1. D.E.V Project
Pre-Calculus 40s
Submitted by:
Mark R.

2. Question #1
Trigonometric Identity

3. Prove this using Trigonometric Identity:
1 − cos( x) 1 + cos( x)
sin( x) sin( x)
+ = +
1 − cos( x) 1 + cos( x)
sin( x) sin( x)

4. The very first thing that you need to do when proving an identity is to
draw the Great Wall of China.
1 − cos( x) 1 + cos( x)
sin( x) sin( x)
+ = +
1 − cos( x) 1 + cos( x)
sin( x) sin( x)

5. Then we are going to work with either the equation on the left
hand side or the right hand side in this case I decided to work
with the right hand side.
NOTE: Most of the time it is better to pick the
most complicated side.
1 + cos( x) sin( x)
+
1 + cos( x)
sin( x)
Since both terms have a different denominator we have to multiply each
term to a term that is equivalent to 1 in order to achieve a common
denominator.
For the first term in the right hand side it would look like this:
1 + cos( x)  1 + cos( x)   1 + cos( x) 
=  
 sin( x)   1 + cos( x) 
sin( x)

6. ( 1 + cos( x) )
2
1 + cos( x)
=
sin( x)(1 + cos( x))
sin( x)
For the second term on the right hand side we would have
something like this:
 sin( x)   sin( x) 
sin( x)
=  
1 + cos( x)  1 + cos( x)   sin( x) 
2
sin( x) sin ( x)
=
1 + cos( x) sin( x)(1 + cos( x))

7. Now, since both terms on the right hand side have the same
denominator we can write them in to a single term. That looks like this:
( 1 + cos( x) )
2
+ sin ( x) 2
sin( x)(1 + cos( x))
Then, simplify both the numerator and
denominator.
1 + 2 cos( x) + cos ( x) + sin ( x)
2 2
sin( x) + sin( x) cos( x)

8. Can you see the Pythagorean Identity on
the numerator?
cos ( x) + sin ( x) = 1
2 2
1 + 2 cos( x) + cos ( x) + sin ( x)
2 2
sin( x) + sin( x) cos( x)
Then just substitute the value of the
Pythagorean identity and simplify.
1 + 2 cos( x) + 1
sin( x) + sin( x) cos( x)
2 + 2 cos( x)
sin( x) + sin( x) cos( x)

9. Factor out (1+cos(x)) on both the numerator and the
denominator and then reduce.
2(1 + cos( x))
sin( x)(1 + cos( x))
2
sin( x)
Can you simplify the term from above using one
of the fundamental identities?
1
= csc( x)
sin( x)
2 csc( x)

10. Now we are finish the right hand side lets go do the left hand
side.
1 − cos( x)
sin( x)
+
1 − cos( x) sin( x)
Same as the right hand side both terms have a different
denominator so we have to multiply each term to a term that is
equivalent to 1 in order to achieve a common denominator.
For the first term in the right hand side it would look like this:
 sin( x)   sin( x) 
sin( x)
=  
1 − cos( x)  1 − cos( x)   sin( x) 
sin 2 ( x)
sin( x)
=
1 − cos( x) 1 − cos( x)(sin( x))

11. For the second term we would have something like this:
1 − cos( x )  1 − cos( x )   1 − cos( x ) 
=  
 sin( x )   1 − cos( x ) 
sin( x )
1 − cos( x ) (1 − cos( x )) 2
=
sin( x )(1 − cos( x ))
sin( x )
Now, since both terms on the left hand side have
the same denominator we can write them in to a
single term. That looks like this:
sin ( x) + (1 − cos( x))
2 2
sin( x)(1 − cos( x))

12. Simplify the numerator
sin 2 ( x) + 1 − 2 cos( x) + cos 2 ( x)
sin( x)(1 − cos( x))
Now can you see the Pythagorean Identity on the
numerator? Plug in the value of the identity and
then simplify.
cos ( x) + sin ( x) = 1
2 2
1 + 1 − 2 cos( x)
sin( x)(1 − cos( x))
2 − 2 cos( x)
sin( x)(1 − cos( x))

13. Now lets bring over the term from the right hand side and
then factor out (1-cos(x)) on the numerator of the term
from the left hand side.
2(1 − cos( x))
= 2 csc( x)
sin( x)(1 − cos( x))
Simplify, and notice one of the fundamental
Identities.
1
2 = csc( x)
= 2 csc( x) sin( x)
sin( x)
2 csc( x) = 2 csc( x)
Q.E.D Always put Q.E.D to indicate you are
done.