Question

How do i prove:
sin(theta)+sin(2theta)/cos(2theta)+cos(theta)+1=tan(theta)

Answer 1

use these identities :
sin(2θ) = 2sin(θ)cos(θ) and
cos(2θ) = 2cos^2 (θ) - 1

Answer 2

okayy. i believe i got it!

Answer 3

so,
(sin(θ)+sin(2θ))/(cos(2θ)+cos(θ)+1)
= (sin(θ) + 2sinθcosθ)/(2cos^2 θ - 1 + cosθ+1)
= (sin(θ) + 2sinθcosθ)/(2cos^2 θ + cosθ)

factor out the numerator and the denominator by take the commons, get
(sin(θ) + 2sinθcosθ)/(2cos^2 θ + cosθ)
= sin(θ) (1+ cosθ)/(cosθ(2cosθ+1))
cancel the (1+cosθ)'s, get
= sin(θ)/cos(θ)
= tan(θ)

Answer 4

i meant the last 3 is
sin(θ) (1+ 2cosθ)/(cosθ(2cosθ+1))
so, cancel the (1+ 2cosθ)

Answer 5

okay. great! thats what i got! thank you. i was wondering if you could also help me with....
prove:
sec x - sin x cot x=sin^2x/cos x

Answer 6

i know we need to change the sec x into (1/cos x) and cot x into (cos x/sin x) but after that, im not sure what to do...

Answer 7

yeah, that's good starting to prove it.

Answer 8

sec x - sin x cot x
= 1/cosx - sin * cosx/sinx (cancel the sin's)
= 1/cosx - cosx ?

Answer 9

that is what i get but how do we get that to sin^2 x?

Answer 10

1/cosx - cosx
= 1/cosx - (cosx * cosx/cosx)
= 1/cosx - cos^2 (x)/cosx
see the denominator's already be same, so just adding the numerator. get
= 1-cos^2 (x) / cosx

Answer 11

finally, are you familiar with this identity :
sin^2 (x) + cos^2 (x) = 1 ?

Answer 12

= 1/cosx - (cosx * cosx/cosx) can you explain this?

Answer 13

and yes i am.

Answer 14

i just want the cos(x) be a fraction, it just a trick, look that
cos(x) = cos(x)/1
if you times the numerator and the denominator by cos(x), get
cos(x) = cos(x)/1 * cosx/cosx = cos(x)/cos^2 x, right ?

Answer 15

damn,.. that's backward, i meant cos^2 x/cosx

Answer 16

okay. i understand.

Answer 17

thank you so much! i really appreciate it!

Answer 18

you're welcome :)