# P2 Question (Trig)

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#1
cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?
0
15 years ago
#2
(Original post by TheWolf)
cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?
cot2x = (1 - tan^2 x)/2tanx
cosec2x = 1/(2sinx cosx)
LHS = 4cot2x cosec2x = (1-tan^2 x)/(sinx cosx tanx) = (1 - tan^2 x)/sin^2 x = cosec^2 x - sec^2 x
cosec^2 x = 1 + cot^2 x
sec^2 x = 1 + tan^2 x
so cosec^2 x - sec^2 x = 1 + cot^2 x - 1 - tan^2 x = cot^2 x - tan^2 x = RHS
0
15 years ago
#3
(Original post by TheWolf)
cot^2x - tan^2x = 4cot2x cosec2x

proof LHS = RHS

can anyone do this?
4cot2xcosec2x = 4(1-tan^2x/2tanx)(1/2sinxcosx)
= (sec^2x/tanx)(1/sinxcosx) = sec^2x/tanxsinxcosx = sec^2x/sin^2x
= 1/sin^2xcos^2x = sin^2x+cos^2x/sin^2xcos^2x
= sin^2x/cos^2x + cos^2x/sin^2x = tan^2x + cot^2x

Hmm...

EDIT: Nope, 1-tan^2x is not sec^2x.
0
15 years ago
#4
4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).
0
#5
(Original post by Jonny W)
4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).
ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts
0
15 years ago
#6
(Original post by TheWolf)
ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts
Once you do a few and get good at them you'll enjoy unravelling them.
0
15 years ago
#7
(Original post by TheWolf)
...ps i hate all trig p2 questions it takes me longer to do than any other parts
The best advice I ever got was to always get rid of the double angles first. Its a lot harder to work towards double angles.
0
15 years ago
#8
(Original post by JaF)
The best advice I eer got was to alwats get rid of the double angles first. Its a lot harder to work towards double angles.
Yeah you need to decide which side is the most complicated and try and simplify it.
0
15 years ago
#9
(Original post by TheWolf)
ahh ok thanks guys:

ps i hate all trig p2 questions it takes me longer to do than any other parts
Write both sides in terms of powers of sin(x) and cos(x) only. Then if necessary use sin^2 + cos^2 = 1.
0
#10
yea i think i have to do trig questions all day tommorow lol thats how **** i am at it, i get one in two of them wrong or cant do it at all
0
15 years ago
#11
(Original post by TheWolf)
yea i think i have to do trig questions all day tommorow lol thats how **** i am at it, i get one in two of them wrong or cant do it at all
After *attempting* unsuccessfully to do the question you posted, i have realised i can't do trig either.
0
15 years ago
#12
(Original post by Jonny W)
4cot(2x) cosec(2x)
= 4cos(2x) / sin^2(2x)
= [cos^2(x) - sin^2(x)] / [sin^2(x) cos^2(x)]
= cosec^2(x) - sec^2(x)
= [1 + cot^2(x)] - [1 + tan^2(x)]
= cot^2(x) - tan^2(x).
Does sin^2(2x) = sin^2(x)cos^2(x) ?

Which rule is that?
0
15 years ago
#13
(Original post by Silly Sally)
Does sin^2(2x) = sin^2(x)cos^2(x) ?

Which rule is that?
sin^2(2x) = (sin2x)^2 = (2sinxcosx)^2 = 4sin^2xcos^2x, the 4 is cancelled by the 4 outside.
0
15 years ago
#14
(Original post by ZJuwelH)
sin^2(2x) = (sin2x)^2 = (2sinxcosx)^2 = 4sin^2xcos^2x, the 4 is cancelled by the 4 outside.
AHHHH!!! Thanks!!! All unconfuddled now!!!
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