anonymous one year ago how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1 @zepdrix

1. zepdrix

$\Large\rm \frac{2}{1+\cos x}-\tan^2\left(\frac{x}{2}\right)=1$So ... we have some cosine business, ya? :) I guess we want the Tangent Half-Angle Identity that involves sines and cosines. I cant remember what they look like, I'mma have to look em up

2. zepdrix

$\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}}$Do you see how this one might end up being really useful for us? It gives us a common denominator, that might be helpful at some point.

3. zepdrix

No no no we want the other one actually :) AHHH sorry sine on the bottom is going to be a lot more useful. But, I'm thinking ahead, this can be a tough problem to try and work out on your own.

4. zepdrix

So using our identity,$\Large\rm \color{orangered}{\tan\left(\frac{x}{2}\right)=\frac{1-\cos x}{\sin x}}$notice that in our problem, the tangent is being squared,$\Large\rm \color{orangered}{\tan^2\left(\frac{x}{2}\right)=\left(\frac{1-\cos x}{\sin x}\right)^2}$This is what we're going to replace our tangent with ^

5. anonymous

Ooh ok

6. zepdrix

So our problem,$\Large\rm \frac{2}{1+\cos x}-\color{orangered}{\tan^2\left(\frac{x}{2}\right)}=1$Becomes,$\Large\rm \frac{2}{1+\cos x}-\color{orangered}{\left(\frac{1-\cos x}{\sin x}\right)^2}=1$We'll write it like this:$\Large\rm \frac{2}{1+\cos x}-\frac{(1-\cos x)^2}{\sin^2 x}=1$ya? :o

7. anonymous

alright so can you make a common denominator ?

8. zepdrix

Yes and we'll use this idea of "difference of squares" to achieve that. Do you remember what that would look like? We're trying to do something like this:$\Large\rm (1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x$We multiply the denominator by it's conjugate to get these squares.

9. anonymous

Ohh ok. yes!

10. zepdrix

so looks like we need to give that first fraction a (1-cos x) on top and bottom.

11. zepdrix

$\large\rm \left(\frac{1-\cos x}{1-\cos x}\right)\frac{2}{1+\cos x}-\frac{(1-\cos x)^2}{\sin^2 x}=1$

12. anonymous

$\frac{ 2- 2cosx }{ (1-cosx)^2 }$ ?

13. zepdrix

Top looks ok.

14. anonymous

Oops wait lol $\frac{ 2- 2cosx }{ 1- \cos^2x }$

15. zepdrix

Ah there we go :)

16. anonymous

I feel dumb lol

17. zepdrix

And then our Pythagorean Identity lets us go one step further, ya?

18. anonymous

yup !

19. zepdrix

$\Large\rm \frac{2-2\cos x}{\sin^2x}-\frac{(1-\cos x)^2}{\sin^2 x}=1$So we have,$\Large\rm \frac{2-2\cos x-(1-\cos x)^2}{\sin^2x}=1$Ok with that last step? They have the same denominator, so I combined them into a single fraction.

20. anonymous

mmm I see! just one thing so will it be: $\frac{ 2 - 2cosx +1 -\cos^2x }{ \sin^2x }$ I hope I didn't make another silly mistake =_= lol

21. zepdrix

Ooo ya we did something silly there,$\Large\rm (1-\cos x)^2\ne 1^2-\cos^2x$

22. zepdrix

gotta FOIL that out.

23. zepdrix

$\Large\rm (1-\cos x)^2=(1-\cos x)(1-\cos x)=?$Should end up with more than just two terms when you do :)

24. anonymous

Oooh lol I was taking it as $(1- cosx) (1 +cosx)$

25. zepdrix

:3

26. anonymous

$1 - 2\cos x + \cos^2x$

27. anonymous

T_T

28. zepdrix

So,$\Large\rm \frac{2-2\cos x-(1-\cos x)^2}{\sin^2x}=1$Becomes,$\Large\rm \frac{2-2\cos x-(1-2\cos x+\cos^2x)}{\sin^2x}=1$Ok good.

29. anonymous

waaaait lemme do it

30. zepdrix

Distribute the negative, cancel some stuff out, combine like-terms, if you're able, what are you left with on top? :)

31. zepdrix

my bad :3 stealing too much of the fun lol

32. anonymous

$\frac{ 2 - 2cosx - 1 + 2cosx - \cos^2x }{ \sin^2x }$ $\frac{ 1- \cos^2x }{ sin^2x }$ = 1

33. anonymous

Yasssss *___*

34. zepdrix

Pythagorean Identity on top one more time? Oooo yay you did it \c:/ wooo good job!

35. anonymous

you made this into a cheesecake LOL thank you! xD

36. zepdrix

np, ooo cheese cake sounds delic +_+

37. anonymous

$\frac{2}{1+cosx}=1+\tan^2\frac{x}{2}=\sec^2\frac{x}{2}=\frac{1}{\cos^2\frac{x}{2}}$$2\cos^2\frac{x}{2}=1+cosx$

38. anonymous

Which is the same

39. zepdrix

Yah that's pretty much the approach infinity was using, probably a lot easier to do it that way I suppose :)

40. anonymous

@Nishant_Garg oh cool! I didn't think of that

41. anonymous

:) When I see 1 and tan^2(x) it just automatically connects for me

42. xapproachesinfinity

@Nishant_Garg same here i like to make it sec whne i see 1+tan i did the same approach with this question in a different post