541:dw|

15. hartnn

sec y =1/ cos y is in numerator, so it would be :|dw:1347862109049:dw|

16. anonymous

No, well what you would have is a multiplication, the cosines would cancel.

17. anonymous

oh hey you know how tan x= sinx/ cos x well what are the rest of those cuz it might help me understand better like how we getting cos y when its sin y

18. hartnn

ok, so u want all trignometric formulas ? for converting one ratio to other ?

19. anonymous

20. hartnn

i will list some : $$\huge sin x =1/cosec x \implies cosec x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\ \huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x$$

21. hartnn

if u want sin x when cos x is given or if u want cos x when sin x is given, u use: $$sin^2x+cos^2x=1$$

22. anonymous

okay

23. anonymous

so if its 1+tan^2x/csc^2x that would simplify to sec^2x/csc^2

24. hartnn

yes. or is it ? $$\large1+\frac{tan^2x}{cosec^2x}\quad?$$

25. hartnn

if it is $$\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x$$

26. anonymous

but you put that cos^2x=1+cot^2x

27. hartnn

no the last one is cosec^2 x =1+cot^2 x not cos^2x=1+cot^2x

28. hartnn

csc=cosec

29. anonymous

oh yeah huh cosec is csc not cos i forgot for a minute lol

30. anonymous

so is it 1+ tan^2x/csc^2x or sec^2x/csc^2 and why

31. hartnn

what exactly is your question ? is this : (1+ tan^2x)/csc^2x

32. hartnn

(1+ tan^2x)=sec^2 x from the formula i gave u.

33. anonymous

no i saying is it my answer or yours and why (idk how to put it on my computer like you did so that was the best i could do)

34. hartnn

to tell u what is your answer, i need to know your exact question, that is why i asked whether it is: (1+ tan^2x)/csc^2x

35. anonymous

oh yeah it's that and i need to simplify it

36. hartnn

ok,for that i have shown u the steps, tell me which step, u didn't understand: $$\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x$$

37. anonymous

oh i get it now earlier you put it wierd so i didn't get it

38. anonymous

okay now this one is cot x tan x so it would be (cosx/sinx)(sinx/cosx) correct?

39. hartnn

yes, or u could use cot x = 1/ tanx and u get tan x / tan x =1

40. anonymous

wait because i have two of each would thay cancel out?

41. hartnn

yes, they would. sin x cos x / sin x cos x =1

42. anonymous

oh okay just checking

43. hartnn

|dw:1347864987907:dw|

44. hartnn

okk.

45. anonymous

how bout cot u sin u would that simplify to (cosu/sinu)sin u= cos u?

46. hartnn

thats absolutely correct! u are improving :)

47. anonymous

hooray improvement!

48. hartnn

what next ?

49. anonymous

1-cos^2 beta/sin beta and that would simplify to ooh okay i got nothin for this one :(

50. hartnn

is the numerator (1-cos^2 beta ) ??

51. anonymous

yes

52. hartnn

from sin^2 beta +cos^2 beta =1 u have 1-cos^2 beta = sin^2 beta so u have : $$\large \frac{sin^2 \beta}{sin \beta}=\frac{sin \beta.sin\beta}{sin \beta}=?$$

53. anonymous

sin beta?

54. hartnn

yup, but did u get all steps ?

55. anonymous

yeah so is that another function 1-cos^2 beta = sin^2 beta

56. hartnn

see the list i have given. it has sin^2 x + cos^2 x =1 isn't it ? from here i got 1-cos^2 x = sin^2 x clear ?

57. anonymous

oh so you just went backwards?

58. hartnn

whenever u see sin^2 x or cos^2 x try to use that formula for simplification.

59. anonymous

okay now i have cos x- cos^3 x

60. anonymous

like would that be cosx- (cosx)(cosx)/cosx?

61. hartnn

nopes. take cos x common from both the terms, what do u get ?

62. hartnn

cosx- (cosx)(cosx)(cosx)

63. anonymous

so you get 1

64. hartnn

no! after taking cos x common, u get cos x (1-(cos x)(cos x) ) =cos x (1- cos^2 x) and u already know what is (1- cos^2 x)

65. anonymous

sin^2x

66. hartnn

yes! so your final answer will be cos x sin^2x but did u get all the steps ?

67. anonymous

yeah because you have coxx-(cosx)(cosx)(cosx) =cosx-1(cosx)=1-cos^2x=sin^2x

68. hartnn

i will draw it :|dw:1347866550463:dw|

69. anonymous

oh okay then now my last question is sin^2 u + tan^2 u +cos^2 u/sec u

70. hartnn

is this entire thing (sin^2 u + tan^2 u +cos^2 u) in the numerator ?

71. anonymous

yes

72. hartnn

again using sin^2 x +cos^2 x =1, u have numerator as (1+tan^2 x) got this ? and (1+tan^2 x) = sec^2 x.....see formula list. so u have now sec^2x/sec x = ??

73. hartnn

replace all of my 'x' with 'u'

74. anonymous

oh yeah so now you have sec x!

75. hartnn

yup :)

76. hartnn

sec u ;)

77. anonymous

oh yeah lol thanks for your help i going to sleep now so goodnight^-^

78. hartnn

a very good night to u and have sweet dreams :)