# solve $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$

I came across a question from another forum - find the $$x$$ in the following diagram: I managed to deduce an equation from the following diagram: which is: $$\dfrac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$$

and I know the answer (from WolframAlpha) is: $$\cos \theta= \dfrac{3}{\sqrt{10}}$$

but I'm not able to deduce the answer myself, any ideas?

By the way, $$x=2\sqrt{5}$$, which can be easily deduced by:

$$\dfrac{3\sqrt{2}}{x}=\cos\theta$$

I also tried to solve the original question geometrically: Somehow, I managed to figure out that $$y=3$$ in the above diagram, but I can't prove it either.

• Did you already find $x$ from the given information? – Toby Mak Mar 2 at 2:48
• @TobyMak, $x=2\sqrt{5}$ – lochiwei Mar 2 at 2:51
• I understand now; you're just trying to simplify $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$ right? – Toby Mak Mar 2 at 2:54
• What program did you use to draw and label the triangles? Or is it done by hand? – JavaMan Mar 2 at 3:39
• @JavaMan I used Paper to draw those diagrams, and yes, by hand. – lochiwei Mar 2 at 3:44

Probably not the most elegant solution.

If you use the tangent half-angle substitution $$x=\tan \left(\frac{\theta }{2}\right)$$, you end with $$\sqrt{10}\, x^4-2 \left(6-\sqrt{10}\right)\, x^3-2 \sqrt{10} \,x^2-2 \left(6+\sqrt{10}\right) x+\sqrt{10}=0 \tag 1$$

Using the method for quartic equations, there are two real roots and one of them is $$x=\sqrt{10}-3\implies \cos(\theta)=\frac{3}{\sqrt{10}}$$ The other roots are really messy.

Defining $$c := \cos\theta$$ and $$s := \sin\theta$$, we can write $$c^2\sqrt{5} = s\left(3 \sqrt{2} + c\sqrt{5}\right) \tag{1}$$ Squaring, re-writing $$s^2 = 1-c^2$$, and re-arranging, $$10 c^4 + 6 c^3 \sqrt{10} + 13 c^2 - 6c \sqrt{10} - 18 = 0 \tag{2}$$

At this point, if we had the presence of mind to identify $$10$$ as $$\sqrt{10}^2$$ and $$13$$ as $$-27 + 4\sqrt{10}^2$$, then (defining $$r:=\sqrt{10}$$) we could gather terms and factor

\begin{align} 0 &= r^2 c^4 +\left(-3r + 9 r\right)c^3 + (-27+4r^2)c^2 +(-12r+ 6r)c - 18 \\[4pt] &= \left(r^2 c^4 - 3rc^3\right) + \left(9 rc^3 -27c^2\right)+\left(4r^2c^2 -12rc\right)+ \left(6rc - 18\right)\\[4pt] &= \left(r c - 3 \right) \left( rc^3 + 9 c^2 +4 r c + 6 \right) \tag{3} \end{align} (That is, we have factored over $$\mathbb{Q}\left[\sqrt{10}\right]$$.) The first factor yields the target root, $$\cos\theta = 3/\sqrt{10}$$. (Note that the second factor obviously has no positive solutions.)

Without such intuition, but with a suspicion that the $$\sqrt{10}$$s were impeding progress to a reasonably-nice solution, we could write $$(2)$$ as $$10 c^4 + 13 c^2-18 = 6c\sqrt{10}\left(1-c^2\right) \tag{4}$$ Now, squaring will eliminate the pesky $$\sqrt{10}$$, and we have $$100 c^8 - 100 c^6 + 529 c^4 - 828 c^2 + 324 = 0 \tag{5}$$ From here, old-fashioned factoring gives

$$\left(10 c^2 - 9\right) (10 c^6 - c^4 + 52c^2 -36 ) = 0\tag{6}$$

Again, the first factor gives the target root, $$\cos\theta=3/\sqrt{10}$$ (as well as a newly-introduced extraneous root, $$\cos\theta=-3/\sqrt{10}$$). It's not clear that the second factor has no valid roots; indeed, Mathematica gives the positive solution $$\cos\theta = 0.80501\ldots$$ (amid otherwise negative or non-real candidates), but we can check that it doesn't satisfy $$(1)$$.