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

**flashylightsmeow** · February 25th 2011, 11:14 AM

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:
$\sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}+1/\sqrt{2}sin{x}=0$

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

**pickslides** · February 25th 2011, 11:22 AM

Originally Posted by flashylightsmeow

$\sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0$

now group some like terms, what do you get?

**TheEmptySet** · February 25th 2011, 11:26 AM

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:
$\sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}-1/\sqrt{2}sin{x}=0$

I then tried squaring both sides and using trig identities to put everything in terms of $cos^2{x}$ . 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.

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

This gives

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

**flashylightsmeow** · February 25th 2011, 11:45 AM

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

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

then grouped together like terms:

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

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

then: using the pythagorean trig identity

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

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

but from here:
$cos^2{x}=-2/8$ which doesn't give a real number AGH!!!!

**flashylightsmeow** · February 25th 2011, 11:47 AM

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!

**flashylightsmeow** · February 25th 2011, 12:36 PM

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 -.

**flashylightsmeow** · February 25th 2011, 02:09 PM

Originally Posted by TheEmptySet

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

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

This gives

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

so from this i get:

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

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

if so where do I go from there?

**TheEmptySet** · February 25th 2011, 02:15 PM

Originally Posted by flashylightsmeow

so from this i get:

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

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

if so where do I go from there?

Yes then

$\cos(x-{\frac{\pi}{24}})=0 \iff x-{\frac{\pi}{24}}=\cos^{-1}(0)=\frac{\pi}{2}+n\pi$ for $n \in \mathbb{Z}$

**flashylightsmeow** · February 25th 2011, 02:19 PM

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

**TheCoffeeMachine** · February 25th 2011, 04:20 PM

Originally Posted by flashylightsmeow

$\cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$

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

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

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

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

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

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

$\sin\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$

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

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

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

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

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

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