JR - Term 1 test 2000 (1 Viewer)

[Jago]

Okay this is just a quick question from that test.

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

 

[velox]

Cos^2 x = 1/2 (1 + cos2x)

 

[Jago]

ooh...double angles

 

[currysauce]

yup
cos2x = 2cos²x - 1

2cos²x = cos2x + 1

cos²x = 1/2 cos 2x + 1/2

cos²3x = 1/2 cos 6x + 1/2

 

[KFunk]

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

cos²(nx) = 1/2[1 + cos(2nx)]

sin²(nx) = 1/2[1 - cos(2nx)]

 

[Slidey]

Haha. I always derive them. Bad memory.

 

[maths > english]

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

 

[Jago]

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]

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]

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.

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

Yep. Derive those, too.

 

[Jago]

which ones are the prodcuts to sums trig laws?

 

[velox]

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]

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)]