# Thread: Finding the exact value of sin 3

1. ## Finding the exact value of sin 3

Question says if...

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

then find the exact value of sin 3

How would one go about doing this?

Question says if...

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

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.

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

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

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