You are here:
Calculus/identity
Advertisement
Question
verify the identity:
-sin2x+sin6x/sin2x-sin6x= tan4x cot2x
Try letting u=2x. The problem is then
(-sinu + sin3u)/(sinu-sin3u) = tan2u cotu.
On the left side, use the formulas
sin3u = sin2u cosu + sinu cos2u,
sin2u = 2sinucosu, and
cos2u = cosČu = sinČu.
On the right side, use the formula
tan2u = 2sinucosu/(cosČu-sinČu).
I suggest using c2 for cos2u, s2 for sin2u, c for cosu, and s=sinu.
Also, note that sČu sin sinČu and cČu is cosČu.
It really will become quite a long equation otherwise.
Related Articles
Answers by Expert:
Scotto
Expertise
Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.
Experience
Experience in the area: I have tutored students in all areas of mathematics for over 25 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Publications
Maybe not a publication, but I have respond to well oveer 7,500 questions on the PC.
Well over 2,000 of them have been in calculus.
Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors
I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients
My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.
©2012 About.com, a part of The New York Times Company. All rights reserved.