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find the value sin36.

Mathematics
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  • DLS
Refer this: http://www.goiit.com/posts/list/trignometry-find-value-of-sin-36-cos-18-1014971.htm
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sin(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

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