Solving Trig Identity: sin5xcos3x=sin4xcos4x+sinxcosx

[Geekchick]

Homework Statement

sin5xcos3x=sin4xcos4x+sinxcosx, solve the identity

Homework Equations

all the identities and formulas mentioned in my last thread.

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]

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]

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 won't work its in the form sinxcosx +sinycosy hmmm.

 

[dynamicsolo]

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]

Thank you so much!