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# Expressing trig identities watch

1. Express:

(cosx - sinx)2 in terms of sin2x
I expanded the brackets and got the answer to just be cos2x, I cant seem to think how I could express it in terms of sin2x.

cos4x - sin4x in terms of cos2x
I got the answer to be cos2(2x) but the book says the answer is just cos2x
2. (Original post by znx)
Express:

(cosx - sinx)2 in terms of sin2x
I expanded the brackets and got the answer to just be cos2x, I cant seem to think how I could express it in terms of sin2x.

cos4x - sin4x in terms of cos2x
I got the answer to be cos2(2x) but the book says the answer is just cos2x
Not quite correct. Expansion is

Secondly, note that . Rather,
3. (Original post by RDKGames)
Not quite correct. Expansion is

Secondly, note that . Rather,
so (cos2x - sin2x)(cos2x + sin2x) = (cos(2x))(cos(2x)) = cos2(2x)?
4. (Original post by znx)
so (cos2x - sin2x)(cos2x + sin2x) = (cos(2x))(cos(2x)) = cos2(2x)?
5. Express:

(cosx - sinx)2 in terms of sin2x

cos4x - sin4x in terms of cos2x

a) You'd get cos^2(x) + sin^2(x) - 2sin(x)cos(x). What do you notice about cos^2(x) + sin^2(x) and the -2sin(x)cos(x). What identities are these?

b) cos4x - sin4x = [cos^2(x) + sin^2(x)][cos^2(x)-sin^2(x]. Look at the 2 brackets. You know the first equals 1 and the second equals cos(2x). So 1cos(2x) = ... ?
6. (Original post by RDKGames)
Oh yeah its just 1, so the answer would just be cos2x, thanks again mate
7. (Original post by RDKGames)
How do you do that fancy text again?
8. (Original post by Bulletzone)
How do you do that fancy text again?
https://www.thestudentroom.co.uk/help/latex
9. (Original post by thekidwhogames)
Express:

(cosx - sinx)2 in terms of sin2x

cos4x - sin4x in terms of cos2x

a) You'd get cos^2(x) + sin^2(x) - 2sin(x)cos(x). What do you notice about cos^2(x) + sin^2(x) and the -2sin(x)cos(x). What identities are these?

b) cos4x - sin4x = [cos^2(x) + sin^2(x)][cos^2(x)-sin^2(x]. Look at the 2 brackets. You know the first equals 1 and the second equals cos(2x). So 1cos(2x) = ... ?
Thank you mate, managed to get both right answers

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