Trigonometry Identity

[fiir]

I'm mathematically retarded, so please excuse my stupidity and tell me how to prove this one: sin2x/1-cos2x = cotx

Thanks .

 

[Deleted member 4993]

I'm mathematically retarded, so please excuse my stupidity and tell me how to prove this one: sin2x/1-cos2x = cotx

Thanks .

First of all - you need to write your problem correctly by grouping operations with parentheses.It should be written as:

sin2x/(1-cos2x) = cotx

Use

cos(2x) = 1- 2sin[sup:2htzp82u]2[/sup:2htzp82u]x

and

sin(2x) = 2 sin(x) cos(x)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[fiir]

What I've been doing so far:

sin2x/ (1-cos2x) = cotx

Left-hand side: sin2x/(1-cos2x) => 2sinxcosx/-2sin^2x= -cotx

Right hand side: cotx

: (

 

[soroban]

Hello, fiir!

You have an extra minus-sign . . .

What I've been doing so far:

\(\displaystyle \frac{\sin2x}{1-\cos2x} \:=\: \cot x\)

Left-hand side: .\(\displaystyle \frac{\sin2x}{1-\cos2x}\quad\Rightarrow \quad \frac{2\sin x\cos x}{-2\sin^2\!x} \:=\: -\cot x\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \uparrow\)
Right hand side: .\(\displaystyle \cot x\)

\(\displaystyle \text{The identity we use is: }\;\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 1 - \cos2\theta \:=\:2\sin^2\!\theta\)