# Trig ratios, double angles

1. Mar 21, 2009

### DERRAN

1. The problem statement, all variables and given/known data
Prove that sin(45+x).sin(45-x) = $$\frac{1}{2}$$cos2x

2. Relevant equations
double angle formulae
reduction formulae
special angles
identities

3. The attempt at a solution
(sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)
=(sin45.cosx-cos45.sinx)2
=($$\frac{1}{\sqrt{2}}$$cosx-$$\frac{1}{\sqrt{2}}$$sinx)2

=$$\frac{1}{2}$$cos2x-sinx.cosx+$$\frac{1}{2}$$sinx2

=$$\frac{1}{2}$$-sinx.cosx

2. Mar 21, 2009

### danago

How did you get from:

(sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)

To:

(sin45.cosx-cos45.sinx)^2

?

3. Mar 21, 2009

### DERRAN

I really don't know. I must of been smoking something(hehehe):rofl:
But thanks any way for pointing out my error I got it now.

it should be

(sin45.cosx+cos45.sinx)(sin45.cosx-cos45.sinx)
=sin245.cos2x-cos245.sin2x
=$$\frac{1}{2}$$cos2x-$$\frac{1}{2}$$sin2x
=$$\frac{1}{2}$$cos2x

Last edited: Mar 21, 2009
4. Mar 21, 2009

### General_Sax

Start from here:

http://http://i623.photobucket.com/albums/tt316/Saxifrage_Russell/PhysicsForumcomMarch21st.png" [Broken]

Last edited by a moderator: May 4, 2017