Hi,
First off I know how to do things such as,
cosx x cos 4x = (1/2) x (cos(x+4x) + cos(x-4x))
= (1/2) x (cos 5x) x (cos(-3x))
However i'm getting very confused over one simple difference which is,
2 sin5x . sin2x, as in what do I with the 2 out the front of this equation?
This is what I tried (I have no idea though if its right or wrong)
(2 sin5x . sin2x) = (1/2)x(2) x (sin(5x+2x) + (cos5x-2x))
= (sin 7x) + (cos3x)
Any help would be great.
The cos product identity is (where * denotes multiplication and / denotes division)
cos(x)*cos(y) = (cos(x+y) + (cos(x-y))/2
The sin product identity is
sin(x)*sin(y) = (cos(x-y) - (cos(x+y))/2
You cannot simply replace cos with sin.
so for your equation
2*(sin(5x)*sin(2x)) we need to use the sin product identity.
We will first tackle the (sin(5x)*sin(2x)) and after we solve for that, we will substitute it into the original equation
sin(5x)*sin(2x)= (cos(5x-2x)-cos(5x+2x))/2 = (cos(3x)-cos(7x))/2
Plug this back into your equation for (sin(5x)*sin(2x)) and you have 2*((cos(3x)-cos(7x))/2) where the 2s cancel out and you have
= cos(3x)-cos(7x)
Please start a new thread for this , if you have problems in starting them , ask itOriginally Posted by kitty
Plz help me to understand double angles
btw here is a general answer to that question, we cant teach you that , but we can give links to useful sites
http://en.wikipedia.org/wiki/Proofs_...ric_identities
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