Find $\sin{\frac{\pi}{10}}$ in form of $a+b\sqrt{k}$

Question

Use De Moivre's Theorem to express $\cos{5\theta}, \sin{5\theta}$ in powers of $\sin{\theta}$ and $\cos{\theta}$.

Show that $$\frac{\cos{5\theta}}{\cos{\theta}}=16\sin^4{\theta}-12\sin^2{\theta}+1$$

Find the solutions of $$16\sin^4{\theta}-12\sin^2{\theta}+1=0,\ for\ 0\le\theta\le\frac{\pi}{2}$$

Use this result to find $\sin{\frac{\pi}{10}}$ in the form $a+b\sqrt{k}$, and give the values of $a,\ b\ and\ k$.

For $\cos{5\theta}$, I have $\cos5{\theta}=\cos^5{\theta}-10\cos^2{\theta}\sin^3{\theta}+5\cos{\theta}\sin^4{\theta}$

and for $\sin{5\theta}$, I have $\sin{5\theta}=5\cos^4{\theta}\sin{\theta}-10\cos^2{\theta}\sin^3{\theta}+\sin^5{\theta}$

Then it's easy to get $\frac{\cos{5\theta}}{\cos{\theta}}=16\sin^4{\theta}-12\sin^2{\theta}+1$.

So on, $\frac{\cos{5\theta}}{\cos{\theta}}=16\sin^4{\theta}-12\sin^2{\theta}+1=0$

Since $\cos{\theta}\ne0$, so $\cos{5\theta}=0$

Then $$5\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\frac{9\pi}{2},$$

$$\therefore \theta=\frac{\pi}{10},\frac{3\pi}{10},\frac{\pi}{2},\frac{7\pi}{10},\frac{9\pi}{10}$$

Since $\cos{\theta}\ne0, \therefore \theta\ne\frac{\pi}{2}, and\ also\ 0\le\theta\le\frac{\pi}{2}$

$$\theta=\frac{\pi}{10}, \frac{3\pi}{10}$$

Then I used the quadratic formula to get $$\sin^2\frac{\pi}{10}=\frac{3\pm\sqrt{5}}{8}$$

But this is wrong somehow. Can anyone give me some hints to finish this question?

Answer 1

You got a correct expression, but it's not what the question asked for. $$\sin^2 \frac{\pi}{10} = \frac{3-\sqrt{5}}{8} \tag{1}$$ is correct (we discard the other root $\frac{3+\sqrt{5}}{8}$ because it is greater than $\sin^2 \frac{\pi}{4} = \frac{1}{2}$). But the question asked for $\sin \frac{\pi}{10}$ in the form $a + b \sqrt{k}$.

To this end, note that $$3-\sqrt{5} = \frac{6-2\sqrt{5}}{2} = \frac{(\sqrt{5})^2 - 2\sqrt{5} + 1}{2} = \frac{(\sqrt{5} - 1)^2}{2}. \tag{2}$$ Hence $$\sin \frac{\pi}{10} = \sqrt{\frac{3 - \sqrt{5}}{8}} = \sqrt{\frac{(\sqrt{5}-1)^2}{16}} = \frac{\sqrt{5} - 1}{4}. \tag{3}$$ This yields $a = -1/4$, $b = 1/4$, $k = 5$.

Answer 2

For another approach to solve for $x := \sin(\pi/10)$ in $16 x^4 - 12 x^2 + 1 = 0$, let us try completing the square in a slightly unusual way: instead of using the first two terms, we will use the outer two terms and observe that $(4x^2 - 1)^2 = 16x^4 - 8x^2 + 1$. Therefore, $$16x^4 - 12x^2 + 1 = (4x^2 - 1)^2 - 4x^2 = (4x^2 - 2x - 1) (4x^2 + 2x - 1).$$ From here, it should be easy to find the roots of the polynomial, and then you just need to determine which one of those roots is actually equal to $\sin(\pi/10)$.

Answer 3

OP’s answer is correct! It can be completed by just taken square root by letting $$ \frac{3-\sqrt{5}}{8}=(a+b \sqrt{5})^2=a^2+5b^2+2ab\sqrt 5 $$ By comparing the rational and irrational parts on both sides, we have $$ \left\{\begin{array}{l} a^2+5 b^2=\frac{3}{8} \cdots(1)\\ 2 a b=-\frac{1}{8} \cdots(2) \end{array}\right. $$ $(1)+(2)\times 3$ gives $$ a^2+5 b^2=-6 a b \Leftrightarrow a=-b \text { or }-5 b \textrm{ (rejected after checking)} $$ Putting back to $(2)$ gives $a=-\frac{1}{4}$ and $b=\frac{1}{4}$ and hence $$ \boxed{\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}} $$

Alternative method

Let $\theta=\frac{\pi}{10} $, then $$ \begin{aligned} & \cos 2 \theta=\sin 3 \theta \\ \Leftrightarrow \quad & 1-2 \sin ^2 \theta=3 \sin \theta-4 \sin ^3 \theta \\ \Leftrightarrow \quad & 4 x^3-2 x^2-3 x+1=0 \cdots (*), \quad \textrm{ where }x=\sin \theta. \end{aligned} $$ Noting that $x=1$ is a root of $(*)$, we can factorise it as $$ (x-1)\left(4 x^2+2 x-1\right)=0 \Leftrightarrow x=1 \text { or } \frac{-1 \pm \sqrt{5}}{4} $$ Since $\sin \frac{\pi}{10}$ is positive other than $1$, we have $$ \boxed{\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}} $$

Answer 4

For diversity, here is a geometrical argument, which should be accesible to a lower level. Consider the isosceles triangle $\Delta ABC$ with side $BC=1$, $AB=AC=x$, and an angle of $36^\circ$ in $A$. The angle bisector in $B$ intersects $AC$ in $D$.

Then shown angles put in evidence the isosceles triangles $\Delta DAB$ and $\Delta BCD$. This gives $AD=BD=BC=1$. On the other side, the side $AC=x$ is cut in $D$ in two segments that respect the proportion $AD:DC=BA:BC=x:1$, which leads to $AD=x\cdot x/(1+x)$, $CD=x\cdot 1/(1+x)$. We obtain an equation for $x$, $$1=AD=\frac {x^2}{1+x}\ ,$$ with the positive solution $x=\frac 12(1+\sqrt 5)$. Draw now the angle bisector from $A$, for each of the formed angles of $18^\circ$ we have then in each of the formed right triangles with hypotenuse $x=AB=AC$: $$ \sin \frac\pi{10} =\sin 18^\circ = \frac{1/2}x=\frac{1/2}{(\sqrt5+1)/2} =\frac 1{\sqrt 5+1} =\frac {\sqrt 5-1}{5-1}= \bbox[yellow]{\ \frac 14(\sqrt 5-1)\ }\ . $$