charfreemannn
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#1
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#1
i need help with this q
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charfreemannn
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don't understand how you turn sec2^a and tan2^a into cos/sin format
Last edited by charfreemannn; 1 month ago
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charfreemannn
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#3
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#3
anyone??????????????????
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Muttley79
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#4
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(Original post by charfreemannn)
anyone??????????????????
What have you tried?
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Faith&hardwork
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#5
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#5
Where do you need help?
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charfreemannn
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#6
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#6
(Original post by achieve526kk)
hope fullt this helps i tried me guess
omg thank u so much. i hate to be that person but do you mind taking another pic? it's a bit blurry and i can't make out the last part. thank u so so much btw
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mqb2766
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#7
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#7
(Original post by achieve526kk)
hope fullt this helps i tried me guess
Pls delete and read the forum guidelines about not posting solutions. The OP has not posted any attempt.
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davros
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#8
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#8
(Original post by charfreemannn)
anyone??????????????????
You need to post your thoughts first - as per forum guidelines. You can start from either side of the identity.
(Original post by achieve526kk)
hope fullt this helps i tried me guess
Please do not post full solutions - especially when the OP has made no attempt at the question. This is against forum rules!
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charfreemannn
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#9
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(Original post by davros)
You need to post your thoughts first - as per forum guidelines. You can start from either side of the identity.

Please do not post full solutions - especially when the OP has made no attempt at the question. This is against forum rules!
I obviously have attempted this question and struggled to find a solution, hence I posted it on this forum? I can't show you what I've done so far as I can't take photos on my laptop.

(Original post by mqb2766)
Pls delete and read the forum guidelines about not posting solutions. The OP has not posted any attempt.
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mqb2766
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#10
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#10
(Original post by charfreemannn)
I obviously have attempted this question and struggled to find a solution, hence I posted it on this forum? I can't show you what I've done so far as I can't take photos on my laptop.
https://www.thestudentroom.co.uk/sho....php?t=4919248
You could describe what you'd tried, what you had difficulties with, ...
That way its possible to offer hints about what you've tried.
Last edited by mqb2766; 1 month ago
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charfreemannn
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#11
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#11
(Original post by mqb2766)
https://www.thestudentroom.co.uk/sho....php?t=4919248
You could describe what you'd tried, what you had difficulties with, ...
That way its possible to offer hints about what you've tried.
I have. Want to offer any help now?
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mqb2766
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(Original post by charfreemannn)
I have. Want to offer any help now?
What identities have you tried / why didnt they work / ...
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charfreemannn
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#13
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#13
(Original post by mqb2766)
What identities have you tried / why didnt they work / ...
tried 1/cos and sin/cos but i don't know what to do with the 2^a bit since i've never come across a trig question with powers before
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mqb2766
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#14
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#14
Its a typo in the question. The left hand side should be
sec(2A) + tan(2A)
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Buddy324
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#15
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#15
1/cos 2A = sec 2A
sin 2A/ cos 2A = tan 2A.
Last edited by Buddy324; 1 month ago
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davros
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#16
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#16
(Original post by charfreemannn)
I obviously have attempted this question and struggled to find a solution, hence I posted it on this forum? I can't show you what I've done so far as I can't take photos on my laptop.
As above, you can type what you've tried, using brackets to make things clear if necessary. You should also note that the question doesn't make sense as written with those "powers" - it should be sec(2A) and tan(2A) as noted buy other posters.. You can now put the LHS over a common denominator and use some well-known identities
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charfreemannn
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#17
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#17
thanks guys i worked it out. didn't realise it was a typo hence i was stuck

(Original post by davros)
As above, you can type what you've tried, using brackets to make things clear if necessary. You should also note that the question doesn't make sense as written with those "powers" - it should be sec(2A) and tan(2A) as noted buy other posters.. You can now put the LHS over a common denominator and use some well-known identities
(Original post by Buddy324)
1/cos 2A = sec 2A
sin 2A/ cos 2A = tan 2A.
(Original post by mqb2766)
Its a typo in the question. The left hand side should be
sec(2A) + tan(2A)
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