I need to calculate the exact value of sin18. How do I do that? I see that $\displaystyle \frac {90}{5}=18$, so I must use some form of the addition theorems. What do I do?
BTW, how do I type the degree symbol in latex?
Read this: Trigonometry--The derivation of Sin 18°
[tex]18^0[/tex] is easiest.
see this link Trigonometry Angles--Pi/5 -- from Wolfram MathWorld
in the link above you can find $\displaystyle \cos 36^o$
but
$\displaystyle \cos 2x = 1-2\sin ^2 x $ so
$\displaystyle \cos 36^o = 1-2\sin ^2 18^o$
Hello This the solution :
$\displaystyle \begin{array}{l}
let:x = 18^\circ \\
5x = 90^\circ \\
3x + 2x = 90^\circ \\
3x = 90^\circ - 2x \\
\sin 3x = \cos 2x \\
\end{array}$
I'have this equation :
$\displaystyle \begin{array}{l}
3\sin x - 4\sin ^3 x = 1 - 2\sin ^2 x \\
let:\sin x = y \\
4y^3 - 2y^2 - 3y + 1 = 0 \\
\left( {y - 1} \right)\left( {4y^2 + 2y - 1} \right) = 0 \\
\end{array}
$
I'deduct:
$\displaystyle \begin{array}{l}
y_1 = 1...NO \\
y_2 = \frac{{ - 1 - \sqrt 5 }}{2} < 0...NO \\
y_3 = \frac{{ - 1 + \sqrt 5 }}{2} = \sin 18^\circ \\
\end{array}
$
sin 18 = cos 72. we have solved this already at http://www.mathhelpforum.com/math-he...tml#post344761
Working out in degrees, we have
Let x =72 degrees, multiply both sides by 5;
5x = 360
2x + 3x = 360
2x = 360-3x
Then, taking cos of both side
cos 2x = cos(360 -3x)
cos 2x = cos 3x
Using an identity for double angle and triple angle, equating them we have
2cos^2 x -1= 4cos^3 x - 3 cos x
4cos^3 x - 2cos^2 x - 3 cos x +1=0
4cos^3 x - 4cos^2 x + 2 cos^2 x - 2 cos x - cos x +1=0
4cos^2x(cos x -1) +2cos x( cos x -1) -1( cos x - 1) =0
(cos x - 1) (4cos^x +2 cos x - 1) =0
Either cos x =0 means x=90 degree
Solve (4cos^x +2 cos x - 1) =0
cos x = [-2 +(plus or minus){sq rt (4+16)] /8
cos x = [-2 +(plus or minus){sq rt (20)] /8
cos x = [-2 +2(plus or minus){sq rt (5)] /8
cos x = [-1 +(plus or minus){sq rt (5)] /4
Neglecting NEGATIVE values of cos x, because x lies in first quadrant
Therefore cos 72 degrees = (sqrt(5) -1)/4
cos 72 = 0.30901699437494742410229341718 . . . . = sin 18
Check this post.
http://www.mathhelpforum.com/math-he...tml#post350000
$\displaystyle \sin54^\circ-\sin18^\circ=\cos36^\circ-\sin18^\circ=1-2\sin^2 18^\circ-\sin18^\circ=\frac{1}{2}$
It is a quadratic equation.