## Rosy95 Group Title PreCal Help Please..I am so stuck on what to do one year ago one year ago

1. Rosy95 Group Title

Use the equation below to find the value of sin x. $\frac{ \tan x }{ \cot x } - \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }$

2. hartnn Group Title

write everything in terms of sin and cos

3. Rosy95 Group Title

That is what I am not understanding. I have a really hard time writting out what it is supposed to be once you start changing everything.

4. Hero Group Title

Hint: Write it this way. Then simply change everything to sin and cos $\tan(x) \div \cot(x) - \sec(x) \div \cos(x) = 2 \div \csc(x)$

5. Rosy95 Group Title

So how would you change it?

6. Hero Group Title

$\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)} - \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}$

7. Rosy95 Group Title

so could the first two fracts cancel out?

8. Hero Group Title

Not exactly

9. Hero Group Title

How do you divide: $\frac{1}{4} \div \frac{7}{6}$

10. Rosy95 Group Title

Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.

11. Hero Group Title

Answer the previous question. How do you divide that?

12. Rosy95 Group Title

Oh wait I remember

13. Rosy95 Group Title

you switch the #'s of the second fraction so it is now 6/7 and you change divide to multiply. you now have 1/4 X 6/7 which is 6/28 simplified to 3/14

14. Hero Group Title

Yes, so do that same thing with the trig equation

15. Rosy95 Group Title

so it is now $\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }$??

16. Hero Group Title

Yes, but do the rest of them. Change all of the divisions into multiplication.

17. Rosy95 Group Title

so it looks like$\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }$

18. Rosy95 Group Title

Did I do that right?

19. Rosy95 Group Title

If so, wouldn't 1/cosx times cosx/1 cancel out?

20. Hero Group Title

no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.

21. Rosy95 Group Title

oh duh ok so then it is $\frac{ \sin^2 x}{ \cos^2x } -\frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }$

22. Hero Group Title

Yes

23. Rosy95 Group Title

so do the mult fractions combine to get -1/cos^2x

24. Rosy95 Group Title

so then what do you do?

25. Hero Group Title

Yes, so what do you have after that?

26. Rosy95 Group Title

$\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }$

27. Hero Group Title

Okay, do you know how to combine fractions on the left side?

28. Rosy95 Group Title

couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side

29. Rosy95 Group Title

What do I do next?

30. Hero Group Title

On the left side, you have to combine fractions if they have the same denominator.

31. Hero Group Title

For example: $\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}$

32. Hero Group Title

The general rule for combining fractions with the same denominator is: $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$

33. Rosy95 Group Title

so it is now $\frac{ \sin^2x-1 }{ \cos^2x }=2sinx$

34. Rosy95 Group Title

or is the 2sin x supposed to be sin^2x

35. Hero Group Title

Very good

36. Hero Group Title

Now change the 1 to $$\sin^2x + \cos^2x$$

37. Hero Group Title

No, you have it right. 2sin(x)

38. Rosy95 Group Title

wait, you put sin^2x + cos^2x instead of one? what is the new equation?

39. Hero Group Title

Yes, you put $$\sin^2x + \cos^2x$$ in place of 1 because they are equal.

40. Hero Group Title

What do you get afterwards?

41. Rosy95 Group Title

$\frac{ \sin^2x-\sin^2x+\cos^2x }{ \cos^2x }=2sinx$ then you can subtract the sin^2x to get $\frac{ \cos^2x }{ \cos^2x }=2sinx$ then if thats the case you would just have 2sinx

42. Hero Group Title

Actually, if you do it correctly you get: $-\frac{ \cos^2x }{ \cos^2x }=2\sin x$

43. Hero Group Title

Which simplifies to $-1 = 2 \sin x$ Remember that we are still solving for x

44. Rosy95 Group Title

so sinx is -2

45. Hero Group Title

You were supposed to get this after substituting the 1: $\frac{ \sin^2x-(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x$ Then get : $\frac{ \sin^2x-\sin^2x-cos^2x }{ \cos^2x }=2\sin x$

46. Rosy95 Group Title

or no sinx would be -1/2

47. Hero Group Title

Ultimately you end up isolating the sin(x) on the right side: $-\frac{1}{2} = \sin x$

48. Hero Group Title

Now all you have to do is take the inverse sin of both sides to get $\sin^{-1}\left(-\frac{1}{2}\right) = x$

49. Hero Group Title

So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.

50. Rosy95 Group Title

-.52

51. Rosy95 Group Title

And thank you for your help :)

52. Hero Group Title

Well, in exact mode, you should have gotten $x = -\frac{\pi}{6}$

53. Hero Group Title

Nevertheless:$-\frac{\pi}{6} \approx -.52$