Double angle identies - Simplify the expression

[calc1]

I've lost my identities again!

I have mixed them up but don't see how to manipulate the expression.

Please help simplify:

cos2(x)/2sin(x)cos(x)

 

[Deleted member 4993]

I've lost my identities again!

I have mixed them up but don't see how to manipulate the expression.

Please help simplify:

cos2(x)/2sin(x)cos(x)

As written - following PEMDAS - your problem is:

\(\displaystyle \dfrac{cos2(x)}{2}*sin(x)*cos(x)\ \)

What do you want to do with that expression? Please be specific and share your work.

 

[calc1]

What do you want to do with that expression? Please be specific and share your work.

Okay. More parentheses. I am in a section of my course on double-angle identities. The only instruction for this exercise is: Simplify
I have tried using different variations of the double-angle formulas but don't see a way to develop the problem.

(cos²(x)-sin²(x))/sin2(x) . . ?

(cos²(x)-sin²(x))/(2sin(x)cos(x)) . . ?

(2cos²(x)-1)/sin2(x) . . ?

(2cos²(x)-1)/(2sin(x)cos(x)) . . ?

(1-2sin²(x))/sin2(x) . . ?

(1-2sin²(x))/(2sin(x)cos(x)) . . ?

It is possible that this problem is accidentally from a future section, Other Advanced Identities, that covers cofunction, power-reducing, and half-angle identities.

Original problem: cos2(x)/(2sin(x)cos(x))

 

[soroban]

Hello, calc1!

You are expected to know this identity: .\(\displaystyle \sin2\theta \,=\,2\sin\theta\cos\theta\)

\(\displaystyle \text{Simplify: }\:\dfrac{\cos2x}{2\sin x\cos x}\)

\(\displaystyle \displaystyle \frac{\cos2x}{2\sin x\cos x} \;=\;\frac{\cos2x}{\sin2x} \;=\;\cot2x\)

 

[calc1]

Thanks soroban. I looked back a my scribblings and of all the permutations, I missed that one! I tend to look for every problem to have a lengthy solution.