# Math Help - numerical identity

1. ## numerical identity

prove that cos 36 - sin 18 = 1/2

2. 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.

3. Originally Posted by pacman
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.
Actually, we do have a geometric solution!! But I don't know how to post picture in the forum.
I try to state it in a abstract way(if you are really interested in it ,try to draw the graph according to my statement..)
Let $\triangle ABC$ be a triangle, $\angle ABC=\angle ACB=72^{\circ}, \angle BAC=36^{\circ}$, $D\in AB,\angle DCB=36^{\circ},BC=1,AB=AC=x$,we connect $C,D$.
It is easy to know $\triangle ABC$~ $\triangle CDB,AD=CD=BC$. Thus $\frac{AB}{BC}=\frac{BC}{BD}\Longleftrightarrow x=\frac{1}{BD}$, so $BD=\frac{1}{x}$. So $AD=AB-BD=x-\frac{1}{x}=BC=1$.Thus $x-\frac{1}{x}=1$.
Solve this equation, we get $x=\frac{1+\sqrt{5}}{2}$. But it is also obvious that $\frac{\frac{1}{2}BC}{AB}=\cos 72^{\circ}=\sin 18^{\circ}$, so we can get $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$!!!
Then all the things can be solved..

4. prove that cos 36 - sin 18 = 1/2
since $cos2A=1-sin^2(A)$

$cos36=1-2sin^2(18)$

Therefore,

$1-2sin^2(18)-sin(18)$
$=1-2((5)^{1/2} - 1)^2/16 - ((5)^{1/2}-1)/4$
$=1-(5-2(5)^{1/2}+1)/8 - ((5)^{1/2})/4 + 1/4$
$=1/2$

use the fact that sin18={sqrt(5)-1}/4

5. bandedkrait, your solution is concise and simplier than mine. Thanks