## anonymous one year ago Trig / Pre Cal/ identities Am I on the right path? Please guide me and do not give me the answers $(\tan x + \sin x)(1-cos x) = \sin^2 x \tan x$ Do I multiply like with foil starting with tan or do I turn tan into $$\frac{sin x}{cox x}$$ and then use foil?

1. Nnesha

foil! :-) i guess that will be great

2. anonymous

Just with tan $\tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x)$ correct so far?

3. Nnesha

yep looks right :-)

4. anonymous

Ok, how do I approch it from here? do I turn all the tan into sin / cos? and work it out? $\tan x -(\tan x)(\cos x) + \sin x -(\sin x)(\cos x)$

5. Nnesha

yes convert 2nd tan to sin /cos and let me find notebook and a pencil :3 hehe

6. Nnesha

make sure question is right :3 :-)

7. anonymous

Yes it is right :-)

8. Nnesha

is it equal to tanxsinx or tan^2xsinx ??

9. anonymous

I have $\tan x -\sin x + \sin x -(\sin x)(\cos x) = \sin^2 x \tan x$ $\tan x -(\sin x)(\cos x) = \sin^2 x \tan x$

10. Nnesha

yeah so for the final answer i got sinx tanx so that's why ..hm are u sure it's sin square ??

11. Nnesha

next step is to change tan to sin/cos

12. anonymous

Yes, it is sin^2 x tan x

13. Nnesha

:( alright

14. Nnesha

alright got it!! :-)

15. Nnesha

hahah i'm soo happy :P

16. anonymous

$$\tan x -(\sin x)(\cos x) = \sin^2 x \tan x$$ $$\frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x$$ ????

17. Nnesha

$$\huge\rm \frac{sin x}{cos x} -(\sin x)(\cos x) =$$ ???? find common denominator :-)

18. anonymous

Got ya :-)

19. anonymous

one sec

20. Nnesha

alright :-)

21. anonymous

I am using cos as the common denominator. that is correct?

22. Nnesha

yes right

23. anonymous

Ok I got $\frac{sin x}{cos x} -(\sin x)(\cos x) = \sin^2 x \tan x$ $\frac{(sin x)(cos x)}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x$ correct???

24. Nnesha

$\frac{(sin x)(\cancel{cos x})}{\cos x} -\frac{((\sin x))}{\cos x} \frac{\cos^2}{cos} = \sin^2 x \tan x$ multiply cosx/1 so when you multiply first fraction by cos both cos will cancel each other out

25. Nnesha

so there isn't supposed to be any cosx $\frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x$ like this

26. anonymous

Oh right!!! Made a boo boo

27. anonymous

One sec, now we should have

28. anonymous

$\frac{(sin x)}{\cos x} -\frac{((\sin x cos^2x))}{cos x} = \sin^2 x \tan x$ $\frac{(sin x)}{\cos x} -\frac{((\sin x (1-\sin^2 x)2x))}{cos x} = \sin^2 x \tan x$ Ok I am lost

29. anonymous

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

30. Nnesha

$\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos$ multiply |dw:1434231940649:dw|

31. Nnesha

$\cos x \times \frac{ \sin x }{ \cos x} - \frac{ sinx \cos x }{ \cos } \times \cos$ $\huge\rm \frac{ \sin x - sinx cosx (cosx) }{ cosx }$you will get this let me if that doesn't make sense

32. anonymous

Yes that makes sence

33. Nnesha

alright now multiply cos x times cos x = ?? $\huge\rm \frac{ \sin x - sinx \color{Red}{cosx (cosx)} }{ cosx }$you will get this let me if that doesn't make sense

34. anonymous

Was this originally $\frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x }$ $\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ 1 } \times \cos$ $\cos x \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos x }{ 1 } \times \cos$ Is that what we did there?

35. Nnesha

we know that common denominator is cos so we multiplied numerator by cos x

36. Nnesha

if the original fraction already have a same denominator as common denominator so u just have to carrry it down|dw:1434232700973:dw|

37. anonymous

If we did that we have $\cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos$ $\sin x - (\sin x)(\cos^2 x)$ ??

38. Nnesha

first of all we know that cosx is common denominator so that should be at the denominator $\frac{ ?????? }{ \cos x }$

39. Nnesha

after that we have to multiply both fraction by cos If we did that we have $\cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos$ $\large\rm \frac{?????????}{cosx }$ ??

40. Nnesha

wait a sec that's not cos^2 yet

41. Nnesha

let me fix it :-)

42. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nixy If we did that we have $\cos \times \frac{ \sin x }{ \cos x} - \frac{ \sin x \cos^2 x }{ \cos x } \times \cos$ $\sin x - (\sin x)(\cos^2 x)$ ?? $$\color{blue}{\text{End of Quote}}$$ we know common denominator is cosx now we have to find numerator $\frac{ ?????? }{ \cos x }$ original equation is $\huge\rm \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 }$ to find out numerator we should multiply both fraction by cosx(which is common denominator ) $\large\rm cosx \times \frac{ sinx }{ cosx } - \frac{ sinx \cos x }{ 1 } \times cosx$

43. Nnesha

and that's how $\large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 }$ that's the numerator part

44. anonymous

Ok, I think we are on the same page now WOW this stuff gets confusing lol Ok I am where you are at

45. Nnesha

what ??

46. Nnesha

do you got that denominator thingy ?

47. anonymous

So now we have $\large\rm\frac{ sinx }{{ 1} } - \frac{ sinx \cos^2 x }{ 1 }$

48. Nnesha

that's the numerator what's our common denominator ?

49. anonymous

From what section?? I am going from here $\large\rm \cancel{cosx} \times \frac{ sinx }{\cancel{ cosx} } - \frac{ sinx \cos^2 x }{ 1 }$

50. Nnesha

sinx/1 = sinx so you can write it as $\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }$

51. anonymous

Yes correct

52. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha sinx/1 = sinx so you can write it as $\huge\rm \frac{ sinx - sinxcos^2x }{ cosx }$ $$\color{blue}{\text{End of Quote}}$$ do you understand this step ??

53. Nnesha

want to make sure!!

54. anonymous

Yes I understand Next ? $\huge\rm \frac{ sinx - sinx(1-sin^2) }{ cosx }$

55. Nnesha

nope next step is to take out common term

56. Nnesha

$\huge\rm \frac{\color{ReD}{ sinx} -\color{ReD}{ sinx}cos^2x}{ cosx }$ factor the numerator :-)

57. Nnesha

let sinx = a cosx = b so you can write it as $\large\rm (a-ab^2)$

58. anonymous

So this $$(a-ab^2) = (ab^2-a) = (ab+a)(ab-a)$$ correct?

59. anonymous

I cannot believe this one problem is like this. It has been an hour and 30 minutes lol

60. Nnesha

nope what is common in these both terms a - ab^2 ??

61. Nnesha

almost done .-.

62. anonymous

so a(1-b^2)

63. Nnesha

yes right!!! so now change it back to sin cos

64. Nnesha

$\huge\rm \frac{\color{ReD}{ sinx}( 1-cos^2x)}{ cosx }$

65. anonymous

and that is the answer :-)

66. Nnesha

separate it $\huge\rm \frac{\color{ReD}{ sinx} }{ cosx } \times (1-cos^2x)$

67. anonymous

WOW, I cannot believe we got a problem like this. Some of these things take a long time.

68. Nnesha

yay! done!

69. Nnesha

practice more and more! then you will be able to finish it just in 5 minutes :3

70. anonymous

You work them one way then come to find out it is wrong then work them another and repeat and finally you get it lol This was the worst one yet!!! thank you for your help!!!!

71. anonymous

I do not have time to practice we have a week to fit a lot in and then we move on. Next week we probably will not even mess with these. idk

72. Nnesha

ohh but still try do 2 question every night :3 = nightmares haha jk

73. Nnesha

gO_Od luck!!!

74. anonymous

Thank you again!!!