# How do you prove #[1 - cos(x)]/[sin(x)] = tan(x/2)#?

##### 3 Answers

#### Explanation:

Take note:

Take note:

God bless....I hope the explanation is useful.

This is a similar process to the other answer,but hopefully this shows a more intuitive approach to determining in what way to manipulate the expressions,

Modifying the right-hand side only,

#tan(x/2)=sin(x/2)/cos(x/2)#

Using these two identities:

#=sqrt((1-cosx)/2)/sqrt((1+cosx)/2)=sqrt(((1-cosx)/2)/((1+cosx)/2))=sqrt((1-cosx)/2(2/(1+cosx)))#

#=sqrt((1-cosx)/(1+cosx))#

At this point, knowing what to do could become confusing. However, whenever we have denominators split into two parts, it's always a good idea to think about conjugates -- especially since

#=sqrt(((1-cosx)(1-cosx))/((1+cosx)(1-cosx)))=sqrt(((1-cosx)^2)/sin^2x)#

Both squares are cancelled by the square root:

#=(1-cosx)/sinx#

Here is a solution that does not ignore the sign question in the identities for

#### Explanation:

Substitute: Let

Our task now is to

show that

Obviously, if we can show that (1-cos2y)/(sin2y) = sinxy/cosy#, hen we can finish.

Let's rewrite the left side using double angle formulas. But wait!. There are 3 ways of rewriting

It might be too much to keep in our head, so let's write down the choices. Then we'll think about what the next steps would look like and try to make a selection.

OR

OR

Thinking about the next steps, it looks like the middle way might work out best.

# = (2sin^2y)/(2sinycosy)#

# = (cancel(2)cancel(siny)siny)/(cancel(2)cancel(siny)cosy)#

# = siny/cosy#

# = tany#

**It is not necessary to change the variable** when we write up the proof, but I think the notation is cleaner and easier to read.

Here's how it looks without changing the variable.

# = (1-(1-2sin^2(x/2)))/(2sin(x/2)cos(x/2))#

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

# = (cancel(2)cancel(sin(x/2))sin(x/2))/(cancel(2)cancel(sin(x/2))cos(x/2))#

# = sin(x/2)/cos(x/2)#

# = tan(x/2)#