Not sure what to do with trigonometry problem $\sin54^\circ\cos108^\circ$

Question

What is the value of the following expression? $$\sin54^\circ\cos108^\circ$$

So what I tried here is:

$\sin54^\circ\cos108^\circ=\\\cos36^\circ\cos(2\cdot54^\circ)=\\\cos36^\circ(\cos^254^\circ-\sin^236^\circ)=\\\cos36^\circ\cos^254^\circ-\cos^354=\\{1\over2}\big[\cos90^\circ +\cos18^\circ \big]\cdot\cos54^\circ-\cos^336^\circ=\\{1\over2}\cos18^\circ\cos54^\circ-\cos^336^\circ$

I don't think I can do with this anything really, seems pointless to continue. Where have I gone wrong? What should've I done instead of this?

Edit: I'm a high school student, I have only done the values from the trigonometry circle. If possible, solve this without using computing of values, since I don't know how to do that at all.

Answer 1

$\sin 54^\circ = \frac 14 + \frac {\sqrt {5}}{4}\\ \cos 108^\circ = \frac 14 - \frac {\sqrt {5}}{4}$

Consider this isosceles triangle:

If we bisect the base angle we create a similar triangle and we create more isosceles triangles.

If we say $AB = 1, BC = x$ then $AD = x,$ and $CD = x^2$

And, $\cos 72 = \frac 12 x$

$x+ x^2 = 1$

Solving the quadratic. $x = \frac {-1 \pm \sqrt 5}{2}$

And since we know that $x > 0$

$\cos 72^\circ = \frac {-1 + \sqrt 5}{4}$

$\cos 108^\circ = - \cos 72$ by symmetry about $90^\circ$

We use double angle or half angle formla to find

$\cos 36^\circ$ or $\cos 144^\circ$

And complimentary / suppimentary formula to find $\sin 54^\circ$

Answer 2

In general, you could find the value of $\sin36^o$ and $\cos36^o$ respectively by looking at a triangle with angles $72^o, 72^o$ and $36^o$, taking the bisector and looking at the similar triangles that appear.

Once you have that you can get the value of the $\sin18^o$

Notice that those calculations will give you the answer by using the fact that $\sin\alpha=\cos(90^o-\alpha)$ and $\cos\alpha=\sin(90^o-\alpha)$

Answer 3

$P= \sin(54^{\circ})\cos(108^{\circ})= \sin(90^{\circ} - 2\cdot 18^{\circ})\cos(90^{\circ}+18^{\circ})= -\cos(2\cdot 18^{\circ})\sin(18^{\circ})= (2\sin^2(18^{\circ}) - 1)\sin(18^{\circ})= 2x^3-x, x = \sin(18^{\circ})$. We have: $\cos(2\cdot 18^{\circ})= \sin(3\cdot 18^{\circ})\implies 1-2x^2= 3x-4x^3\implies 4x^3-2x^2-3x+1 = 0\implies (x-1)(4x^2+2x-1) = 0\implies 4x^2+2x - 1 = 0$ since $x \neq 1$. Thus $4x^3 +2x^2-x = 0\implies x^3 = \dfrac{x-2x^2}{4}\implies P = 2x^3-x = \dfrac{x-2x^2}{2}-x= -\dfrac{2x+4x^2}{4}= -\dfrac{1}{4}$ .

Answer 4

Notice $$\begin{align} \sin 54^\circ \cos 108^\circ &= \cos 36^\circ \cos 108^\circ = \frac12(\cos(108-36)^\circ + \cos(108+36)^\circ)\\ &= \frac12(\cos 72^\circ + \cos 144^\circ)\end{align}$$ Now look at a regular pentagon with vertices at $(\cos \theta,\sin\theta)$ for $\theta = 0^\circ, \pm 72^\circ, \pm 144^\circ$. Because of symmetry, the center of mass of the pentagon is located at origin. If one look at the $x$-coordinates, this give us $$1 + 2\cos 72^\circ + 2\cos 144^\circ = 0 \quad\implies\quad \cos 72^\circ + \cos 144^\circ = -\frac12$$ This leads to $$\sin 54^\circ \cos 108^\circ = \frac12\times -\frac12 = -\frac14$$

Answer 5

set $\cos 36^\circ = x \gt 0$

by standard trig identities $$ \cos 72^\circ = 2x^2 -1 \\ \cos 108^\circ = 4x^3 -3x $$ and $$ \cos 72^\circ + \cos 108^\circ = 0 $$ i.e. $$ 4x^3 + 2x^2 -3x -1 = 0 $$ giving $$ (x+1)(4x^2 - 2x- 1) = 0 $$ since evidently $x \ne -1$, $x$ must be the positive root of $$ 4x^2 - 2x- 1 = 0 $$ it is simplest to see $x$ as a half of the golden section, i.e. $x = \frac{\phi}2$ where $\phi$ satisfies $\phi^2 = \phi + 1$

now $$ \sin 54^\circ \cos 108^\circ = - \cos 36^\circ \cos 72^\circ = - x(2x^2 -1) = -\frac14\bigg(\phi(\phi^2 -2) \bigg) = -\frac14 $$

Answer 6

$$\sin54\cos108=\cos36\cos(180-72)=-\cos36\cos72$$

Now for $\sin x\ne0,$

$\cos x\cos2x=\dfrac{\sin2x\cos2x}{2\sin x}=\dfrac{\sin4x}{4\sin x}$

Here $x=36^\circ,\sin4x=\sin x$

Also, observe that there are some other values of $x$ for which $\dfrac{\sin4x}{\sin x}=\dfrac14$ holds true