JR - Term 1 test 2000 (1 Viewer)

Jago

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Okay this is just a quick question from that test.

Q4. How do they go from step 1 to step 2?
 

Jago

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ooh...double angles
 

currysauce

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yup
cos2x = 2cos²x - 1

2cos²x = cos2x + 1

cos²x = 1/2 cos 2x + 1/2

cos²3x = 1/2 cos 6x + 1/2
 

KFunk

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It's handy to keep the general form at the back of your mind for integrals:

cos<sup>2</sup>(nx) = 1/2[1 + cos(2nx)]

sin<sup>2</sup>(nx) = 1/2[1 - cos(2nx)]
 

Slidey

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Haha. I always derive them. Bad memory.
 

maths > english

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lol, i could never remember them either, especially the products to sums trig laws
 

Jago

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i always try to remember the formulas just before exams, and i promptly forget the minute i step out of the exam room
 

Pace_T

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Jago said:
i always try to remember the formulas just before exams, and i promptly forget the minute i step out of the exam room

that's better than not remembering them when walking INTO the exam room, lol.
i have to derive them.
it takes up so much time :(
 

Slidey

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What? Deriving them should be very fast?

A hint when deriving the cos^2(x) identity is to remember that cos^2x-sin^2x=1-2sin^2x=2cos^2x-1, so you can skip the line cos^2x-sin^2x.

maths > english said:
lol, i could never remember them either, especially the products to sums trig laws
Yep. Derive those, too. :D
 

Jago

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which ones are the prodcuts to sums trig laws?
 

velox

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stuff that we dont have to worry about, although we could have a question which derives them as a result, so a little familiarity with the results would do no harm.
 

FinalFantasy

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Jago said:
which ones are the prodcuts to sums trig laws?
do u mean things like
sinA+sinB=2sin[1\2(A+B)]cos[1\2(A-B)]
sinA-sinB=2cos[1\2(A+B)]sin[1\2(A-B)]
cosA+cosB=2cos[1\2(A+B)]cos[1\2(A-B)]
cosA-cosB=-2sin[1\2(A+B)]sin[1\2(A-B)]
 

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