# Thread: [SOLVED] Proving a Trigonometric Identity

1. ## [SOLVED] Proving a Trigonometric Identity

my question is this...

$\displaystyle \cos x - \cos y=-2 \sin\left (\frac{x + y}{2}\right)\sin\left (\frac{x - y}{2}\right)$

how do you make LHS equal RHS???

i end up with the RHS equaling

$\displaystyle \frac{\cos x \sin y - \sin x \cos y}{2}$

or

$\displaystyle \cos x \sin y - \sin x \cos y$ without the divided by 2

2. ya know that $\displaystyle \cos(a+b)=\cos a\cos b-\sin a\sin b,$ and $\displaystyle \cos(a-b)=\cos a\cos b+\sin a\sin b,$ thus by just subtractin' these we get $\displaystyle \cos(a-b)-\cos(a+b)=2\sin a\sin b.$

put $\displaystyle x=a-b$ and $\displaystyle y=a+b$ then solve for $\displaystyle (a,b)$ and you'll get the desired identity.

3. where do u subtract the first two identities from????

and how did u get $\displaystyle 2\sin a\sin b$ ?

are u working with the LHS or RHS??? I am sorry but i don't understand and i have looked it over again and again.

could you show me please i find it a little hard to understand what u mean

4. Originally Posted by mathnewb
where do u subtract the first two identities from????

and how did u get $\displaystyle 2\sin a\sin b$ ?

are u working with the LHS or RHS??? I am sorry but i don't understand and i have looked it over again and again.

could you show me please i find it a little hard to understand what u mean
You have been told exactly how to do it.

You need to go back to post #2 and read it carefully. Try writing it out. Think carefully about what you're writing.

5. i understand all steps up to when im supposed to make x=a-b and y=a+b, however, i dont understand what im supposed to solve for (a,b)??

that part confuses me?!

6. Originally Posted by mathnewb
i understand all steps up to when im supposed to make x=a-b and y=a+b, however, i dont understand what im supposed to solve for (a,b)??

that part confuses me?!
$\displaystyle x = a - b$ .... (1)

$\displaystyle y = a + b$ .... (2)

Solve equations (1) and (2) simultaneously:

$\displaystyle a = \frac{x + y}{2}$ .... (3)

$\displaystyle b = \frac{y - x}{2}$ .... (4)

You know $\displaystyle \cos (a - b) - \cos (a + b) = 2 \sin a \sin b$.

Substitute (1) and (2) into the left hand side. Substitute (3) and (4) into the right hand side. You get the identity you were asked to prove.

7. could someone please post the full steps and show conclusion where RHS=LHS please!

i am struggling a lot!!!!

8. no! you try it!

what you suggest gives me to understand that you want your meal directly into your mouth!

9. i did try it

this is what i end up with in the end...

in the beginning i changed x/2 to a and y/2 to b

this is the last part i end up with

RHS=(-2cosb^2)+(2cosa^2)

how can i make this equal cosx-cosy?

10. Originally Posted by mathnewb
i did try it

this is what i end up with in the end...

in the beginning i changed x/2 to a and y/2 to b

this is the last part i end up with

RHS=(-2cosb^2)+(2cosa^2)

how can i make this equal cosx-cosy?
I told you how to do it in post #6 (this built on what you had already been told in post #2).

Although I'm sure it's not the case, your posts do not suggest that you have read or understood anything that has been posted in reply to your problem.

Please .... clearly state what it is in post #6 that you do not understand and cannot do. Because the complete suggestion has more or less been given to you.

11. ok.....

when u solve the LHS you get 2sinxsiny....i got that

the RHS i sub into and get -2sinxsiny

how do i get rid of negative on RHS???

12. nvm i got it

thx