What is the proof for sin(45+x).sin(45-x) = \frac{1}{2}cos2x?

[DERRAN]

Homework Statement

Prove that sin(45+x).sin(45-x) = $$\frac{1}{2}$$cos2x

Homework Equations

double angle formulae
reduction formulae
special angles
identities

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

 

[danago]

How did you get from:

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

To:

(sin45.cosx-cos45.sinx)^2

?

 

[DERRAN]

How did you get from:

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

To:

(sin45.cosx-cos45.sinx)^2

?

I really don't know. I must of been smoking something(hehehe)
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

 

[General_Sax]

Start from here:

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