## Atkinsoha Group Title PreCalc & Trig Help! Simplify the expression using a sum or difference formula: sin(x+y)sec x sec y one year ago one year ago

1. Atkinsoha Group Title

^opps, show all steps too please.

2. satellite73 Group Title

$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$ then multiply out

3. Atkinsoha Group Title

how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.

4. satellite73 Group Title

you can look in your text for a proof of that result, but that is not what you are being asked to do. you are being asked to multiply out, the point of the exercise it to use that identity, not to derive it

5. satellite73 Group Title

in other words, you are being asked a question to see if you know the identity multiplying out is now a matter of algebra

6. Atkinsoha Group Title

I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..

7. satellite73 Group Title

oh i have confused you that is not the answer at all that is the identity $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$

8. Atkinsoha Group Title

Yes, you have confused me..

9. satellite73 Group Title

now your job is to multiply $\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)$\]

10. Atkinsoha Group Title

Maybe I should tell you this.. I've been gone from school for a few weeks now (death in the family) I have no idea how to do any of this..

11. satellite73 Group Title

you get $\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)$ as a first step

12. Atkinsoha Group Title

okay.. then were the heck do you go?

13. satellite73 Group Title

that was multiplying out using the distributive law then $$\sec(x)=\frac{1}{\cos(x)}$$ so you can cancel some stuff

14. satellite73 Group Title

lets look at the first term $\sin(x)\cos(y)\sec(x)\sec(y)$ that is equal to $\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}$

15. satellite73 Group Title

cancel the $$\cos(y)$$ top and bottom and get $\frac{\sin(x)}{\cos(x)}$ which you can rewrite as $\tan(x)$

16. Atkinsoha Group Title

okay. that makes some sense..

17. Atkinsoha Group Title

basically the same thing for the other side too then?

18. satellite73 Group Title

did you know $$\sec(x)=\frac{1}{\cos(x)}$$? this is really algebra at this step

19. satellite73 Group Title

yes, same thing for the second term

20. satellite73 Group Title

only this time you will get $$\tan(y)$$

21. Atkinsoha Group Title

Yes I did. Just didn't understand the problem.. so you get down to tan x=tanx?

22. Atkinsoha Group Title

ohh, yeah. tan x + tan y

23. satellite73 Group Title

that is it, yes

24. Atkinsoha Group Title

25. satellite73 Group Title

26. Atkinsoha Group Title

Okay, that makes sense. Do you have time to help me with one more problem?

27. satellite73 Group Title

sure

28. Atkinsoha Group Title

okay, let me take a screen shot.

29. satellite73 Group Title

k

30. Atkinsoha Group Title

31. Atkinsoha Group Title

I kind of understand it.. but not really :/

32. satellite73 Group Title

ok we can do this, but it is going to take a minute ready?

33. Atkinsoha Group Title

Yes.

34. satellite73 Group Title

we are going to use the same formula that we used above $\sin(u+v)=\sin(u)\cos(v)+\cos(u)\sin(v)$ at the moment, we know only two of those four numbers

35. Atkinsoha Group Title

yes, we know sin u and cos v

36. satellite73 Group Title

we need $$cos(u)$$ do you know how to find it?

37. Atkinsoha Group Title

Don't you have to make a triangle or something?

38. satellite73 Group Title

yes exactly

39. Atkinsoha Group Title

Okay, I'm not sure exactly how to come about finding it on a triangle

40. satellite73 Group Title

|dw:1362154017344:dw|

41. satellite73 Group Title

third side via pythagoras $\sqrt{61^2-11^2}=60$

42. Atkinsoha Group Title

Wouldn't it be 50?

43. satellite73 Group Title

no $\cos(u)=\frac{60}{61}$

44. satellite73 Group Title

45. Atkinsoha Group Title

Okay. so we still ned to find the sin one right?

46. satellite73 Group Title

right but actually i made a mistake $\cos(u)=-\frac{60}{61}$ because you are in quadrant 2

47. satellite73 Group Title

now we need $$\sin(v)$$ and it is the same idea as before |dw:1362154367573:dw|

48. satellite73 Group Title

the third side is $\sqrt{41^2-40^2}=9$so $$\sin(v)=\frac{9}{41}$$

49. satellite73 Group Title

now plug the numbers directly in to the formula

50. Atkinsoha Group Title

okay, then you just solve it right? multiplying and adding fractions basically?

51. satellite73 Group Title

$\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{61}\times(- \frac{60}{61})$

52. satellite73 Group Title

yeah it is arithmetic from here on in gotta run, good luck

53. Atkinsoha Group Title

thank you!

54. Atkinsoha Group Title

wait quick

55. satellite73 Group Title

56. Atkinsoha Group Title

for multiplying you flip the second fraction?

57. satellite73 Group Title

$\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{41}\times(- \frac{60}{61})$

58. satellite73 Group Title

oh no

59. satellite73 Group Title

flip nothing multiply straight across

60. Atkinsoha Group Title

okay. ill give it a try... thanks!

61. satellite73 Group Title

yw