## Rosy95 2 years ago PreCal Help Please..I am so stuck on what to do

1. Rosy95

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

write everything in terms of sin and cos

3. Rosy95

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

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

So how would you change it?

6. Hero

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

7. Rosy95

so could the first two fracts cancel out?

8. Hero

Not exactly

9. Hero

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

10. Rosy95

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

11. Hero

Answer the previous question. How do you divide that?

12. Rosy95

Oh wait I remember

13. Rosy95

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

Yes, so do that same thing with the trig equation

15. Rosy95

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

16. Hero

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

17. Rosy95

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

18. Rosy95

Did I do that right?

19. Rosy95

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

20. Hero

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

21. Rosy95

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

Yes

23. Rosy95

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

24. Rosy95

so then what do you do?

25. Hero

Yes, so what do you have after that?

26. Rosy95

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

27. Hero

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

28. Rosy95

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

29. Rosy95

What do I do next?

30. Hero

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

31. Hero

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

32. Hero

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

33. Rosy95

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

34. Rosy95

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

35. Hero

Very good

36. Hero

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

37. Hero

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

38. Rosy95

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

39. Hero

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

40. Hero

What do you get afterwards?

41. Rosy95

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

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

43. Hero

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

44. Rosy95

so sinx is -2

45. Hero

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

or no sinx would be -1/2

47. Hero

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

48. Hero

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

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

50. Rosy95

-.52

51. Rosy95

And thank you for your help :)

52. Hero

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

53. Hero

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