## Sunday, April 19, 2015

### Vietnamese Mathematical Olympiad (Trigonometric) Problem of 1962

Solve the equation $\sin^6 x+\cos^6 x=\dfrac{1}{4}$.

This is one of the brilliant Mathematics Olympiad Contest Problems because we can show to the students how there are plenty of ways to attacking a good problem and how one approach is different from the other and how heuristic skill enable us to find solution quickly that save us time for more challenging problems!

Approach (I):

One may "reduce" the power of sixth on each sine and cosine function by taking $a=\sin^2 x$ and $b=\cos^2 x$ (whenever $a=\sin^2 x$ and $b=\cos^2 x$ have been introduced, we need to be careful to always replace $a+b$ by $1$ since $a+b=\sin^2 x+\cos^2 x=1$) so now we have:

$(\sin^2 x)^3+(\cos^2 x)^3=\dfrac{1}{4}$

$a^3+b^3=\dfrac{1}{4}---(1)$

The LHS can easily be factored as if we use the binomial theorem to expand $(a+b)$ to the third power, we get

$(a+b)^3=a^3+3ab(a+b)+b^3\implies a^3+b^3=(a+b)((a+b)^2-3ab)$

Apply this to equation (1) we get

$(a+b)((a+b)^2-3ab)=\dfrac{1}{4}$

Since $a+b=1$ the above equation becomes

$(1)((1)^2-3ab)=\dfrac{1}{4}$

$3ab=\dfrac{3}{4}$

At this point we need to replace $a$ and $b$ back to sine and cosine function to get

$(\sin^2 x)(\cos^2 x)=\dfrac{1}{4}$

$(\sin x\cos x)^2=\dfrac{1}{4}$ but we have from double angle identity that $\sin 2x=2\sin x \cos x$

$\therefore \dfrac{\sin 2x}{2}=\pm\dfrac{1}{2}$

$\sin 2x=\pm 1$

Therefore, $x=45^{\circ}+180^{\circ}n,\,135^{\circ}+180^{\circ}n$, where $n\in \mathbb{Z}$.

Approach (II):

Notice that we can relate $\sin^2 x$ with $\cos^2 x$ by the identity $\sin^2 x+\cos^2 x=1$ and if we replace $\sin^2 x=1-\cos^2 x$ into the original given equation, we see that we have

$(\sin^2 x)^3+\cos^6 x=\dfrac{1}{4}$

$(1-\cos^2 x)^3+\cos^6 x=\dfrac{1}{4}$

Use the binomial theorem to expand $(1-\cos^2 x)^3$, ah! We can already "see" something very promising as it will give us a pretty quadratic equation, let's see:

$1^3-3\cos^2 x+3(\cos^2 x)^2-(\cos^2 x)^3+\cos^6 x=\dfrac{1}{4}$

$\dfrac{3}{4}-3\cos^2 x+3(\cos^2 x)^2-\cancel{\cos^6 x}+\cancel{\cos^6 x}=0$

$(\cos^2 x)^2-\cos^2 x+\dfrac{1}{4}=0$

The above equation is equivalent to $\left(\cos^2 x-\dfrac{1}{2}\right)^2=0$ and this simplifies to $\cos x=\pm \dfrac{1}{\sqrt{2}}$ and hence the solutions to the given original equation are $x=45^{\circ}+180^{\circ}n,\,135^{\circ}+180^{\circ}n$, where $n\in \mathbb{Z}$.

The last method that I am going to show you is the clever heuristic method but I will save it on another blog post because I want you to try it out first, who knows, perhaps you are capable and finally bring yourself and me the pleasant surprise!