Question

verify each identity.

Answer 1

hello

Answer 2

hi, i have to put the picture up first of the problem

Answer 3

@daphne101

Answer 4

@Directrix

Answer 5

You should post the half angle formula for sine Then, let's see what you can do for the first step after that, okay?

Answer 6

\[\frac{ 1-\cos \alpha }{ 2 }\]

Answer 7

@Directrix

Answer 8

What you have is in a square root, with a \(\pm\) before it and is equal to \(\sin \dfrac{\alpha}{2} \).

Answer 9

yes it's supposed to be qith a square root

Answer 10

Since the first thing on the numerator is 1 and you need to end up with tan x, multiply the fraction by tan x/tan x. Keep i mind that tan x = sin x/cos x

Answer 11

can you write it for me or draw its hard for me to understand it written out

Answer 12

Since you have sin^2 x/2, then the right side becomes only the fraction you wrote without the square root.

Answer 13

\(\sin^2 \dfrac{t}{2} = \dfrac{\tan t - \sin t}{2\tan t}\)

Answer 14

We use the identity:

\(\sin \dfrac{\alpha}{2} = \pm \sqrt{\dfrac{1 - \cos \alpha}{2}} \)

Answer 15

what next?

Answer 16

We replace the left side of your problem with the square of the right of the identity.

\(\left( \pm \sqrt{\dfrac{1 - \cos t}{2}} \right)^2 = \dfrac{\tan t - \sin t}{2\tan t} \)

Answer 17

You follow so far?

Answer 18

yes

Answer 19

Since the left side is the square of a square root, just drop the square, the square root, and the plus/minus.

Answer 20

\({\dfrac{1 - \cos t}{2}} = \dfrac{\tan t - \sin t}{2\tan t} \)

Ok?

Answer 21

i'm not sure if i follow that