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

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.

• Can you convert cosine to sine and then mod 180? Oct 15, 2018 at 21:29
• What do you mean with mod? Oct 15, 2018 at 21:31
• Sine (190)=sin (10) Oct 15, 2018 at 21:32
• It is known that $\sin 18 = \frac{\sqrt{5}-1}{4}$. So you can get $\cos 18$. $36$ and $54$ are multiples of $18$, so you should be able to compute it. Oct 15, 2018 at 21:33
• @Gibbs I think I'm supposed to use only the known angles from the trigonometry circle, I'm a high school student, never did the computing of those values. Oct 15, 2018 at 21:40

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

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 $$\sin 54^\circ$$

• I just commented on asdf's answer saying something about that method. Oct 15, 2018 at 21:38
• I guess this explains how you got to the angle, I understand it, but if there's another method of solving the problem without knowing those angles I'll answer it tomorrow here. Oct 15, 2018 at 22:17

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 $$\sin 54^\circ \cos 108^\circ = \frac12\times -\frac12 = -\frac14$$

• excellent use of the geometry to minimze computation! Oct 15, 2018 at 23:40
• Oct 16, 2018 at 18:42

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

• How is $\sin18^\circ = \cos(2\cdot18^\circ)$? Oct 15, 2018 at 22:00
• @Aleksa $\cos(2\theta) = 1 - 2\sin^2\theta \implies -\cos(2 \cdot 18^\circ) = (2\sin^2(18^\circ) - 1)$. Oct 15, 2018 at 22:05
• Oops somehow I saw an equal sign where it wasn't, my bad. Oct 15, 2018 at 22:07

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

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

• An elegant solution. Oct 16, 2018 at 22:52

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 $$\cos\alpha=\sin(90^o-\alpha)$$

• Yeah I believe that is a way to do it, but I'm pretty sure this I am supposed to solve that by only using the known values of $30^\circ$, $60^\circ$, $90^\circ$ etc. Oct 15, 2018 at 21:37
• @Aleksa That'd be quite difficult, I think, whereas knowing a few more values solves it directly
– asdf
Oct 15, 2018 at 21:40

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