Think About Trigonometry

Geometry Level 1

$\Large \left(\sqrt{2+\sqrt{2}}\right)^{x} + \left(\sqrt{2-\sqrt{2}}\right)^{x} = 2^{x}$

Find the sum of all real $x$ that satisfy the equation above.

The answer is 2.

4 solutions

Utsav Banerjee
May 23, 2015

$\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \Rightarrow 2(\cos\frac{\pi}{8})^2=1+\cos\frac{\pi}{4} \Rightarrow \cos\frac{\pi}{8}=\sqrt{\frac{2+\sqrt{2}}{4}}$

$\Rightarrow \sin\frac{\pi}{8}=\sqrt{\frac{2-\sqrt{2}}{4}}$

The given equation can be re-written as:

$\Large \left(\sqrt{\frac{2+\sqrt{2}}{4}}\right)^{x}+\left(\sqrt{\frac{2-\sqrt{2}}{4}}\right)^{x}=1$

$\Rightarrow (\cos\frac{\pi}{8})^{x}+(\sin\frac{\pi}{8})^{x}=1$

Let $f(x)=(\cos\frac{\pi}{8})^{x}+(\sin\frac{\pi}{8})^{x}$ . Then, $x=2$ is clearly a solution for the above equation $f(x)=1$ . Let us prove that $x=2$ is the only solution.

$f'(x)=(\cos\frac{\pi}{8})^{x}\ln(\cos\frac{\pi}{8})+(\sin\frac{\pi}{8})^{x}\ln(\sin\frac{\pi}{8})$

Since $\cos\frac{\pi}{8}<1$ & $\sin\frac{\pi}{8}<1$ , we have $f'(x)<0 \,\forall\, x\ge0$ , that is, $f(x)$ is a strictly decreasing function $\,\forall\,x\ge0$ . Hence, $f(x)$ will not be equal to $1$ for any other value of $x \in [0,\infty)-{2}$ .

Also, since $\cos\frac{\pi}{8}<1$ & $\sin\frac{\pi}{8}<1$ , for $x<0$ we have $(\cos\frac{\pi}{8})^{x}>1$ & $(\sin\frac{\pi}{8})^{x}>1$ . So, $f(x)>2\,\forall\,x<0$ .

Therefore, the only solution for $f(x)=1$ is $x=2$ , and the answer is $\boxed{2}$ .

As rightly pointed out by the challenge master, please note that it is not necessary to show that the radicals are values of trigonometric ratios, or to calculate the exact angle. Any equation of the form $a^x+b^x=1$ will have $x=2$ as the only solution if $a<1$ , $b<1$ and $a^2+b^2=1$ , and it can be proved using the above method!

Moderator note:

Your last step isn't quite right. You have only shown that $x=2$ exist, but you failed to show that it is the only solution.

Hint: Show that $f(x) = \left( \cos \frac \pi8 \right)^x + \left( \sin \frac \pi8 \right)^x$ is a decreasing function for all $x> 0$ and show that it's impossible to have a negative solution.

Furthermore, you don't actually need to show that those radicals are values of a certain trigonometric function. You just need to show that their absolute values are both less than 1 and their squares sum up to 1. Do you understand why?

Additional hint: Do you think that it's necessary to find the exact angles to solve this equation?

$\displaystyle \left(\sqrt{1234+\sqrt{5678}}\right)^x + \left(\sqrt{1234-\sqrt{5678}}\right)^x = (2\cdot 1234)^x$

Bonus question: Solve the equation above.

Thank you so much for pointing this out! I have updated the solution now. The approach using trigonometrical ratios is not necessary.

Also, any equation of the form $f(x)=a^x+b^x=1$ will have $x=2$ as the only solution if $|a|<1$ , $|b|<1$ and $a^2+b^2=1$ . This is because $f(x)>2\,\forall\,x<0$ , $f(2)=1$ and $f(x)$ is a strictly decreasing function $\,\forall\,x\ge0$ .

Therefore, the equation $\left(\sqrt{\frac{1234+\sqrt{5678}}{2\times1234}}\right)^{x}+\left(\sqrt{\frac{1234-\sqrt{5678}}{2\times1234}}\right)^{x}=1$ will also have $x=2$ as the only solution.

Utsav Banerjee - 6 years ago

Wonderful work, Utsav Banerjee! Small error though: it should be $(2\times 1234)^2$ in the radicals.

One thing to note that for these kind of equations, we can show that there exist a real $y$ such that $\cos y =\frac12 \sqrt{2 + \sqrt 2}$ and $\sin y =\frac12 \sqrt{2 - \sqrt 2}$ (values are interexchangeable) by proving that $\cos^2 y + \sin^2 y = 1$ .

Another bonus question: For positive numbers $A,B$ , would there still be only one solution for the equation below?

$\displaystyle \left( -\sqrt{A+ \sqrt B} \right)^x + \left(\sqrt{A- \sqrt B} \right)^x = (2A)^x$

Brilliant Mathematics Staff - 6 years ago

Istiak Reza
Jun 26, 2015

2±✓2 = 2( 1±1/✓2)....We know, cos (pi/4)=1✓2....substituting and using the identity 1+cos 2A=2cos^2 A and 1 - cos 2A=2sin^2 A, we get (2 cos pi/8)^x+(2sin pi/8)^x=2^x....Solving we get 2^x=0 which has no real solution and (cos pi/8)^x+(sin pi/8)^x=1=(cos pi/8)^2+(sin pi/8)^2.....equating cos and sin we get x=2 which is the only real solution...

Ubaidullah Khan
Jun 1, 2015

I tried some other method................................................................................................................ Just in a single moment, I square both sides and I got the answer quickly..See the following:-
Hence the answer.....is 4

Moderator note:

You have only shown that $x=2$ is a solution but you didn't show that it is the only solution. See Utsav Banerjee's solution for a proper approach.

this solution ccan be re-inforced by applying femat's last theorem. that x can never be greater than 2....

abhideep singh - 5 years, 7 months ago

Fermat's Last Theorem holds only for natural numbers, and here there are radicals.

Archishman Mukherjee - 5 years ago

Abhideep Singh
Nov 13, 2015

applied femat's last theorem. Either 1 or 2 can be only possible values. check for 1 and 2. Only solution is 2.

Fermat's Last Theorem only works for positive integers. So here, it does not work.

Rishik Jain - 4 years, 11 months ago

Think About Trigonometry

The answer is 2.

4 solutions

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