How to solve 1+cos2X/cotx?

Question

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

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Answer 1 · 2018-01-27T20:51:17

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

It is right

We will use

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:

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