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?

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- Aug 14th 2009, 05:44 AMchengbinEvaluate sin18
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? - Aug 14th 2009, 05:53 AMmr fantastic
Read this: Trigonometry--The derivation of Sin 18°

[tex]18^0[/tex] is easiest. - Aug 14th 2009, 06:01 AMchengbin
- Aug 14th 2009, 06:01 AMAmer
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

$\displaystyle \cos 36^o = 1-2\sin ^2 18^o$ - Aug 14th 2009, 06:31 AMOpalg
- Aug 15th 2009, 12:29 AMdhiab
**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}

$ - Aug 15th 2009, 03:27 AMpacman
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 - Aug 15th 2009, 04:47 AMluobo
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. - Aug 15th 2009, 05:37 AMpacman
**luobo**, nice insight. here is the graph of your equation . . . . - Aug 15th 2009, 05:39 AMpacman
dhiab, nice solution . . . . here is the graph of your equation, 4y^3 - 2y^2 - 3y + 1 = 0.