Exact value

what is the exact value of sin (pi/5) ?

Formula:

\(\begin{array}{|rcll|} \hline \sin(5\varphi) = 16 \sin^5(\varphi) - 20 \sin^3(\varphi) + 5\sin(\varphi) \\ \hline \end{array}\)

Ansatz:

\(\small{ \begin{array}{|lrcl|} \hline (1): & \sin(360^{\circ}) = 0 = \sin(5\cdot 72^{\circ}) = 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ})\\ & 16 \sin^5(72^{\circ}) - 20 \sin^3(72^{\circ}) + 5\sin(72^{\circ}) &=& 0 \quad & | \quad : \sin(72^{\circ}) \\ & 16 \sin^4(72^{\circ}) - 20 \sin^2(72^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(72^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\\\ (2): & \sin(180^{\circ}) = 0 = \sin(5\cdot 36^{\circ}) = 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) \\ & 16 \sin^5(36^{\circ}) - 20 \sin^3(36^{\circ}) + 5\sin(36^{\circ}) &=& 0 \quad & | \quad : \sin(36^{\circ}) \\ & 16 \sin^4(36^{\circ}) - 20 \sin^2(36^{\circ}) + 5 &=& 0 \quad &| \quad x = \sin^2(36^{\circ}) \\ & 16 x^2 - 20 x + 5 &=& 0 \\ \hline \end{array} }\)

The quadratic polynomial \(16 x^2 - 20 x + 5\) has following roots: \(<~ \sin^2(72^{\circ}),\ \sin^2(36^{\circ}) ~>\)

The general quadratic equation is: \( {\displaystyle ax^{2}+bx+c=0.}\)

The quadratic formula is: \({\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}\)

\(\begin{array}{|rcll|} \hline x &=& \displaystyle { \frac {20\pm {\sqrt {20^{2}-4\cdot 16 \cdot 5}}}{2\cdot 16}} \\\\ &=& \frac {20\pm {\sqrt { 400 - 320 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 80 }}} {32} \\\\ &=& \frac {20\pm {\sqrt { 16\cdot 5 }}} {32} \\\\ &=& \frac {20\pm 4 {\sqrt { 5 } }} {32} \\\\ &=& \frac {5\pm {\sqrt { 5 } }} {8} \\\\ \sqrt{x} &=& \sqrt{ \frac {5\pm {\sqrt { 5 } }} {8} } \\ \hline \end{array} \)

Because \(\sin(72^{\circ}) > \sin(36^{\circ})\)
we have:

\(\begin{array}{|rcll|} \hline \sin(72^{\circ}) = \sqrt{ \frac {5 + {\sqrt { 5 } }} {8} } \\ \sin(36^{\circ}) = \sqrt{ \frac {5 - {\sqrt { 5 } }} {8} } \\ \hline \end{array}\)

Again Lisa, you need to use brackets better

(sin pi)/5 = 0

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sin(pi/5)

Well wolfram alpha giives an exact value here.

http://www.wolframalpha.com/input/?i=sin(pi%2F5)

Not sure how to work that out.........

sin(Pi rad) /5 = sin(180 deg) /5 =0

sin (Pi/5 rad) = sin(180/5 deg) = sin(36 deg) = 0.587785252292

Be exact with using brackets, they are important!

\(\sqrt{\dfrac{5}{8}-\dfrac{\sqrt5}{8}}\)??

it equals 

\(\sqrt{\left(\dfrac{\sqrt5-1}{2}\right)\left(\dfrac{\sqrt5}{4}\right)}\)

and the golden ratio = (sqrt5 - 1)/2, and golden ratio is often related to pentagons.

So maybe we can work that exact value out of pentagons? :)

Thank you all very much for your help guys! But I must solve it using the half- angle formulas. Sorry I forgot to mention that, I was in a rush. Also with correct brackets it is sin (pi/5)

Also the problem says that given sin (pi/10) =( -1 + sqrt 5)/4, find sin(pi/5)

Thank you all very much for your help guys! But I must solve it using the half- angle formulas. Sorry I forgot to mention that, I was in a rush. Also with correct brackets it is sin (pi/5)

Also the problem says that given \(\sin ( \frac{\pi } {10}) = \frac{( -1 + \sqrt{ 5}) }{4}\), find \(\sin ( \frac{\pi } {5})\)

Formula:

\(\begin{array}{|rcll|} \hline \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \cos{\varphi} \\ \cos{\varphi} &=& \sqrt{1-\sin^2{\varphi}} \\\\ \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \sqrt{1-\sin^2{\varphi}} \\ \hline \end{array}\)

We set:

\(\begin{array}{|rcll|} \hline \sin{2\varphi} &=& \sin( \frac{\pi } {5})\\\\ \sin{\varphi} &=& \sin( \frac{\pi } {10})\\ &=& \frac{ \sqrt{ 5}-1 }{4} \\ \hline \end{array} \)

We have:

\(\begin{array}{|rcll|} \hline \sin(2\varphi) &=& 2\cdot \sin{\varphi} \cdot \sqrt{1-\sin^2{\varphi}} \\ \sin( \frac{\pi } {5}) &=& 2\cdot \frac{ \sqrt{ 5}-1 }{4} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (\sqrt{5}-1)^2 }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (5-2\sqrt{5}+1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- 5+2\sqrt{5}-1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{10+2\sqrt{5} }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2 }{4} \cdot \frac{(10+2\sqrt{5}) }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (5-2\sqrt{5}+1)\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 60+12\sqrt{5}-20\sqrt{5}-4\cdot 5 } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 40-8\sqrt{5} } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 8\cdot 5-8\sqrt{5} } {8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{8\cdot 5}{8\cdot 8} -\frac{8\sqrt{5}}{8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{5}{8} -\frac{\sqrt{5}}{8} }\\ \hline \end{array} \)

Please Lisa can you be careful to include ALL the information in your original question AND be careful to include all neccessasry brackets as well as any extra brackets that will help people to understand your question properly in the first place.

Otherwise people waste a lot of their time and next tme they may be less likely to offer you any help at all.

Thankyou everyone for your great answers :)

The same answer.

Using the same formula: \(\sin(\varphi) = 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} } \)

Given \(\sin ( \frac{\pi } {10}) = \frac{( -1 + \sqrt{ 5}) }{4} \)

Find \(\sin( \frac{\pi } {5})\)

Formula:
\(\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \cos{ \frac{\varphi}{2} } \\ \cos{\frac{\varphi}{2}} &=& \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\\\ \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \hline \end{array}\)

We set:

\(\begin{array}{|rcll|} \hline \sin{\varphi} &=& \sin( \frac{\pi } {5})\\\\ \sin{\frac{ \varphi } {2} } &=& \sin( \frac{\pi } {10})\\ &=& \frac{ \sqrt{ 5}-1 }{4} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sin(\varphi) &=& 2\cdot \sin{ \frac{\varphi}{2} } \cdot \sqrt{1-\sin^2{\frac{\varphi}{2}}} \\ \sin( \frac{\pi } {5}) &=& 2\cdot \frac{ \sqrt{ 5}-1 }{4} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{1- \left(\frac{ \sqrt{ 5}-1 }{4}\right)^2 } \\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (\sqrt{5}-1)^2 }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- (5-2\sqrt{5}+1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{16- 5+2\sqrt{5}-1) }{16} }\\ \sin( \frac{\pi } {5}) &=& \frac{ \sqrt{ 5}-1 }{2} \cdot \sqrt{\frac{10+2\sqrt{5} }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2 }{4} \cdot \frac{(10+2\sqrt{5}) }{16} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (\sqrt{5}-1)^2\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ (5-2\sqrt{5}+1)\cdot (10+2\sqrt{5}) } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 60+12\sqrt{5}-20\sqrt{5}-4\cdot 5 } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 40-8\sqrt{5} } {64} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{ \frac{ 8\cdot 5-8\sqrt{5} } {8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{8\cdot 5}{8\cdot 8} -\frac{8\sqrt{5}}{8\cdot 8} }\\ \sin( \frac{\pi } {5}) &=& \sqrt{\frac{5}{8} -\frac{\sqrt{5}}{8} }\\ \hline \end{array} \)