## natasha.aries Group Title Can someone PLEASE help me with 14d. and 15a. I have my pre-cal final, and I still have no idea how to do some ?s! one year ago one year ago

1. natasha.aries

2. caesar27

14. C 15.B

3. natasha.aries

what?

4. natasha.aries

does 15 d have an actual answer?

5. natasha.aries

i mean a lol

6. TweT226

tan^2(x) - sin^2(x) sin^2(x)/cos^2(x) - sin^2(x) [sin^2(x) - cos^2(x)sin^2(x)]/cos^2(x) sin^2(x)(1 - cos^2(x))/cos^2(x) sin^2(x)sin^2(x)/cos^2(x) tan^2(x)sin^2(x)

7. Hero

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

8. natasha.aries

but in this kind of problem both sides have to equal to one another, so in this casem @TweT226 is right. but why did you subtract it by 1?

9. natasha.aries

but thank you for all the work @Hero your work was beautiful!

10. natasha.aries

but the answer has to be equal to tan^2xsin^2x

11. natasha.aries

but that is the answer, tan^xsin^2x

12. natasha.aries

haha yes youre right. lol thank you!

13. natasha.aries

do u know how in his way why he subtracted by one though?

14. TweT226

mine or hero?

15. natasha.aries

yours

16. cesckay

14d: work from the right hand side

17. cesckay

15a. $\sin ( x - \Theta) = sinx \cos \theta - \sin \theta cosx$ then solve and put theta = pi $\cos \Pi = -1$ $\sin \Pi = 0$ Put that into the equation and obtain -sinx

18. natasha.aries

i just figured out 15a! thanks!

19. natasha.aries

do u know how to change something in polar form? like sqrt(3) + i?

20. cesckay

14d. i will solve this from the right hand side of the equation: $\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)$ $= \frac{ \sin^2x }{ \cos^2x } \times ( 1 - \cos^2x)$

21. cesckay

solve the equation and you will obtain the expression for the RHS

22. natasha.aries

i dont understand why you did (1−cos2x)

23. cesckay

the identity of sin2x = 1-cos2x

24. natasha.aries

so whenever you see that u change it to (1−cos2x)?

25. natasha.aries

cause i thought that only applied when you added sin2x + cos2x

26. cesckay

well you can determine what you will need to convert depending on whats on the other side of the equation. right??

27. cesckay

solve and let me know how it goes. OK :)

28. natasha.aries

okay! hold on!

29. natasha.aries

no it still didnt realy work out... here let me try again

30. cesckay

alright, i will finish the solution

31. natasha.aries

can i take a picture of what im doing so u can tell me what im doing wrong

32. cesckay

$\frac{ \sin^2x-\cos^xsin^2x }{ \cos^2x }$ $\frac{ \sin^2x }{ \cos^2x } - \frac{ \cos^2xsin^2x }{ \cos^2x }$ $\frac{ \sin^2x }{ \cos^2x }= \tan^2x$ and : $-\frac{ \cos^2xsin^2x }{ \cos^2x }= -\sin^2x$

33. cesckay

sure you can! lemme check

34. natasha.aries

okay cool! awh thanks!

35. natasha.aries

hold onn!

36. natasha.aries

37. cesckay

I worked on from the RHS.. its much easier to figure out from the RHS but I will try to do it from the LHS as well. it should be the same answer. In the meantime, you can check the solutions i put up

38. natasha.aries

okay thank youuu

39. natasha.aries

when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?

40. natasha.aries

why is it sin2x−cosxsin2xcos2x

41. cesckay

from the top, we can see that its just a multiplication of sin2X and whats in the bracket

42. natasha.aries

yeah so shouldnt it be cos2x

43. cesckay

hang on, I'll upload the solution

44. cesckay

45. cesckay

i know right.. :D got it from my computer.. mind reading it backwards?? :/

46. natasha.aries

loll yeah ill try haha

47. natasha.aries

or u could take a picture on ur phone or something and email it to yo self! and did u find a common denominator?

48. cesckay

ook yeah try this

49. cesckay

LaTex is horrible, i'm much of a "down on paper" breed//

50. natasha.aries

ugh ijust dont understand!!! im sorry :(

51. cesckay

Oh.. ok lets try that again:

52. cesckay

Solve from the Right Hand Side.. Did you get the solutions up to a certain point?

53. natasha.aries

yes, and each time i do it i stop at that point

54. cesckay

where exactly?

55. natasha.aries

at the sin2x - sin2xcos2x / cos2x

56. natasha.aries

idk what to d after that because i get tan2x- sin2xcos2x

57. cesckay

Oh there' the mistake right there.. you have to divide sin2x by Cos2x as well as divide by -sin2xcos2x by cos2x

58. natasha.aries

yeah but when i do it i dont have both of those cases thats all i have left

59. cesckay

Ok i will post the solution from that point

60. cesckay

$\frac{ \sin^2x-\sin^2xcos^2x }{ \cos^2x }$

61. cesckay

$\frac{ \sin^2x }{ \cos^2x }-\frac{ \sin^2xcos^2x }{ \cos^2x }$

62. cesckay

solve that..

63. natasha.aries

thanks @Hero !!!

64. natasha.aries

when i solve that yeah i get the answer but i dont see how u got that :(

65. natasha.aries

lol i agree with you but my teacher wont give me credit unless i do it that way

66. natasha.aries

she told us if we do it any other way we wont get credit

67. cesckay

You are supposed to get credits for either solving from LHS or RHS

68. natasha.aries

yeah i agree im just confused

69. natasha.aries

i mean i honestly dont mind this way, just this problem..