# Thread: Proving Sum and difference identities

1. ## Proving Sum and difference identities

Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$

Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$

I have no idea what to do next to prove...

Thank you..

2. Originally Posted by mj.alawami
Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$

Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$

I have no idea what to do next to prove...

Thank you..
$\sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)$

$\sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)$

$\cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)$

$\cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)$

so we get:

$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$

Factorising gives us

$\frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}$

Cancelling 2cos(v) and because anything times 0 is 0 we get:

$\frac{sin(u)}{cos(u)} = tan(u)$

3. Originally Posted by e^(i*pi)
$\sin({u+v}) = sin(u)cos(v) + sin(v)cos(u)$

$\sin({u-v}) = sin(u)cos(v) - sin(v)cos(u)$

$\cos({u+v}) = cos(u)cos(v) - sin(v)sin(u)$

$\cos({u-v}) = cos(u)cos(v) + sin(v)sin(u)$

so we get:

$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$

Factorising gives us

$\frac{sin(u)(2cos(v)) + sin(v)(0)}{cos(u)(2cos(v)) + sin(v)(0)}$

Cancelling 2cos(v) and because anything times 0 is 0 we get:

$\frac{sin(u)}{cos(u)} = tan(u)$

How did you factorize it because i am getting lost?

4. Originally Posted by mj.alawami
Question:
$\frac{sin(u+v)+sin(u-v)}{cos(u+v)+cos(u-v)} =tan u$
$\frac{\sin(u+v)+\sin(u-v)}{\cos(u+v)+\cos(u-v)}$

$=\frac{(\sin u\cos v+\cos u\sin v)+(\sin u\cos v-\cos u\sin v)}{(\cos u\cos v-\sin u\sin v)+(\cos u\cos v+\sin u\sin v)}$

$=\frac{2\sin u\cos v}{2\cos u\cos v}$

$=\frac{\sin u}{\cos u}=\tan u.$

Attempt:
I simplifyied using the equation and i reached to a point where my answer is
$tan(u+v) + tan(u-v)$

I have no idea what to do next to prove...
This is not correct. For example,

$\tan\left(\frac\pi6+\frac\pi6\right)+\tan\left(\fr ac\pi6-\frac\pi6\right)=\sqrt3\neq\frac{\sqrt3}3=\tan\fra c\pi6.$

5. Originally Posted by mj.alawami
How did you factorize it ?
It came from combining like terms, I could have rearranged them first in an extra step which would be going from:

$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$

to

$\frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}$

I could also have then simplified each set of brackets to give:

$(sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)$

$(sin(v)cos(u) - sin(v)cos(u)) = 0$

$(cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)$

$(-sin(u)sin(v) + sin(u)sin(v)) = 0$

this means the 'main' equation will be :

$\frac{2sin(u)cos(v)}{2cos(u)cos(v)}$

As $2cos(v)$ is on the top and bottom it can be cancelled to give ${sin(u)}{cos(u)} = tan(u)$

6. Originally Posted by e^(i*pi)
It came from combining like terms, I could have rearranged them first in an extra step which would be going from:

$\frac{sin(u)cos(v) + sin(v)cos(u) + sin(u)cos(v) - sin(v)cos(u)}{cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)}$

to

$\frac{(sin(u)cos(v) + sin(u)cos(v))+ (sin(v)cos(u) - sin(v)cos(u))}{(cos(u)cos(v) + cos(u)cos(v)) - (sin(u)sin(v) + sin(u)sin(v))}$

I could also have then simplified each set of brackets to give:

$(sin(u)cos(v) + sin(u)cos(v)) = 2sin(u)cos(v)$

$(sin(v)cos(u) - sin(v)cos(u)) = 0$

$(cos(u)cos(v) + cos(u)cos(v)) = 2cos(u)cos(v)$

$(-sin(u)sin(v) + sin(u)sin(v)) = 0$

this means the 'main' equation will be :

$\frac{2sin(u)cos(v)}{2cos(u)cos(v)}$

As $2cos(v)$ is on the top and bottom it can be cancelled to give ${sin(u)}{cos(u)} = tan(u)$
Thanks alott it sunk in very easily