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

• terryds
In summary: You're almost there. Just multiply the two expressions inside the parentheses and simplify.You're almost there. Just multiply the two expressions inside the parentheses and simplify.##sin\ 2x = 2\ sin\ x\ cos\ x\ = \sqrt{1-e^{i\pi/3}} \sqrt{1+e^{i\pi/3}} = \sqrt{1-e^{i\pi/3}+e^{i\pi/3}-e^{2i\pi/3}} = \sqrt{1+cos\frac{\pi}{3} }####sin\ 2x = 2\ sin\ x\ cos\ x\ = \sqrt{1-e^{i\

## 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..

terryds said:

## 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.

Math_QED said:
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?

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

Math_QED said:
Well, in the context of real numbers, there does not seem to be an x that satisfies the equation.
So, no option satisfies??

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##.

blue_leaf77 said:
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

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

blue_leaf77 said:
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?

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.

blue_leaf77 said:
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)...

terryds said:
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)...
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.

blue_leaf77 said:
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 :(

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##.

terryds said:

## 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.

SteamKing said:
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##.

blue_leaf77 said:
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##.
terryds said:
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.

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

blue_leaf77 said:
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??

I mean complex exponential form, ##re^{i\theta}##.
terryds said:
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.

blue_leaf77 said:
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 ??

blue_leaf77 said:
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.

-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
blue_leaf77 said:
-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!

terryds said:
##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##.

terryds

## What is trigonometry and why is it important?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is important because it has a wide range of applications in fields such as engineering, physics, and astronomy.

## What is the cosine function and how is it used in trigonometry?

The cosine function is a trigonometric function that relates the length of the adjacent side of a right triangle to the length of the hypotenuse. In trigonometry, it is commonly used to find missing sides or angles of a triangle.

## How do I solve a trigonometry problem involving the cosine function?

To solve a trigonometry problem involving the cosine function, you can use the identity cos^2x + sin^2x = 1 to rewrite the equation in terms of either sine or cosine. Then, use algebraic techniques to isolate the variable and solve for its value.

## What does the equation cos^3x+1/cos^3x=0 mean?

This equation means that the sum of the cube of the cosine of an angle and 1, divided by the cube of the cosine of the same angle, is equal to 0. In other words, the cosine of the angle must be equal to -1 in order for the equation to hold true.

## What are some common mistakes to avoid when solving trigonometry problems?

Some common mistakes to avoid when solving trigonometry problems include forgetting to convert angles from degrees to radians, using the incorrect trigonometric identity, and making algebraic errors. It is important to always double check your work and make sure your calculations are accurate.