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Annoying C3 TRIG +rep watch

1. i cant seem to do trig identies when its got something to do with half angle formulae

1) prove that cot 2x + cosec2x = cot x

2) using half angle formulae show (1 - cos x) / (sin x) = tan (x / 2)
solve (1 - cos x) / (sin x) = root3 sin x

3) 2cosx + 1 = sin (x / 2) solve this

really grateful if someone could help me on these
2. 1) Write LHS as sub in double angle identities, put fractions over a common denominator (i'll post my solution if you want)

2) & 3) Think of and as and
3. (Original post by devan5)
prove that cot 2x + cosec2x = cot x

4. for the first one i've got up to [ 2sinxcosx(1-tan^2x) + 2tanx ] / 2tanx2sinxcosx where from here?
5. (Original post by devan5)
for the first one i've got up to [ 2sinxcosx(1-tan^2x) + 2tanx ] / 2tanx2sinxcosx where from here?
me too
6. 1st one:

cos2x/sin2x + 1/sin2x
= (cos^2x - sin^x)/2sinxcosx + 1/2sinxcosx
= (cos^2x + cos^2x)/2sinxcosx which cancels to give cotx

Ask if you don't understand the steps.
7. (Original post by devan5)
i cant seem to do trig identies when its got something to do with half angle formulae

1) prove that cot 2x + cosec2x = cot x

2) using half angle formulae show (1 - cos x) / (sin x) = tan (x / 2)
solve (1 - cos x) / (sin x) = root3 sin x

3) 2cosx + 1 = sin (x / 2) solve this

really grateful if someone could help me on these
Where'd you get this question from ?
8. (Original post by FZka)
Where'd you get this question from ?
A2 core for edexcel question is from june 2004 paper

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