Hello here I am again, asking maths.

Do I expand everything by writing cos(2theta + theta) and sin(theta + 2theta) then expand the 2theta bits out again?

I've gotten as far as (cos theta - sin theta)(1+cos2theta) = sin2theta (cos theta +sin theta), do I keep expanding?

I suppose I'm on the wrong direction but I just panic seeing 3theta?! Am I supposed to be grouping everything into one variable theta = something then do inverse trig to find the answer?

Do I expand everything by writing cos(2theta + theta) and sin(theta + 2theta) then expand the 2theta bits out again?

I've gotten as far as (cos theta - sin theta)(1+cos2theta) = sin2theta (cos theta +sin theta), do I keep expanding?

I suppose I'm on the wrong direction but I just panic seeing 3theta?! Am I supposed to be grouping everything into one variable theta = something then do inverse trig to find the answer?

(edited 9 months ago)

Original post by Bookworm524

Hello here I am again, asking maths.

Do I expand everything by writing cos(2theta + theta) and sin(theta + 2theta) then expand the 2theta bits out again?

I've gotten as far as (cos theta - sin theta)(1+cos2theta) = sin2theta (cos theta +sin theta), do I keep expanding?

Do I expand everything by writing cos(2theta + theta) and sin(theta + 2theta) then expand the 2theta bits out again?

I've gotten as far as (cos theta - sin theta)(1+cos2theta) = sin2theta (cos theta +sin theta), do I keep expanding?

Maybe ask in the morning - I think the able mathematicians would probably be asleep rn.

Try the sum to product trig identities if youve done them. Theyre a combination of the usual addition identities. The hint about there being two sets of solutions steers you in the direction of thinking about trying to factorise (so product rather than sum) and not necessarily putting every trig term in terms of just (x) - maybe leave some as a multiple angle if its appropriate.

(edited 9 months ago)

Original post by mqb2766

Try the sum to product trig identities if youve done them. Theyre a combination of the usual addition identities. The hint about there being two sets of solutions steers you in the direction of thinking about trying to factorise (so product rather than sum) and not necessarily putting every trig term in terms of just (x) - maybe leave some as a multiple angle if its appropriate.

Oh dear sum to product identities are the ones I skimmed over just before the A Level exams, and I've forgotten how to use them! Thanks and I'll have a go!

Original post by Bookworm524

Oh dear sum to product identities are the ones I skimmed over just before the A Level exams, and I've forgotten how to use them! Thanks and I'll have a go!

Theyre just a simple combination of the usual addition identities. Tbh, you could have used it with the other question youve just posted as

cos(B-C / 2) - sin(A/2) = cos(B-C / 2) - cos(90 - A/2) = ...

and use the sum to product for cos()-cos() and the triangle A+B+C=180 as rdkgames said and its pretty much a one liner. Though there are a few ways to get there.

(edited 9 months ago)

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