## anonymous 3 years ago 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!

1. anonymous

2. anonymous

14. C 15.B

3. anonymous

what?

4. anonymous

does 15 d have an actual answer?

5. anonymous

i mean a lol

6. anonymous

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. anonymous

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. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

haha yes youre right. lol thank you!

13. anonymous

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

14. anonymous

mine or hero?

15. anonymous

yours

16. anonymous

14d: work from the right hand side

17. anonymous

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. anonymous

i just figured out 15a! thanks!

19. anonymous

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

20. anonymous

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. anonymous

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

22. anonymous

i dont understand why you did (1−cos2x)

23. anonymous

the identity of sin2x = 1-cos2x

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

okay! hold on!

29. anonymous

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

30. anonymous

alright, i will finish the solution

31. anonymous

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

32. anonymous

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

33. anonymous

sure you can! lemme check

34. anonymous

okay cool! awh thanks!

35. anonymous

hold onn!

36. anonymous

37. anonymous

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. anonymous

okay thank youuu

39. anonymous

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

40. anonymous

why is it sin2x−cosxsin2xcos2x

41. anonymous

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

42. anonymous

yeah so shouldnt it be cos2x

43. anonymous

hang on, I'll upload the solution

44. anonymous

45. anonymous

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

46. anonymous

loll yeah ill try haha

47. anonymous

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. anonymous

ook yeah try this

49. anonymous

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

50. anonymous

ugh ijust dont understand!!! im sorry :(

51. anonymous

Oh.. ok lets try that again:

52. anonymous

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

53. anonymous

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

54. anonymous

where exactly?

55. anonymous

at the sin2x - sin2xcos2x / cos2x

56. anonymous

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

57. anonymous

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

58. anonymous

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

59. anonymous

Ok i will post the solution from that point

60. anonymous

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

61. anonymous

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

62. anonymous

solve that..

63. anonymous

thanks @Hero !!!

64. anonymous

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

65. anonymous

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

66. anonymous

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

67. anonymous

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

68. anonymous

yeah i agree im just confused

69. anonymous

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