Successfully reported this slideshow.
Upcoming SlideShare
×

# DEV

637 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### DEV

1. 1. D.E.V Project Pre-Calculus 40s Submitted by: Mark R.
2. 2. Question #1 Trigonometric Identity
3. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.