Question

simple trig identity!

prove
sinx/1+cosx= 1/sinx - 1/tan x

Answer 1

so i think you write not right equal

Answer 2

you should do this on your own. it will help you learn about manipulation of trigonometric identities.

Answer 3

honestly, i would if i knew how. i have 2 assignments of an online functions course left to do and once i'm done, i never have to do math again, so all i really need to do is finish them. i've done what i can on my own and now i'm asking for help.

Answer 4

Note
\[ 1-\cos^2x = \sin^2x \]

And think about what you can multiply the top and bottom of the LHS (or RHS) to use this fact.

Answer 5

tan(x)= sin(x)/cos(x)

Answer 6

My method is more interesting ¬_¬

Answer 7

\[a^2 - b^2 = (a+b)(a-b) \]

May also be useful.

Answer 8

I like Newton's method

Answer 9

this is all just confusing me more. it's so much easier to understand when i see a solution.

Answer 10

Sorry, I can't do your problem for you. Ask one of the other people.

Answer 11

so far i took the right side and got 1/sinx-cosx/sinx
which = 1-cosx/sinx

but i dont know if that is correct, or what to do next

Answer 12

That is correct, and useful. How can you get from that to the LHS?

Answer 13

that's what i don't know.

Answer 14

i some how need to flip it and chance the - to a +

Answer 15

Not technically true. Because on the LHS, the sin(X) is on the top, but on the RHS, the sin(x) is on the bottom. See my above advice.

Answer 16

but there are no ^2 in this proof

Answer 17

OK.

So if you take your LHS and RHS in the new form you have.

\[\frac{\sin x}{1+\cos x} = \frac{1-\cos x}{\sin x} \]

Multiplying both sides by (1+cosx)(sinx) gives a result that is clearly true. (s^2 - 1-c^2)

HOWEVER it is best to go directly from LHS -> RHS (or vice versa) and not assume the result and work from that. So can you see how something similar could get from one side to the otehr directly?