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Geekchick
Aug17-08, 01:07 AM
1. The problem statement, all variables and given/known data

sin5xcos3x=sin4xcos4x+sinxcosx, solve the identity

2. Relevant equations

all the identities and formulas mentioned in my last thread.

3. The attempt at a solution

Alright so I thought I could use the product to sum formula on the left side which ended up being 1/2sin10x then I used the sum/difference formula on the right side and got sin5x. Now according to my calculator these are not the same, so where have I gone wrong?

Defennder
Aug17-08, 04:48 AM
Applying the product to sum formula to the left side is correct. But you don't end up with sin(10x). Where did you get that from?

Geekchick
Aug17-08, 12:55 PM
alright what i did was this.

original problem sin5xcos3x=sin4xcos4x+sinxcosx

Left side sin5xcos3x

Product to sum formula sinxcosx=1/2[sin(x+y)+sin(x-y)]
plug in the numbers

1/2[sin(5x+3x)+sin(5x-3x)]
simplify
1/2(sin8x+sin2x)
simplify
1/2sin10x

right side sin4xcos4x+sinxcosx

sum/difference formula sin(x+y) = sinxcosy + cosxsiny
plug in the numbers
oh,never mind that wont work its in the form sinxcosx +sinycosy hmmm.

dynamicsolo
Aug17-08, 02:08 PM
You're more than halfway home. You have on the left side


1/2[sin(5x+3x)+sin(5x-3x)]
simplify
1/2(sin8x+sin2x)

This is fine. You can leave the left side now.


1/2(sin8x+sin2x)
simplify
1/2sin10x

Alas, angles don't add this way...


right side sin4xcos4x+sinxcosx

Here, don't apply the product-sum identity to the second term (well, you could, but it's overly fussy). What is sin x·cos x equal to (using another identity)? That will get you one of the pieces you're after.

Do use the product-sum identity to the first term -- now you'll be done...

Geekchick
Aug17-08, 06:18 PM
Thank you so much!