# Trigonometry Question from a Math Contest

1. Jan 5, 2005

### MathExpert

sin^8(x) + cos^8(x) = 97 / 128

2. Jan 5, 2005

### Zurtex

The only way I can think off the top of my head is say it is equivalent to:

$$128 \left[ \sin^8x + \left(1 - \sin^2x \right)^4 \right] - 97 = 0$$

And really hope it is solvable.

3. Jan 5, 2005

### gonzo

I'd just keep pluggin in trig identities until you reduce things either all to single powers or maybe a quadratic equation of cosines. It shouldn't be too hard, you can use

2cos^2(x) = 1+cos(2x)
2sin^2(x) = 1 - cos(2x)

4. Jan 5, 2005

### krab

It would be hard to prove because it's not true.

Edit: Sorry. I assumed you meant it as being true for all x. Do you it is to be solved for x?

Last edited: Jan 5, 2005
5. Jan 5, 2005

### dextercioby

It can be reduced to two biquadratic equations,which are solvable.
$$1=[(\sin^{2}x+\cos^{2}x)^{2}]^{2}=(\sin^{4}x+\cos^{4}x+2\sin^{2}x\cos^{2}x)^{2}$$
$$=\sin^{8}x+\cos^{8}x+4\sin^{4}x \cos^{4}x+2\sin^{4}x \cos^{4}x+4\sin^{2}x \cos^{2}x (\cos^{4}x+\sin^{4}x)$$ (1)

From (1) it follows:
$$1=\sin^{8}x+\cos^{8}x+6\sin^{4}x \cos^{4}x+4\sin^{2}x\cos^{2}x[(\sin^{2}x+\cos^{2}x)^{2}-2\sin^{2}x\cos^{2}x]$$
$$=\sin^{8}x+\cos^{8}x-2\sin^{4}x \cos^{4}x+4\sin^{2}x\cos^{2}x$$ (2)

From (2) it follows:
$$\sin^{8}x+\cos^{8}x=1+2\sin^{4}x\cos^{4}x-4\sin^{2}x\cos^{2}x$$ (3)

I use:
$$\sin 2x=2\sin x\cos x$$(4)
to write
$$+2\sin^{4}x\cos^{4}x=+\frac{1}{8}(\sin 2x)^{4}$$ (5)
$$-4\sin^{2}x\cos^{2}x=-(\sin 2x)^{2}$$ (6)

Therefore,combining (5),(6) and (3),one gets:
$$\sin^{8}x+\cos^{8}x=1+\frac{1}{8}(\sin 2x)^{4}-(\sin 2x)^{2}$$ (7)

The initial equation
$$\sin^{8}x+\cos^{8}x=\frac{97}{128}$$ (8)
,using (7),becomes
$$1+\frac{1}{8}(\sin 2x)^{4}-(\sin 2x)^{2}=\frac{97}{128}$$ (9)
,which can be put in the form
$$\frac{1}{8}}(\sin 2x)^{4}-(\sin 2x)^{2}+\frac{31}{128}=0$$ (10)

Multiplying (10) by $+8$ and making the obvious substitution
$$(\sin 2x)\rightarrow y$$ (11)
,the equation (8) is brought to the canonical form of an biquadratic algebraic eq. in the variable "y":
$$y^{4}-8y^{2}+\frac{31}{16}=0$$(12)
We make the substitution:$y^{2}\rightarrow z$ (13)
,and the eq.(12) becomes
$$z^{2}-8z+\frac{31}{16}=0$$ (14)
,with the solutions
$$z_{1}=4+\frac{\sqrt{225}}{4}=\frac{1}{4}$$ (15)
$$z_{2}=4-\frac{\sqrt{225}}{4}= \frac{31}{4}$$(16)

Comparing (11),(13) and (16),we conclude that $z_{2}$ is not a viable solution.
To be continued.

Last edited: Jan 5, 2005
6. Jan 5, 2005

### dextercioby

From (13) and (15),we obtain:
$$y^{2}=+\frac{1}{4}$$ (17)
,with the solutions:
$$y_{1}=-\frac{1}{2}$$ (18)
$$y_{2}=+\frac{1}{2}$$ (19)

It can be easily checked that both solutions (18) and (19) are found in the interval (-1,+1),so they provide viable solutions to the problem.
Putting together (18),(19) and (11),we found two transcendental equations
$$\sin 2x=-\frac{1}{2}$$ (20)
$$\sin 2x=+\frac{1}{2}$$ (21)

We solve these eq.in the interval $(-\frac{\pi}{2},+\frac{\pi}{2})$(for the 'sine' argument,which is 2x).In this case,we can apply the inverse function 'arcsine' to find
$$x_{1}=\frac{1}{2}\arcsin(-\frac{1}{2})=-\frac{\pi}{12}$$ (22)
$$x_{2}=\frac{1}{2}\arcsin(+\frac{1}{2})=+\frac{\pi}{12}$$ (23)

Daniel.

EDIT:Thank you,Krab,i wouldn't have seen it,years after...

Last edited: Jan 5, 2005
7. Jan 5, 2005

### krab

Looks like it's $\pi/12$

8. Jan 5, 2005

### dextercioby

Let's take $\pi\sim 3.14$
$$\sin\frac{\pi}{12}\sim 0.25869$$

$$\sin^{8}\frac{\pi}{12}\sim 2\cdot 10^{-5}$$(1)

$$\cos\frac{\pi}{12}\sim 0.96596$$

$$\cos^{8}\frac{\pi}{12}\sim 75800\cdot 10^{-5}$$ (2)

Add (1) and (2) and get:
$$\sin^{8}\frac{\pi}{12}+\cos^{8}\frac{\pi}{12}\sim 75802\cdot 10^{-5} \sim\frac{97}{128}$$ (3)

Daniel.

Last edited: Jan 5, 2005
9. Jan 5, 2005

### Hurkyl

Staff Emeritus
Let s = sin^2 x, c = cos^2 x be a solution to the equation.

As Zurtex Started, this equation is a quartic in s:

97/128 = s^4 + (1-s)^4

The symmetry suggests making the substitution s = t + 1/2

Note that t is in the range -1/2 <= t <= 1/2, if we assume to look for real solutions.

97/128 = (1/2 + t)^4 + (1/2 - t)^4
= 2t^4 + 3t^2 + 1/8 = 2t^4 + 3t^2 + 16/128

so

2t^4 + 3t^2 - 81/128 = 0

yielding the solution:

t^2 = -3/4 +/- (1/4) sqrt(9 + 81/16)
= -3/4 +/- (1/4) sqrt(225/16)
= -3/4 +/- (1/4) (15/4)
= -12/16 +/- 15/16
= (-12 +/- 15) / 16

t^2 = -27 / 16 or 3 / 16

However, we know that 0 <= t^2 <= 9/4, so we have the single solution:

t^2 = 3/16

so

t = sqrt(3) / 4

s = t + 1/2 = (2 + sqrt(3)) / 4

Now, s = sin^2 x, so we have:

sin x = (1/2) sqrt(2 + sqrt(3))

Where both square roots could be either positive or negative.

For some reason, I feel that knowing a lot of solutions to the original equation should help me -- if x is a solution, then so is -x, ix, and 90 - x, from which you can make lots more solutions -- but I couldn't get such an approach to help.

Last edited: Jan 5, 2005
10. Jan 5, 2005

### krab

But it's not meant to be 1; it's meant to be 97/128, which it is.

BTW, I found your error. In your equation 1, your last 2 should be a 4. Apart from this, the technique is really nice.

11. Jan 5, 2005

### dextercioby

Hurkyl,i found two solutions in the reals,in the interval $(-\frac{\pi}{4},+\frac{\pi}{4})$,because i found it easier to use 'arcsin'.
Solving the eq.
$$\sin 2x=\pm \frac{1}{2}$$
on the whole real exis would yield all real solutions to the equation.

If u wanna inlcude complex solutions as well,u should use not circular,but hyperbolic trigonometry (identities and stuff).

Daniel.

12. Jan 6, 2005

### Curious3141

The solutions have already been given, but perhaps the most "direct" way to do this would be like so :

Let $c = \cos \theta$, $s = \sin \theta$, $S = \sin 2\theta$

$$s^8 + c^8 = 97/128$$

$$(c^4 + s^4)^2 - 2(c^2s^2)^2 = \frac {97}{128}$$

$$((c^2 + s^2)^2 - 2c^2s^2)^2 - 2(\frac{1}{4}S^2)^2 = \frac {97}{128}$$

$$(1 - \frac{1}{2}S^2)^2 - 2(\frac{1}{4}S^2)^2 = \frac {97}{128}$$

$$(1 - \frac{1}{2}S^2)^2 - \frac{1}{8}S^4 = \frac {97}{128}$$

Expanding the LHS, rearranging and simplifying,

$$\frac{1}{8}S^4 - S^2 + \frac{31}{128} = 0$$

which is a quadratic equation in S^2. Solving that will give the required general solution in the reals :

$$\theta = n\pi (+/-) \frac{\pi}{12}$$ or $$n\pi (+/-) \frac{5\pi}{12}$$

Last edited: Jan 6, 2005
13. Jan 7, 2005

### tongos

oh boy, anyone in highschool taking the amc 12 in a few weeks?