# trig reduction

• September 25th 2013, 10:39 AM
Bonganitedd
trig reduction
Sin(-180-x)
This is my attempt
First I used sum/difference formulae
Sin(A-B) =sinAcosB-cosAsinB
=sin(-180)cos(x)-cos(-180)sin(x)
=-sin(180)cos(x) - - cos(180)sin(x)
=0.cos(x) + (-1)]sin(x)
=-sin(x)
My another option was to multiply by (-1)
After multiply I have sin (180+x) which I knw Sin(x) is negative in third quadrant(not sure about this method though)
According to my textbook answer is Sin (x)
Where did I lost it??
• September 25th 2013, 10:58 AM
Plato
Re: trig reduction
Quote:

Originally Posted by Bonganitedd
Sin(-180-x)

The $\sin(x)$ is an odd function.

This $\sin(-180-x)=-\sin(180+x)=-[\sin(x)\cos(180)+\sin(180)\cos(x)]$
• September 25th 2013, 11:03 AM
Bonganitedd
Re: trig reduction
I now see. Thank you.
• September 25th 2013, 11:56 PM
Prove It
Re: trig reduction
Quote:

Originally Posted by Plato
The $\sin(x)$ is an odd function.

This $\sin(-180-x)=-\sin(180+x)=-[\sin(x)\cos(180)+\sin(180)\cos(x)]$

Or even \displaystyle \begin{align*} -\sin{ \left( 180^{\circ} + x \right) } = - \left[ -\sin{ (x) } \right] \end{align*} from the symmetry of the unit circle...