A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0Refer this: http://www.goiit.com/posts/list/trignometryfindvalueofsin36cos181014971.htm

mayankdevnani
 2 years ago
Best ResponseYou've already chosen the best response.1sin(5x) = 16sin^5(x)  20sin^3(x) + 5sin(x). With x = 36°, we see that: sin[5(36°)] = 16sin^5(36°)  20sin^3(36°) + 5sin(36°) ==> 16sin^5(36°)  20sin^3(36°) + 5sin(36°) = sin(180°) ==> 16sin^5(36°)  20sin^3(36°) + 5sin(36°) = 0, since sin(180°) = 0. Then, if we let u = sin(36°), the equation becomes: 16u^5  20u^3 + 5u = 0. By factoring: 16u^5  20u^3 + 5u = 0 ==> u(16u^4  20u^2 + 5) = 0 ==> u = 0 and 16u^4  20u^2 + 5 = 0. However, since sin(36°) > 0 since 36° lies in Quadrant I, u = 0 is discarded. We are now left to solve: 16u^4  20u^2 + 5 = 0. (Note that this is just a quadratic equation in u^2) By the Quadratic Formula: (b^2  4ac = (20)^2  4(16)(5) = 80): u^2 = [b ± √(b^2  4ac)]/(2a) = (20 ± √80)/32 = (20 ± 4√5)/32 = (5 ± √5)/8. By taking the positive square root (again, u = sin(36°) > 0): u = √[(5 ± √5)/8] = √(10 ± 2√5)/4. At this point, we need to decide which sign we should be pick. To do this, we note that: sin^2(36°) < sin^2(45°) = 1/2. Since the positive square root gives us sin^2(36°) > 1/2 as: (5 + √5)/8 > (5  1)/8 = 1/2, we pick the negative sign. Therefore, sin(36°) = √(10  2√5)/4. http://answers.yahoo.com/question/index?qid=20110212184922AAbvfnJ
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.