# Trig eqn cos(x+TT/6) + cos(x-TT/4) = 0

• Feb 25th 2011, 11:14 AM
flashylightsmeow
Trig eqn cos(x+TT/6) + cos(x-TT/4) = 0
Hi there, I was wondering if someone would be able to help me solve this equation :

cos(x+TT/6) + cos(x-TT/4) = 0

It's driving me insane. I use compound angle theorem to expand the components and end up with:
$\displaystyle \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}+1/\sqrt{2}sin{x}=0$

I then tried squaring both sides and using trig identities to put everything in terms of $\displaystyle cos^2{x}$. And got something that I wont waste your time with. Please help!
• Feb 25th 2011, 11:22 AM
pickslides
Quote:

Originally Posted by flashylightsmeow
$\displaystyle \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0$

now group some like terms, what do you get?
• Feb 25th 2011, 11:26 AM
TheEmptySet
Quote:

Originally Posted by flashylightsmeow
Hi there, I was wondering if someone would be able to help me solve this equation :

cos(x+TT/6) + cos(x-TT/4) = 0

It's driving me insane. I use compound angle theorem to expand the components and end up with:
$\displaystyle \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0$

I then tried squaring both sides and using trig identities to put everything in terms of $\displaystyle cos^2{x}$. And got something that I wont waste your time with. Please help!

Since the right hand side is zero why not use the sum to product identities.

$\displaystyle \displaystyle \cos(a)+\cos(b)=2\cos\left( \frac{a+b}{2}\right)\cos\left( \frac{a-b}{2}\right)$

This gives

$\displaystyle 2\displaystyle \cos\left( \frac{2x-\frac{\pi}{12}}{2}\right)\cos\left( \frac{\frac{5\pi}{12}}{2}\right)=0$
• Feb 25th 2011, 11:45 AM
flashylightsmeow
Quote:

Originally Posted by pickslides
now group some like terms, what do you get?

So i squared everything and multiplied through by 4 to get rid of the denominators and got

$\displaystyle 6cos^2{x}-2sin^2{x}+4cos^2{x}+4sin^2{x}=0$

then grouped together like terms:

$\displaystyle 10cos^2{x}+2sin^2{x}=0$

but from here
$\displaystyle 10cos^2{x}+2sin^2{x}=0$

then: using the pythagorean trig identity

$\displaystyle 10cos^2{x}+2(1-cos^2{x})=0$

$\displaystyle 8cos^2{x}+2=0$

but from here:
$\displaystyle cos^2{x}=-2/8$ which doesn't give a real number AGH!!!!
• Feb 25th 2011, 11:47 AM
flashylightsmeow
ah i didn't even look at it like that. I'll give it a shot, and if it doesnt work, I'll be crawling back!
• Feb 25th 2011, 12:36 PM
flashylightsmeow
Quote:

Originally Posted by pickslides
now group some like terms, what do you get?

Hi Pickslides, I made an error with the sign on the last term, it should be a + instead of -.
• Feb 25th 2011, 02:09 PM
flashylightsmeow
Quote:

Originally Posted by TheEmptySet
Since the right hand side is zero why not use the sum to product identities.

$\displaystyle \displaystyle \cos(a)+\cos(b)=2\cos\left( \frac{a+b}{2}\right)\cos\left( \frac{a-b}{2}\right)$

This gives

$\displaystyle 2\displaystyle \cos\left( \frac{2x-\frac{\pi}{12}}{2}\right)\cos\left( \frac{\frac{5\pi}{12}}{2}\right)=0$

so from this i get:

$\displaystyle 2cos(x-{\pi/24})cos({5\pi/24})=0$

Can i deduce from this expression that $\displaystyle cos(x-{\pi/24})$ must equal 0?

if so where do I go from there?
• Feb 25th 2011, 02:15 PM
TheEmptySet
Quote:

Originally Posted by flashylightsmeow
so from this i get:

$\displaystyle 2cos(x-{\pi/24})cos({5\pi/24})=0$

Can i deduce from this expression that $\displaystyle cos(x-{\frac{\pi}{24}})$ must equal 0?

if so where do I go from there?

Yes then

$\displaystyle \cos(x-{\frac{\pi}{24}})=0 \iff x-{\frac{\pi}{24}}=\cos^{-1}(0)=\frac{\pi}{2}+n\pi$ for $\displaystyle n \in \mathbb{Z}$
• Feb 25th 2011, 02:19 PM
flashylightsmeow
OH yea! Man I am bad at maths. Just a quick question, is it possible to use the method i initially used using cos(x+y) - cos(x-y)? I don't get where i went wrong :S
• Feb 25th 2011, 04:20 PM
TheCoffeeMachine
Solution by exploiting symmetry
Quote:

Originally Posted by flashylightsmeow
$\displaystyle \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$

By $\displaystyle \cos\left(\pi+\theta \right) = -\cos{\theta}$, we have $\displaystyle \cos\left(x-\frac{\pi}{4}\right) = \cos\left(\pi+x-\frac{\pi}{4}-\pi\right) = -\cos\left(x-\frac{5\pi}{4}\right)$.

Thus $\displaystyle \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0 \Rightarrow \cos\left(x+\frac{\pi}{6}\right) = -\cos\left(x-\frac{\pi}{4}\right) = \cos\left(x-\frac{5\pi}{4}\right).$

So $\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right)$. If $\displaystyle \cos{\theta} = \cos{\alpha}$, then $\displaystyle \theta = 2n\pi + \alpha$ or $\displaystyle \theta = 2n\pi -\alpha$, thus:

$\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi + x - \frac{5\pi}{4} \Rightarrow 0 = 2n\pi-\frac{17\pi}{12}{\pi}~$ (dead-end).

Or we have $\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi -x + \frac{5\pi}{4} \Rightarrow x = n\pi+\frac{13\pi}{24} ~ \checkmark.$

Most problems of this kind can be solved by some consideration of symmetry; a twin of your problem is:

Quote:

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$
$\displaystyle \cos{\theta} = \sin\left(\frac{\pi}{2}-\theta\right)$, thus $\displaystyle \cos\left(x-\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}+\frac{\pi}{4}-x\right) = \sin\left(\frac{3\pi}{4}-x\right)$, hence

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) = -\cos\left(x-\frac{\pi}{4}\right) \Rightarrow \sin\left(x+\frac{\pi}{6}\right) = -\sin\left(\frac{3\pi}{4}-x\right) = \sin\left(x-\frac{3\pi}{4}\right).$

So $\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right)$. If $\displaystyle \sin{\theta} = \sin{\alpha}$, then $\displaystyle \theta = 2n\pi+\alpha$ or $\displaystyle (2n+1)\pi-\alpha$, thus

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi+x-\frac{3\pi}{4} \Rightarrow 0 = 2n\pi-\frac{3\pi}{4}-\frac{\pi}{6}$ (dead-end).

Or $\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = (2n+1)\pi-x+\frac{3\pi}{4} \Rightarrow x = (2n+1)\frac{\pi}{2}+\frac{7\pi}{24} ~ \checkmark.$