## anonymous one year ago Emergency help on an Algebra II exam! Prove: sin θ - sin θ•cos2 θ = sin3 θ. You must show all work.

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1. nincompoop

it looks like this is an exam

2. anonymous

It's a practice exam, but it's worth a good bit of points and while I can understand the proofs written out in my lessons, this particular problem is stumping me. I've been working on these tests for hours :(

3. anonymous

So I know that identity of sin^2theta + cos^2theta = 1, or...something pretty close to that...

4. nincompoop

show me the attempts you have made.

5. anonymous

Okay. So the sin theta should cancel, right? cos^2 theta = sin^3 theta ****(that 3 is a cube, not sin(3))**** I don't know what to do with that!

6. nincompoop

how does it cancel?

7. anonymous

Bc it's just subtracted. Like 1 - 1 = 0.

8. anonymous

sin theta - sin theta = 0.

9. nincompoop

no

10. nincompoop

sin is multiplied to cos sin θ - (sin θ•cos2 θ) = sin3 θ ORDERS of operation

11. anonymous

Parentheses first, then

12. nincompoop

btw, is that $$cos^2 \theta$$ or $$cos (2 \theta)$$

13. anonymous

The first, it's squared.

14. anonymous

Okay, so...then do we distribute the negative into the parentheses??

15. anonymous

I'm thinking maybe we subtract sin theta from both sides to give us (sin theta * cos^2 theta) = sin^2 theta

16. nincompoop

is that $$sin^3 \theta$$ or $$sin (3 \theta)$$

17. anonymous

Cubed, the first one.

18. nincompoop

you need to start typing these things appropriately because they mean different things

19. anonymous

I'm sorry, I didn't realize it before I posted bc I just copy and pasted it to preserve the theta symbols :(

20. nincompoop

like this? PROVE $$sin ~\theta - (sin~ \theta ~cos^2 \theta) = sin^3 \theta$$

21. anonymous

There weren't any parentheses, but otherwise, yes.

22. nincompoop

so tell me what should be the best identity we can use?

23. anonymous

The only one I know of is the one I typed above, the sin^2 theta + cos^2 theta = 1.

24. anonymous

We didn't focus a lot on identities, so I suppose I may be forgetting some, but they aren't in my notes =/

25. nincompoop

what did you focus on?

26. nincompoop

let us use trigonometric expansion and make use of some algebra techniques like factoring, yes?

27. nincompoop

do you think you can factor the left-hand-side (LHS) of the equation for me?

28. anonymous

Sorry! Lost connection!

29. anonymous

I can tryyy, hold on

30. nincompoop

|dw:1440734256689:dw|

31. anonymous

sin theta(1 - cos^2 theta) = sin^3 theta

32. nincompoop

we can skip writing the theta for now and just add it in the end so it is easier, agreed?

33. anonymous

Okay! sin(1 - cos^2) = sin^3

34. nincompoop

do you have a note for your pythagorean identities? if not it is very simple $$\large sin^2 + cos^2 = 1$$ keep in mind for future reference that $$sin = y$$ and $$cos = x$$

35. nincompoop

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36. anonymous

Yep, I remember that part from the unit circle section we just did.

37. nincompoop

you probably already learned the Pythagorean Theorem which states that the length of the longest leg $$\large r$$ is equal to the square root of the sum of the two shorter legs |dw:1440734959343:dw|

38. nincompoop

do you think we can manipulate our identity algebraically so it would look familiar as the problem you have?

39. anonymous

Let me see...

40. anonymous

y(1 - x^2) = y^3 when you say cos = x and sin = y...

41. nincompoop

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42. anonymous

I'm sorry this is taking me so long to understand ._.

43. nincompoop

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44. nincompoop

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45. nincompoop

what part did I lose you?

46. nincompoop

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47. anonymous

Where did the sin^3 go?

48. nincompoop

that's just the left-hand-side (LHS) since the board is too small for me to fit everything in

49. anonymous

So it's 1^2 + sin^3 on the right hand?

50. nincompoop

no, the right-hand-side stays the same since it is what we're trying to accomplish

51. anonymous

...so then where is the sin^3 if you said it's not on the left hand? D:

52. nincompoop

what we are basically doing is trying to manipulate and re-arrange the left-hand-side of your problem so it would EQUAL to the right-hand side of the problem

53. nincompoop

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54. nincompoop

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55. anonymous

Okay, I'm mostly following

56. nincompoop

57. nincompoop

from the pythagorean identity, we derived that $$1-cos^2 = sin^2$$ because from our original identity (derived from the unit circle) $$sin^2 + cos^2 = 1$$ we can re-arrange it to $$sin^2 + cos^2 \color{red}{-cos^2} =1 \color{red}{-cos^2}$$ simplifying to: $$sin^2 = 1-cos^2$$

58. anonymous

I don't think any of this is getting through anymore, it's really late here and I'm thinking my exam has probably timed out by now ._. I'm tracking but I don't see where this is actually leading, we keep ending up back at this identity and I'm not sure what to do with it

59. nincompoop

then it would suffice to say that $$sin - sin~cos^2 = sin^3$$ because $$sin (1-cos^2) = sin^3$$ because the left-hand side is equivalent to $$sin (sin^2)$$ therefore $$sin^3 = sin^3$$

60. anonymous

I'll definitely reference this when I retake. Thank you so so so much for all of your time and help <3

61. anonymous

You were very patient, I hope it comes back to you tenfold :x

62. nincompoop

you're not even telling me what part you are not following

63. nincompoop

up to what part you are able to grasp and what part seems a little fuzzy still

64. anonymous

I'm sorry, it's just because I don't really know where I'm lost, I'm just really tired >.<

65. nincompoop

come back when your brain is fully rested. there is no point in studying when your brain is not working to its potential.

66. nincompoop

go to sleep

67. anonymous

Exactly :( I hope you have a good night, thanks so much for the guidance, I'm sure it'l help a lot more in the morning

68. anonymous

You, too!