Question

find the value sin36.

Answer 1

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

Answer 2

ok

Answer 3

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

Answer 4

ok @sunnymony