Products-to-sums identities (1 Viewer)

[yellowhighlighterr]

For this one, I am struggling with b. Could I have some help please?

 

[Masaken]

for part (b), i would write the LHS as 4cosa(sin5acos3a). that allows you to use product to sum with the bit that can be changed isolated in brackets, so you get 2cosa(sin8a + sin2a) [you have a 2 out front because you multiplied 4 by 1/2 as per the given product to sum identity]. from there, expand everything out, so you get 2sin8acosa + 2sin2acosa. again, apply the product to sum identity for each of the terms given, so you would get (with the 2 being cancelled out for both because it's multiplied by 1/2) = (sin 9a + sin 7a) + (sin 3a + sina) = RHS.

 

[yellowhighlighterr]

for part (b), i would write the LHS as 4cosa(sin5acos3a). that allows you to use product to sum with the bit that can be changed isolated in brackets, so you get 2cosa(sin8a + sin2a) [you have a 2 out front because you multiplied 4 by 1/2 as per the given product to sum identity]. from there, expand everything out, so you get 2sin8acosa + 2sin2acosa. again, apply the product to sum identity for each of the terms given, so you would get (with the 2 being cancelled out for both because it's multiplied by 1/2) = (sin 9a + sin 7a) + (sin 3a + sina) = RHS.

aaak ok. thanks m8

 

[yellowhighlighterr]

But @Masaken why do you take the cosalpha out only? can't you take any of the other terms out like sin5alpha? Would that be right?

 

[howcanibesmarter]

She took it out and put it into brackets so u can visualise the product to sum, its a multiplication so if u take out any it doesn't really matter, its more rearranging if anything