Double Angle Identity Mystery: Solving 2/(tanx+cotx)=sinx

[nzashadow]

Homework Statement

2/(tanx+cotx)=sinx

Homework Equations

Double Angle Identities, Pythagorean Identities

The Attempt at a Solution

2/(tanx+cotx)=

2/[(sinx/cosx)+(cosx/sinx)]=

2/[((sinx)^2+(cosx)^2)/(sinxcosx)]=

2sinxcosx/[(sinx)^2+(cosx)^2]=

2sinxcosx = sin(2x) =/= sinx

The book I'm getting this problem in says the answer is sinx, however I get sin(2x) which does not equal sinx. I am certain I am right, though I may also be making a stupid mistake :P

 

[L2E]

Your answer seems correct went over it a couple of times, It seems like an error with your textbook.

 

[Delta2]

Seems correct to me too, maybe the textbook has tan(x/2)+cot(x/2) instead?

 

[abhishek ghos]

you are right

 

[nzashadow]

nah it says tanx+cotx, went over it many times.

thanks guys