# Double and half angle identities?

• Feb 20th 2011, 07:06 PM
homeylova223
Double and half angle identities?
Can anyone help me verify these two identities

1. csc2x= 1/2 secx cscx I am not sure how to start this one

My second one I am not sure if I have done it right?

(sinx+cosx)^2-1= sin2x
(sinx+cosx) (sinx+cosx)-1
sinx^2+sinxcosx+sinxcosx+cosx^2-1
1+2sincosx-1=sin2x
2sinxcosx=sin2x
sin2x=sin2x(Wondering) Did I do this right?
• Feb 20th 2011, 07:10 PM
TheCoffeeMachine
Quote:

Originally Posted by homeylova223
(sinx+cosx)^2-1= sin2x
(sinx+cosx) (sinx+cosx)-1
sinx^2+sinxcosx+sinxcosx+cosx^2-1
1+2sincosx-1=sin2x
2sinxcosx=sin2x
sin2x=sin2x(Wondering) Did I do this right?

Right! (Yes)

A hint for the first one -- it says:

$\displaystyle \frac{1}{\sin{2x}} = \frac{1}{2\sin{x}\cos{x}}$
• Feb 20th 2011, 07:12 PM
pickslides
Quote:

Originally Posted by homeylova223
Can anyone help me verify these two identities

1. csc2x= 1/2 secx cscx I am not sure how to start this one

$\displaystyle \csc 2x = \frac{1}{\sin 2x} = \frac{1}{2\sin x \cos x} = \frac{1}{2}\csc x \sec x$