## anonymous one year ago 1 last trig problem

1. anonymous

2. jim_thompson5910

Try rewriting everything in terms of either sine or cosine

3. jim_thompson5910

eg: sec(x) = 1/cos(x)

4. anonymous

ahhhhh ok

5. anonymous

give me one sec

6. jim_thompson5910

also, when it comes to proving identities, you should only work on one side

7. jim_thompson5910

it might be easier to work on the right side

8. anonymous

$q \cot q/\cos^{-1} -\sin(\theta)/expsec/\csc ^2\frac{ f1/\sec}{ ? }$

9. anonymous

10. anonymous

one sec, let me try again

11. anonymous

ok I got my answer, but when I try to put it in the equation on this website it messes up

12. anonymous

so i started working on the right side

13. jim_thompson5910

After trying out the right side, it's somewhat more complicated than I thought. So I'm going to work on the left side instead $\Large \frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large {\color{red}{\frac{\sec(\beta)}{\sec(\beta)}*}}\frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-{\color{red}{\frac{\sin(\beta)}{\sin(\beta)}}}*\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ Take a few minutes and look all that over. Then let me know when you're ready for the next step.

14. anonymous

wait I think I need to prove the right side of the equation not the left

15. jim_thompson5910

you can start on either side

16. anonymous

ok i get it

17. jim_thompson5910

just as long as you only work with one side

18. anonymous

alright is that it

19. anonymous

@jim_thompson5910

20. jim_thompson5910

look at the "Pythagorean Identities" on page 2

21. jim_thompson5910

what is sec^2 equal to?

22. anonymous

tan^2(theta)+1

23. jim_thompson5910

and what is sin^2 equal to?

24. anonymous

1-cos^2(theta)

25. jim_thompson5910

So, $\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$ $\Large \frac{\tan^2(\beta)+1-(1-\cos^2(\beta))}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}$

26. jim_thompson5910

simplify the numerator for the denominator, use the fact that sin*sec = sin*(1/cos) = sin/cos = ???

27. anonymous

Ok I did it, can you show me what you got

28. jim_thompson5910

when you simplified the numerator, what did you get?

29. anonymous

i wish I could show you, trust me, but I cant

30. jim_thompson5910

draw it out

31. jim_thompson5910

or type what you can

32. anonymous

I got back to where we started,

33. jim_thompson5910

$\Large \tan^2(\beta)+1-(1-\cos^2(\beta))$ $\Large \tan^2(\beta)+1-1+\cos^2(\beta)$ which becomes what?

34. anonymous

what the other side of the equation is

35. jim_thompson5910

yeah ultimately the goal is to get the left side equal to the right side

36. anonymous

37. jim_thompson5910

what is sin(x)*sec(x) equal to?

38. jim_thompson5910

hint: look at "Tangent and Cotangent Identities" and "Reciprocal Identities" on that pdf

39. anonymous

ok I got it

40. anonymous

it is sinxsec

41. jim_thompson5910

sin(x)*sec(x) = sin(x)*(1/cos(x)) = sin(x)/cos(x) = ????

42. anonymous

tan

43. jim_thompson5910

yep

44. anonymous

alright thanks