Question

How to solve 1+cos2X/cotx?

`(1+cos2X)/cotX` I got 2cosxsinx but that doesn't sound right please help!

Answer 1

`(1+cos(2x))/cot(x)=2cos(x)*sin(x)=sin(2x)`

Explanation:

It is right

We will use

• `cot(x)=cos(x)/sin(x)`

• `cos(2x)=2cos^2(x)-1` *

Then

`LHS=(1+cos(2x))/cot(x)`

`=(1+(2cos^2(x)-1))/cot(x)`

`=(2cos^2(x))/cot(x)`

`=(2cos^2(x))/(cos(x)/sin(x))`

`=(2cos^2(x)*sin(x))/cos(x)`

`=2cos(x)*sin(x)`

`=RHS`

Which also can be expressed as `sin(2x)`

*
A little note why `cos(2x)=2cos^2(x)-1` is true

It can be shown from the trigonometric addiction formulas,
here we use:

`cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b)`

Several good proof of the trigonometric addiction formulas are available on the internet

A good intuition is given by the diagram:

![http://trigonography.com/2015/09/28/angle-sum-and-difference-for-sine-and-cosine/]()

And now the working out

We have the trigonometric formula

`cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b)`

Substitute `a=b`

`cos(a+a)=cos(a)*cos(a)-sin(a)*sin(a)`

`cos(2a)=cos^2(a)-sin^2(a)`

Use `sin^2(a)=1-cos^2(a)`

`cos(2a)=cos^2(a)-(1-cos^2(a))`

`cos(2a)=2cos^2(a)-1`

Which was what we wanted to show