For maths-based tomfoolery enter here!

Royality

Royality

Royality

Soldato
Joined
7 Dec 2003
Posts
3,528
Location
Normandy
Who here likes A-level Identities. That's right - everyone!

Please can you help me.

With the parameters of 0<x<2pi what are the solutions to the riddle below:

2cotx + 2cosec^2x = 5

I've deduced one of the answers to be 1.11 (which is correct). However the other answers of 2.82, 4.25 and 5.96 are eluding me. :(

I've an idea of what's going wrong but my fiddling is too time consuming. I'm sure I'll understand more when the dam is broken etc.

Could anyone explain to me where I'm going wrong (and possibly stay in this thread for the rest of the evening. ;))
 
from sin^2 + cos^2 = 1 you can get

1 + cot^2 = cosec^2 by dividing through bin sin^2

back in to the original equation...

2cotx + 2 (1 + cot^2x) = 5

expand the brackets and make it equal to 0

2cot^2x + 2cotx - 3 = 0

solve like a normal quadratic :)
 
marc_howarth said:
from sin^2 + cos^2 = 1 you can get

1 + cot^2 = cosec^2 by dividing through bin sin^2

back in to the original equation...

2cotx + 2 (1 + cot^2x) = 5

expand the brackets and make it equal to 0

2cot^2x + 2cotx - 3 = 0

solve like a normal quadratic :)
Click to expand...

Sorry! Annoyingly I got the equation wrong. Real equation is:

5cotx + 2cosec^2x = 5

Oh and the rearranging I can do, it's the solving of the resulting equations to find solutions I can't do (i.e to get cotx = 1/2 and -3 to give values of 2.82, 4.25, 5.96).
 
ok well never mind i did think that quadratic was a bit of a *****

so if cotx = 1/2 and -3

tan x = 2 and -1/3

bring out the calculator here and do arctan of those values to get

tan x = 2
x = 1.11

tanx = -1/3
x = -0.35

but now you have to remember that tan repeats over and over again with a gap of 180 degrees between each one. this question is in radians however so a gap of Pi between each tan repetition

simply add Pi to those values of x untill you're out of range to get the solutions :)
 
marc_howarth said:
ok well never mind i did think that quadratic was a bit of a *****

so if cotx = 1/2 and -3

tan x = 2 and -1/3

bring out the calculator here and do arctan of those values to get

tan x = 2
x = 1.11

tanx = -1/3
x = -0.35

but now you have to remember that tan repeats over and over again with a gap of 180 degrees between each one. this question is in radians however so a gap of Pi between each tan repetition

simply add Pi to those values of x untill you're out of range to get the solutions :)
Click to expand...

Okay, that helped a lot.

Hehe, I have more problems though... :(

Prove that cosec A + cot A = 1/(cosec A - cot A)

I decided to divide left from right to equal 1.

Can you help with this, i'd greatly appreciate it; why is it that maths books never include blooming solutions? :o
 
marc_howarth said:
have a feeling this could be a long night....

\gets working :p
Click to expand...

Wow I actually did the one above.

Apparently if you divide a fraction by a fraction the answer = (numerator numerator) x (denominator denominator) / (numerator denominator) x (denominator numerator)

i.e (5/4)/(16/12) = (5x12)/(4x16) = 60/64 = 15/16

How weird is it that I never realised that... :D
 
ok this isn't a proper proof but it's the general idea

same identity as the last question

cosec^2 - cost^2 = 1

if you multiply up the (cosec A - cos A) you get

1 = (cosec A - cos A)(cosec A + cot A)

expand out that bracket

1 = cosec^2 - cot^2

1 = 1

really you should start LHS or RHS and make it equal to the opposite side but i'll leave that to you :)
 
Royality said:
Okay, that helped a lot.

Hehe, I have more problems though... :(

Prove that cosec A + cot A = 1/(cosec A - cot A)

I decided to divide left from right to equal 1.

Can you help with this, i'd greatly appreciate it; why is it that maths books never include blooming solutions? :o
Click to expand...

If you're proving that LHS = RHS you have to manipulate one side to match the other, you can't divide one side by the other. Start from the left or right and use identities until it matches the other.

I'd give it a bash for you but I'm in an incredibly lazy mood I'm afraid :p
 
Dave said:
If you're proving that LHS = RHS you have to manipulate one side to match the other, you can't divide one side by the other. Start from the left or right and use identities until it matches the other.

I'd give it a bash for you but I'm in an incredibly lazy mood I'm afraid :p
Click to expand...

If you divide one side from another the answer will equal 1 when two identities are the same (like dividing 2 by 2). Hence you can use that method... that's what my book says at least! :p

I just can't get my head round these bloody identities, it's a miracle I got 56 in my C3 exam first attempt in January; unfortunately I need around 85 to get an A now. :(
 
good luck with ** exams all

trig identities aren't really that bad, most of them can be derieved from

sin^2 + cos^2 = 1

some double angle formula can be worked out from the formulaes given in the formula booklet, i guess the hardest part is knowing which one to use when but like always..... practise.... :o
 
If this isn't taught until C3 then I bemoan the state of mathematics education in schools today. This kind of question should be obvious to anyone who's sat the first half of A Level Maths.
 
Royality said:
If you divide one side from another the answer will equal 1 when two identities are the same (like dividing 2 by 2). Hence you can use that method... that's what my book says at least! :p

I just can't get my head round these bloody identities, it's a miracle I got 56 in my C3 exam first attempt in January; unfortunately I need around 85 to get an A now. :(
Click to expand...

Hmm that's iffy, on the IB you'd get no marks for that :p

If it's got a "3 bar equals sign" I've always been taught to derive one side by manipulating the other rather than mess around with the original identity.
 
You must log in or register to reply here.
Back
Top Bottom