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C3 Trigonometry Help! watch

1. 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
they are:

1) sec²θ(cot²θ-cos²θ ) ≡cot²θ

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

Any ideas??
2. 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 try it and think about it.
don't expand the bracket! just try simplfying what is inside the bracket.
3. no... just started doing it!
doin whole a level in 1 year
4. (Original post by 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
im just starting c4 next week lol
5. Re-write sec²θ for question 1.
6. then expand with the brackets ?
7. and Re-write sin2θ and cos2θ for question two
8. (Original post by notan123)
then expand with the brackets ?
Try and see what happens.
9. 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
10. (Original post by 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
Read the rules Mr.

anyway, he said "help" not "answer"
11. (Original post by 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
Read the rules Ms.

anyway, he said "help" not "answer"
12. (Original post by 12ian34)
Read the rules Mr.

anyway, he said "help" not "answer"
Im a girl and i didnt read it properly Sorry
13. for q2 would i write sin2θ as 2sinθcosθ and cos2θ as cos²θ - sin²θ
14. (Original post by 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

(Original post by nushii)
Im a girl and i didnt read it properly Sorry
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!
15. (Original post by notan123)
no... just started doing it!
doin whole a level in 1 year
Me too! Well, I'm doing 3 but maths' the nicest (minus D1 that is!)

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