Question

Need help with trig identities

Answer 1

The question is in the attached file

Answer 2

hello anyone there?

Answer 3

well can you simplify for c, first ?

Answer 4

k one sec

Answer 5

Is there a difference between

Answer 6

one sec idk why my equation is working

Answer 7

isnt*

Answer 8

yeah, the tool of equation and draw are not working since 2 days ago

Answer 9

is there difference between tan^2 x and tan x^2

Answer 10

tan^2 (x) = (tan(x))^2

Answer 11

oh ok

Answer 12

I get (2-(1+tan^2(x))(1/2x))/((1+tan^2(x))(1/2x))) when i try to simplify

Answer 13

idk if you can actually make that out

Answer 14

the basicly, (a-b)/c = a/c - b/c , right ?

Answer 15

yup

Answer 16

so,
(2 - sec^2 (1/2 x))/sec^2 (1/2 x))
= 2/sec^2 (1/2 x)) - sec^2 (1/2 x))/sec^2 (1/2 x))
= 2/sec^2 (1/2 x)) - 1

agree ?

Answer 17

yes

Answer 18

ok, by using the identity of sec = 1/cos
now the equation above can be :

2/sec^2 (1/2 x)) - 1
= 2cos^2 (1/2 x)) - 1
ok ?

Answer 19

oh ok ic yes

Answer 20

then use this identity, that :
cos(2θ) = 2cos^2 (θ) - 1

Answer 21

lol u said it b4 me

Answer 22

ok ty i think i got it now

Answer 23

you're welcome :)