Math Help - Finding the exact value of sin 3

1. Finding the exact value of sin 3

Question says if...

$cos 72 = \frac{\sqrt5 - 1}{4}$

then find the exact value of sin 3

How would one go about doing this?

2. Originally Posted by username11111
Question says if...

$cos 72 = \frac{\sqrt5 - 1}{4}$

then find the exact value of sin 3

How would one go about doing this?
Do you mean sin 72?

And I assume angles are being measured in degrees?

3. Originally Posted by mr fantastic
Do you mean sin 72?

And I assume angles are being measured in degrees?
No, it's cos 72.

...and yes, it's degrees.

4. Originally Posted by username11111
No, it's cos 72.

...and yes, it's degrees.
*Sigh* Didn't you notice what I highlighted in red? It means that I asked do you mean sin 72 rather than sin 3.

5. Originally Posted by mr fantastic
*Sigh* Didn't you notice what I highlighted in red? It means that I asked do you mean sin 72 rather than sin 3.
No. You're given the exact value of cos 72 and asked to find the exact value of sin 3.

6. Re: Finding the exact value of sin 3

Originally Posted by mr fantastic
Do you mean sin 72?

And I assume angles are being measured in degrees?
It realy is cos 72.

cos 72 = sin 18, and sin 18 = (51/2-1)/4

7. Re: Finding the exact value of sin 3

sin 3 = sin 18 - 15 = sin 18 cos 15 - cos 18 sin 15

8. Re: Finding the exact value of sin 3

Start with \displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } = \sin{ \left( 90^{\circ} - 72^{\circ} \right) } = \sin{\left(18^{\circ}\right)} \end{align*}, so \displaystyle \begin{align*} \sin{\left(18^{\circ}\right)} = \frac{\sqrt{5} - 1}{4} \end{align*}. We then have

\displaystyle \begin{align*} \cos{\left(18^{\circ}\right)} &= \sqrt{ 1 - \left[ \sin{ \left( 18^{\circ} \right) } \right]^2} \\ &= \sqrt{1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2} \\ &= \sqrt{ 1 - \frac{5 - 2\sqrt{5} + 1}{16} } \\ &= \sqrt{ 1 - \frac{6 - 2\sqrt{5}}{16} } \\ &= \sqrt{ \frac{16 - \left(6-2\sqrt{5}\right)}{16} } \\ &= \frac{\sqrt{10 + 2\sqrt{5}}}{4} \end{align*}.

Now note that

\displaystyle \begin{align*} \sin{\left(3^{\circ}\right)} &= \sin{\left(18^{\circ} - 15^{\circ}\right)} \\ &= \sin{\left(18^{\circ}\right)}\cos{\left(15^{\circ} \right)} - \cos{\left(18^{\circ}\right)}\sin{\left(15^{\circ} \right)} \end{align*}

To evaluate \displaystyle \begin{align*} \sin{\left(15^{\circ}\right)} \end{align*} and \displaystyle \begin{align*} \cos{\left(15^{\circ}\right)} \end{align*}, apply the half angle formula to \displaystyle \begin{align*} \sin{\left(30^{\circ}\right)} \end{align*} and \displaystyle \begin{align*} \cos{\left(30^{\circ}\right)} \end{align*}. Good luck

9. Re: Finding the exact value of sin 3

cos15=cos(60-45)=cos60cos45+sin60sin45=(1/2)(1/root2)+(root3/2)(1/root2)=(1+root3)/(2root2) whiich simplifies to (root2+root6)/4
Can do similar for sin15

10. Re: Finding the exact value of sin 3

sin 18 = (51/2-1)/4
sin 15 = sin 45-30 = sin 45 cos 30 - cos 45 sin 30
cos 18..
cos 15

11. Re: Finding the exact value of sin 3

The final awnser looks like this:
sin 3 degrees =
( sqrt( 3 - sqrt(5) + ( 3sqrt(3) - sqrt(15) )/2 ) - sqrt( 5 + sqrt(5) - ( 5sqrt(3) + sqrt(15) )/2 ) )/4

12. Re: Finding the exact value of sin 3

Originally Posted by DanvanVuu
The final awnser looks like this:
sin 3 degrees =
( sqrt( 3 - sqrt(5) + ( 3sqrt(3) - sqrt(15) )/2 ) - sqrt( 5 + sqrt(5) - ( 5sqrt(3) + sqrt(15) )/2 ) )/4
Here it is from latex.codecogs:

[IMG]http://latex.codecogs.com/png.latex?\displaystyle%20\begin{align*}\frac{\sqr t{3%20-%20\sqrt{%205%20}%20+%20\frac{\left%28%203%20sqrt{ 3}%20-%20sqrt{15}%20\right%29}{2}}%20-%20\sqrt{5%20+%20\sqrt{%205%20}%20-%20\frac{\left%28%205%20sqrt{3}%20+%20sqrt{15}%20\ right%29}{2}}}{4}\\%20\end{align*}[/IMG]

13. Re: Finding the exact value of sin 3

Write 72=45+(30-3) and apply the formulas for the cosine and the sine of a sum of two angles. This will give you an equation involving sin(3) and cos(3), to which you apply the Pythagorean theorem to express cos(3) in terms of sin(3); this will result in a second-order equation in sin(3), whose positive solution you seek.