ynj, thanks.
i did not know that it has a geometric solution, i failed to scan the forum.
i do it this way, though very long.
prove that cos 36 - sin 18 = 1/2. I will evaluate the left hand side of the equation; since cos 36 = sin 54, then
sin 54 - sin 18 = sin (36 + 18) + sin 18; using the addition law for sine
[sin (A + B) = (sin A)(cos B) - (cos A)(sin B);
(sin 36)(cos 18) - (cos 36)(sin 18) - sin 18 = ?
Ah, i will use the double-angle identity
sin 2A = 2(sin A)(cos A) and cos 2A = 1 - 2 sin^2 A.
And also cos^2 A = 1 - sin^2 A.
We have,
2(sin 18)(cos 18)(cos 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?
2(sin 18)(cos^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?
2(sin 18)(1 - sin^2 18) + (1 - 2 sin^2 18)(sin 18) - sin 18 = ?
2(sin 18) - 2(sin^3 18) +
sin 18 - 2(sin^3 18) -
sin 18 = ?
2(sin 18)
- 2(sin^3 18) - 2(sin^3 18) = ?
2(sin 18) - 4(sin^3 18) = ?
from
http://www.mathhelpforum.com/math-he...tml#post350585
sin 18 = (5^(1/2) - 1)/4, substituting to the above expression and
expanding it [2(sin 18) - 4(sin^3 18)];
2[(5^(1/2) - 1)/
4] - 4[(5^(1/2) - 1)/4]^3 = ?
[(5^(1/2) - 1)/2] -
4[(5^(3/2)- 3(5)(1) + 3(5)^(1/2) - 1)/
4^3] = ?
(5^(1/2) - 1)/2 - [
5(5^(1/2) - 15 +
3(5)^(1/2) - 1)/
4^2] = ?
(5^(1/2) - 1)/2 - [
8(5^(1/2) -
16)/
16] = ?
[(5^(1/2)]/2 - 1/2 -
[(5^(1/2)]/2 + 1 = ?
-1/2 + 1 = ?
Thus, cos 36 - sin 18 = 1/2.