1. ## trigggg

1.Find the value of Cos 3π/5 + cos π/5.
2.According to cos55=a, find the value of (cos20) /2 as type of a.
Thanks for helping

2. ## Re: trigggg

Originally Posted by eylulderya
1.Find the value of Cos 3π/5 + cos π/5.
Do you not have a calculator? You are not going to be able to find and "exact" value for this so the best way to do it is just use a calculator.

2.According to cos55=a, find the value of (cos20) /2 as type of a.
Thanks for helping
20= 2(10) so cos(20)= cos(2(10)= cos^(10)- sin^2(10).

I used 10 instead of 20 because 10= 55- 45. cos(10)= cos(55)cos(45)+ sin(55)sin(45) and sin(10)= sin(55)cos(45)- cos(55)sin(45).

3. ## Re: trigggg

Not be able to find an exact value? I beg to differ HallsOfIvy...

Start with a regular pentagon. Since the angle sum of an "n" sided polygon is \displaystyle \begin{align*} \left( n - 2 \right) 180^{\circ} \end{align*}, that means the angle sum of a pentagon is \displaystyle \begin{align*} \left( 5 - 2 \right) 180^{\circ} = 3 \cdot 180^{\circ} = 540^{\circ} \end{align*}. Since it's a regular pentagon, all angles are equal, and so each angle is \displaystyle \begin{align*} \frac{540^{\circ}}{2} = 108^{\circ} \end{align*}.

Now join one vertex to all others, splitting the pentagon into three isosceles triangles. Since the left and right hand triangles are isosceles, the two remaining angles are equal. Thus each of those angles is \displaystyle \begin{align*} 36^{\circ} \end{align*}. Since the top vertex is \displaystyle \begin{align*} 108^{\circ} \end{align*}, that means the remaining unknown angle in it must be \displaystyle \begin{align*}36^{\circ} \end{align*}. Since the middle triangle is also isosceles, its remaining angles must be \displaystyle \begin{align*} 72^{\circ} \end{align*}.

If we call one of the side lengths of the pentagon \displaystyle \begin{align*} l \end{align*} and the drawn-in diagonals \displaystyle \begin{align*} a \end{align*}, then if look at the right hand triangle, then by the Cosine Rule we have

\displaystyle \begin{align*} l^2 &= a^2 + l^2 - 2\,a\,l \cos{ \left( 36^{\circ} \right) } \\ 0 &= a^2 - 2\,a\,l \cos{ \left( 36^{\circ} \right) } \\ 2\,a\,l \cos{ \left( 36^{\circ } \right) } &= a^2 \\ 2\,l \cos{ \left( 36^{\circ} \right) } &= a \end{align*}

So now we can focus on the middle triangle.

If we bisect one of the \displaystyle \begin{align*} 72^{\circ} \end{align*} angles, the smallest triangle is now another isosceles triangle (because its remaining angle must be \displaystyle \begin{align*} 72^{\circ} \end{align*}. Thus the length we have just drawn in is also \displaystyle \begin{align*} l \end{align*}. The larger triangle is also clearly isosceles, so its remaining length is also \displaystyle \begin{align*} l \end{align*} as shown in the diagram.

The only unknown length is the smallest length of the smallest triangle, but it is clearly equal to \displaystyle \begin{align*} a - l = 2\,l\cos{ \left( 36^{\circ} \right) } - l \end{align*}. Since the smallest triangle is similar to the largest, their corresponding sides are in proportion, thus

\displaystyle \begin{align*} \frac{2\,l\cos{ \left( 36^{\circ} \right) } - l }{l} &= \frac{l}{2\,l\cos{ \left( 36^{\circ} \right) }} \\ 2\,l\cos{ \left( 36^{\circ} \right) } \left[ 2\,l \cos{ \left( 36^{\circ} \right) } - l \right] &= l^2 \\ 4\,l^2 \cos^2{ \left( 36^{\circ} \right) } - 2\,l ^2 \cos{ \left( 36^{\circ} \right) } &= l^2 \\ 4\cos^2{ \left( 36^{\circ} \right) } - 2\cos{ \left( 36^{\circ} \right) } &= 1 \\ \cos^2{ \left( 36^{\circ} \right) } - \frac{1}{2} \cos{ \left( 36^{\circ} \right) } &= \frac{1}{4} \\ \cos^2{ \left( 36^{\circ} \right) } - \frac{1}{2} \cos{ \left( 36^{\circ} \right) } + \left( -\frac{1}{4} \right) ^2 &= \frac{1}{4} + \left( - \frac{1}{4} \right) ^2 \\ \left[ \cos{ \left( 36^{\circ} \right) } - \frac{1}{4} \right] ^2 &= \frac{5}{16} \\ \cos{ \left( 36^{\circ} \right) } - \frac{1}{4} &= \frac{\sqrt{5}}{4} \textrm{ clearly the positive solution is the only possible one...} \\ \cos{ \left( 36^{\circ } \right) } &= \frac{ 1 + \sqrt{5}}{4} \end{align*}

If written in radians, \displaystyle \begin{align*} \cos{ \left( \frac{\pi}{5} \right) } &= \frac{1 + \sqrt{5}}{4} \end{align*}.

Now since \displaystyle \begin{align*} \cos{ \left( 3\theta \right) } &\equiv 4\cos^3{ \left( \theta \right) } - 3\cos{ \left( \theta \right) } \end{align*} that means

\displaystyle \begin{align*} \cos{ \left( \frac{3\pi}{5} \right) } &= 4 \cos^3{ \left( \frac{\pi}{5} \right) } - 3\cos{ \left( \frac{\pi}{5} \right) } \\ &= 4 \left( \frac{1 + \sqrt{5}}{4} \right) ^3 - 3 \left( \frac{1 + \sqrt{5}}{4} \right) \end{align*}

and thus

\displaystyle \begin{align*} \cos{ \left( \frac{\pi}{5} \right) } + \cos{ \left( \frac{3\pi}{5} \right) } &= \frac{1 + \sqrt{5}}{4} + 4 \left( \frac{1 + \sqrt{5}}{4} \right) ^3 - 3 \left( \frac{1 + \sqrt{5}}{4} \right) \\ &= 4 \left( \frac{1 + \sqrt{5}}{4} \right) ^3 - 2 \left( \frac{1 + \sqrt{5}}{4} \right) \\ &= \frac{ \left( 1 + \sqrt{5} \right) ^3}{16} - \frac{1 + \sqrt{5}}{2} \end{align*}

which can probably be simplified more...

4. ## Re: trigggg

I'm impressed!

5. ## Re: trigggg

Just a side note: I read from an article that regular 3,5,17,257,65537-gon can be drawn with only compass and ruler because they are all Fermat Primes.
The reason for me bringing this up is that, it might be possible to find the exact value of cos(pi/17), cos(pi/257) and cos(pi/65537) then.