Simplify: (cos2x+1)/sin2x?

Question

Answer must be in terms of one trigonometric function.

Answer 1

2cos^2x-1+1/2sinxcosx
=2cos^2x/2sinxcosx
=cosx/sinx
=cotx

Answer 2

9

Answer 3

there is a double angle formula someplace that you should find. i think sin 2x = 2sin x cos x or something like that, then you should be able to simplify

Answer 4

Do you mean cos^2x and sin^2x? There is a trigonometric identity that says: sin^2x+cos^2x=1 Therefore cos^2x=1-sin^2x cos^2x+1-sin^2x = cos^2x+cos^2x = 2cos^2x Answer: 2cos^2x

Answer 5

Reminder:

cos2x = [cos(x)]^2 - [sin(x)]^2

(Note: I'm writing it like that instead of cos^2(x) to avoid confusion. I know on paper it's written like cos^2(x) but they are essentially the same thing).

Anyway, sin2x = 2sin(x)cos(x)

So we just substitute:

[(cos(x))^2 - (sin(x))^2 - 1]/[2*sin(x)cos(x)]

[(cos(x))^2 - ( -1 * ((1 - sin(x))^2)/[2*sin(x)cos(x)]

(cos(x))^2 - (-1 * (cos(x))^2) / [2 * sin(x) cos(x)]
2(cos(x))^2 / [2 * sin(x) cos(x)]

2 cos(x)/2 sin(x)
cos(x)/sin(x)
cot(x)

Answer 6

sure.
remind me later, 'mkay?