## 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 Group Title

2. caesar27 Group Title

14. C 15.B

3. natasha.aries Group Title

what?

4. natasha.aries Group Title

does 15 d have an actual answer?

5. natasha.aries Group Title

i mean a lol

6. TweT226 Group Title

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 Group Title

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

8. natasha.aries Group Title

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 Group Title

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

10. natasha.aries Group Title

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

11. natasha.aries Group Title

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

12. natasha.aries Group Title

haha yes youre right. lol thank you!

13. natasha.aries Group Title

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

14. TweT226 Group Title

mine or hero?

15. natasha.aries Group Title

yours

16. cesckay Group Title

14d: work from the right hand side

17. cesckay Group Title

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 Group Title

i just figured out 15a! thanks!

19. natasha.aries Group Title

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

20. cesckay Group Title

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 Group Title

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

22. natasha.aries Group Title

i dont understand why you did (1−cos2x)

23. cesckay Group Title

the identity of sin2x = 1-cos2x

24. natasha.aries Group Title

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

25. natasha.aries Group Title

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

26. cesckay Group Title

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

27. cesckay Group Title

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

28. natasha.aries Group Title

okay! hold on!

29. natasha.aries Group Title

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

30. cesckay Group Title

alright, i will finish the solution

31. natasha.aries Group Title

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

32. cesckay Group Title

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

33. cesckay Group Title

sure you can! lemme check

34. natasha.aries Group Title

okay cool! awh thanks!

35. natasha.aries Group Title

hold onn!

36. natasha.aries Group Title

37. cesckay Group Title

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 Group Title

okay thank youuu

39. natasha.aries Group Title

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

40. natasha.aries Group Title

why is it sin2x−cosxsin2xcos2x

41. cesckay Group Title

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

42. natasha.aries Group Title

yeah so shouldnt it be cos2x

43. cesckay Group Title

hang on, I'll upload the solution

44. cesckay Group Title

45. cesckay Group Title

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

46. natasha.aries Group Title

loll yeah ill try haha

47. natasha.aries Group Title

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 Group Title

ook yeah try this

49. cesckay Group Title

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

50. natasha.aries Group Title

ugh ijust dont understand!!! im sorry :(

51. cesckay Group Title

Oh.. ok lets try that again:

52. cesckay Group Title

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

53. natasha.aries Group Title

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

54. cesckay Group Title

where exactly?

55. natasha.aries Group Title

at the sin2x - sin2xcos2x / cos2x

56. natasha.aries Group Title

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

57. cesckay Group Title

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

58. natasha.aries Group Title

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

59. cesckay Group Title

Ok i will post the solution from that point

60. cesckay Group Title

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

61. cesckay Group Title

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

62. cesckay Group Title

solve that..

63. natasha.aries Group Title

thanks @Hero !!!

64. natasha.aries Group Title

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

65. natasha.aries Group Title

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

66. natasha.aries Group Title

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

67. cesckay Group Title

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

68. natasha.aries Group Title

yeah i agree im just confused

69. natasha.aries Group Title

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