C3 Trigonometry Help!

I've seen 2 questions in the text book but i cannot do them!
You have to do something like the left hand side = and then make it come up as the right hand side :confused:

they are:

1) sec²θ(cot²θ-cos²θ ) ≡cot²θ

2)( sin2θ+sinθ )/( 1+cos2θ+cosθ )≡tanθ

Any ideas??

Reply 1

Remarqable M

15

are you revising c3-retake? lol good boy.
hopefully no retake for me..

i think you need to take L.H.S and use the trignometric identities :smile:

try it and think about it.
don't expand the bracket! just try simplfying what is inside the bracket.

Reply 2

notan123

OP

no... just started doing it!
doin whole a level in 1 year

Reply 3

Remarqable M

15

notan123

no... just started doing it!
doin whole a level in 1 year

WoW! thats intensive man! i hope you get the grades you're after :smile:

im just starting c4 next week lol

Reply 4

gottastudy

Re-write sec²θ for question 1.

Reply 5

notan123

OP

then expand with the brackets ?

Reply 6

12ian34

and Re-write sin2θ and cos2θ for question two

Reply 7

12ian34

notan123

then expand with the brackets ?

Try and see what happens.

Reply 8

nushii

1.) LHS:
sec²θ(cot²θ-cos²θ ) ≡cot²θ
1/cos²θ x (cos²θ /sin²θ - cos²θ )
cos²θ /sin²θcos²θ - cos²θ/cos²θ
1/sin²θ – 1
cosec²θ -1 = cot²θ

using identity: sin²θ + cos²θ = 1
Divide by sin²θ would give:
sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
1 + cot²θ = cosec ²θ
cosec²θ – 1 = cot²θ

Reply 9

12ian34

nushii

1.) LHS:
sec²θ(cot²θ-cos²θ ) ≡cot²θ
1/cos²θ x (cos²θ /sin²θ - cos²&#952 :wink:

cos²θ /sin²θcos²θ - cos²θ/cos²θ
1/sin²θ – 1
cosec²θ -1 = cot²θ

using identity: sin²θ + cos²θ = 1
Divide by sin²θ would give:
sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
1 + cot²θ = cosec ²θ
cosec²θ – 1 = cot²θ

Read the rules Mr.

anyway, he said "help" not "answer"

Reply 10

12ian34

nushii

1.) LHS:
sec²θ(cot²θ-cos²θ ) ≡cot²θ
1/cos²θ x (cos²θ /sin²θ - cos²&#952 :wink:

cos²θ /sin²θcos²θ - cos²θ/cos²θ
1/sin²θ – 1
cosec²θ -1 = cot²θ

using identity: sin²θ + cos²θ = 1
Divide by sin²θ would give:
sin²θ/sin²θ + cos²θ/sin²θ = 1/sin²θ
1 + cot²θ = cosec ²θ
cosec²θ – 1 = cot²θ

Read the rules Ms.

anyway, he said "help" not "answer"

Reply 11

nushii

12ian34

Read the rules Mr.

anyway, he said "help" not "answer"

Im a girl and i didnt read it properly Sorry :frown:

Reply 12

notan123

OP

for q2 would i write sin2θ as 2sinθcosθ and cos2θ as cos²θ - sin²θ

Reply 13

12ian34

notan123

for q2 would i write sin2θ as 2sinθcosθ and cos2θ as cos²θ - sin²θ

You're on the right track with the sin2θ, but cos2θ should be written in one of the other ways. There are three ways to expand cos2θ.

This site might help a bit more, but only look if you can't think of another way:

http://library.thinkquest.org/20991/alg2/trigi.html

nushii

Im a girl and i didnt read it properly Sorry :frown:

Ha sorry about that, I edited it about 30 seconds after I posted that comment once i noticed!
Also, about the rules, I mean, you shouldn't really give the whole answer, just nudge them so they figure it out themselves, but don't worry about it!