Solve Trigonometry Problem: Cos^3x+1/Cos^3x=0

[terryds]

Homework Statement

If $cos^3x+\frac{1}{cos^3x} = 0$, then sin 2x equals ...

A. -1
B. -√3
C. -√2
D. 1
E. 2

Homework Equations

Basic trigonometry

The Attempt at a Solution

[/B]
Let y = cos^3 x

y+1/y = 0
y^2 + 1 = 0
y = √(-1)

Then, I don't know what to do.. Help please..

 

[member 587159]

Homework Statement

If $cos^3x+\frac{1}{cos^3x} = 0$, then sin 2x equals ...

A. -1
B. -√3
C. -√2
D. 1
E. 2

Homework Equations

Basic trigonometry

The Attempt at a Solution

[/B]
Let y = cos^3 x

y+1/y = 0
y^2 + 1 = 0
y = √(-1)

Then, I don't know what to do.. Help please..

I think you can solve this one for cosx. Make the expression into one fraction and remember a/b = 0 <=> a = 0 when b =/= 0.

 

[terryds]

I think you can solve this one for cosx. Make the expression into one fraction and remember a/b = 0 <=> a = 0 when b =/= 0.

$\frac{cos^6x+1}{cos^3x}=0$
$cos^6x = -1$
$cos x =\sqrt[6]{-1}$
$sin 2x = 2 sin x cos x \\ sin 2x = 2 \sqrt{1-\sqrt[3]{-1}}\sqrt[6]{-1}$

How to solve it?

 

[member 587159]

Well, in the context of real numbers, there does not seem to be an x that satisfies the equation.

 

[terryds]

Well, in the context of real numbers, there does not seem to be an x that satisfies the equation.

So, no option satisfies??

 

[blue_leaf77]

Then $x$ must be complex. From the original equation, out of the two possibilities you can choose $\cos^3 x = i = e^{i\pi/2}$. From this, find $\cos^2 x$ and $\cos x$, then $\sin x$.

 

[terryds]

Then $x$ must be complex. From the original equation, out of the two possibilities you can choose $\cos^3 x = i = e^{i\pi/2}$. From this, find $\cos^2 x$ and $\cos x$, then $\sin x$.

It seems a bit complicated to take the cubic root from exponential form.
Is there any easier way??
During test, calculator is not allowed.. So I try not using calculator since this is my preparation

 

[blue_leaf77]

No, it's fairly easy actually.
From $\cos^3 x = e^{i\pi/2}$, find $\cos x$ in terms of complex exponential.

 

[terryds]

No, it's fairly easy actually.
From $\cos^3 x = e^{i\pi/2}$, find $\cos x$ in terms of complex exponential.

$\cos x = e^{i\pi/6} \\ sin x = \sqrt{1-e^{i\pi/3}} \\ sin 2x = 2 sin\ x\ cos\ x\ = 2\sqrt{1-e^{i\pi/3}} e^{i\pi/6}$

Now, it's complicated.
How to solve it?

 

[blue_leaf77]

Not that complicated if you have put some more creativity and trials, now in
$$
sin x = \sqrt{1-e^{i\pi/3}} \\
$$
express $1-e^{i\pi/3}$ in $x+iy$ form.

 

[terryds]

Not that complicated if you have put some more creativity and trials, now in
$$
sin x = \sqrt{1-e^{i\pi/3}} \\
$$
express $1-e^{i\pi/3}$ in $x+iy$ form.

Hmmm... I'm new to complex number... and, I don't know how to convert $1-e^{i\pi/3}$ to rectangular form since there is a constant (in this case, 1)...
Please help..

 

[blue_leaf77]

Hmmm... I'm new to complex number... and, I don't know how to convert $1-e^{i\pi/3}$ to rectangular form since there is a constant (in this case, 1)...
Please help..

In rectangular form of a complex number $x+iy$, both $x$ and $y$ are real and every complex number can be transformed to this form. To do what you need to do, transform $e^{i\pi/3}$ into rectangular form using Euler formula.

 

[terryds]

In rectangular form of a complex number $x+iy$, both $x$ and $y$ are real and every complex number can be transformed to this form. To do what you need to do, transform $e^{i\pi/3}$ into rectangular form using Euler formula.

Alright,
so it becomes like this
$1-e^{i\pi/3} = 1 - cos \frac{\pi}{3} + i\ sin \frac{\pi}{3} = \frac{1}{2} + \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1+\sqrt{3}i) \\ sin\ x = \sqrt{\frac{1}{2} (1+\sqrt{3}i)}$

and I convert cos x to rectangular form,

$cos\ x = \frac{1}{2}\sqrt{3}+\frac{1}{2}i$

$sin\ 2x = 2\ sin\ x\ cos\ x\ = 2 (\sqrt{\frac{1}{2} (1+\sqrt{3}i)} )( \frac{1}{2}\sqrt{3}+\frac{1}{2}i)$

Still seems complicated :(

 

[blue_leaf77]

You make a mistake $e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. After incorporating this correction, you will get $\sin^2 x$ in rectangular form, then cast it back into polar form. Take its square root to obtain $\sin x$.

 

[SteamKing]

Homework Statement

If $cos^3x+\frac{1}{cos^3x} = 0$, then sin 2x equals ...

A. -1
B. -√3
C. -√2
D. 1
E. 2

Homework Equations

Basic trigonometry

The Attempt at a Solution

[/B]
Let y = cos^3 x

y+1/y = 0
y^2 + 1 = 0
y = √(-1)

Then, I don't know what to do.. Help please..

After looking at all the complex algebra and trig which has been expended on solving this problem in the posts above, you can eliminate some of the answers A. - E., at least in terms of real solutions for x.

The problem statement wants to know what the value of sin (2x) is, and like all sine functions, its amplitude is bound to the interval [-1, 1], so that means that any choices whose value lies outside this range cannot be correct. The remaining choices can be easily checked to see if they can satisfy the original cosine equation.

Given the comments made so far in this thread, I can't help but wonder if something has been left out of the original problem statement.

 

[blue_leaf77]

like all sine functions, its amplitude is bound to the interval [-1, 1],

If the argument is complex, then it can be bigger than unity. For example $\cos (iy) = \cosh y \geq 1$.

 

[terryds]

You make a mistake $e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. After incorporating this correction, you will get $\sin^2 x$ in rectangular form, then cast it back into polar form. Take its square root to obtain $\sin x$.

Alright,
so it becomes like this
$1-e^{i\pi/3} = 1 - cos \frac{\pi}{3} + i\ sin \frac{\pi}{3} = \frac{1}{2} + \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1+\sqrt{3}i) \\ sin\ x = \sqrt{\frac{1}{2} (1+\sqrt{3}i)}$

and I convert cos x to rectangular form,

$cos\ x = \frac{1}{2}\sqrt{3}+\frac{1}{2}i$

$sin\ 2x = 2\ sin\ x\ cos\ x\ = 2 (\sqrt{\frac{1}{2} (1+\sqrt{3}i)} )( \frac{1}{2}\sqrt{3}+\frac{1}{2}i)$

Still seems complicated :(

I forgot to put parentheses
$1-e^{i\pi/3} = 1 - (cos \frac{\pi}{3} + i\ sin \frac{\pi}{3}) = \frac{1}{2} - \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1-\sqrt{3}i) \\ sin\ x = \sqrt{\frac{1}{2} (1-\sqrt{3}i)}$
$1-e^{i\pi/3} = 1 - (cos \frac{\pi}{3} + i\ sin \frac{\pi}{3}) = \frac{1}{2} - \frac{1}{2}\sqrt{3}i = \frac{1}{2} (1-\sqrt{3}i) \\ sin\ x = \sqrt{\frac{1}{2} (1-\sqrt{3}i)} \\ sin\ 2x = 2\ sin\ x\ cos\ x\ = 2 (\sqrt{\frac{1}{2} (1-\sqrt{3}i)} )( \frac{1}{2}\sqrt{3}+\frac{1}{2}i)$

Stuck again.

 

[blue_leaf77]

Like I said, cast $\frac{1}{2} (1-\sqrt{3}i)$ back to polar form.

 

[terryds]

Like I said, cast $\frac{1}{2} (1-\sqrt{3}i)$ back to polar form.

Polar form?? Do you mean something like r (cos alpha + i sin alpha) ?
$r = \sqrt{{\frac{1}{2}}^2+({\frac{1}{2}}\sqrt{3})^2} = \sqrt{1} = 1 \\ \alpha = tan^{-1} (\frac{-\sqrt{3}}{1}) \\ \alpha = \frac{5}{3}\pi \\ \frac{1}{2}(1-\sqrt{3}i) = 1 (cos \frac{5}{3}\pi+i sin\frac{5}{3}\pi) \\$

Taking the root of it?? Why should I make it in polar form??

 

[blue_leaf77]

I mean complex exponential form, $re^{i\theta}$.

Taking the root of it?? Why should I make it in polar form??

You don't need to actually but doing so will simplify your calculation.

 

[terryds]

I mean complex exponential form, $re^{i\theta}$.

So,
$\frac{1}{2}(1-\sqrt{3}i) = \frac{1}{2} - \frac{1}{2} \sqrt{3}i = e^{i\frac{5\pi}{3}} \\$
$sin\ x= e^{i\frac{5\pi}{6}} \\ sin\ 2x = 2\ sin\ x\ cos\ x = 2 e^{i\frac{5\pi}{6}} e^{i\frac{\pi}{6}} = 2e^{i\pi} = -2$
Is it correct ??

 

[SteamKing]

If the argument is complex, then it can be bigger than unity. For example $\cos (iy) = \cosh y \geq 1$.

Yeah, but I still wonder about this problem, especially since the OP claims he's never worked with complex numbers much.

 

[blue_leaf77]

-2 is surely one of the possibilities, but it's not among the choices. You can end up with a different solution if you use another angle instead of $5\pi/3$, hint: consider a negative angle.

 

[terryds]

-2 is surely one of the possibilities, but it's not among the choices. You can end up with a different solution if you use another angle instead of $5\pi/3$, hint: consider a negative angle.

$sin\ x = e^{\frac{-\pi}{6}i}$
$sin\ 2x = 2 e^{\frac{-\pi}{6}i} e^{\frac{\pi}{6}i} = 2$
The answer is E. 2... Thanks a lot, pal!

 

[haruspex]

$sin\ x = e^{\frac{-\pi}{6}i}$
$sin\ 2x = 2 e^{\frac{-\pi}{6}i} e^{\frac{\pi}{6}i} = 2$
The answer is E. 2... Thanks a lot, pal!

Having got to $\cos (x)=e^{i\frac{\pi}6}$, I think the easiest way is to look at sin²(2x). $\sin^2(x)\cos^2(x)=\cos^2(x)-\cos^4(x)=e^{i\frac{\pi}3}-e^{2i\frac{\pi}3}=\cos(\pi/3)+i\sin(\pi/3)-\cos(2\pi/3)-i\sin(2\pi/3)=2\cos(\pi/3)=1$.