Prelim 2016 Maths Help Thread (7 Viewers)

InteGrand · Apr 24, 2016

Rathin said:
Apply what rule?

The rule sin(A)cos(A) = (1/2)sin(2A), which is what you used to get up to what you were up to.

Green Yoda · Apr 24, 2016

InteGrand said:
The rule sin(A)cos(A) = (1/2)sin(2A), which is what you used to get up to what you were up to.

I still don't understand, sorry.
Do I do 1/2*1/2*cos(x)sin(x)cos(2x)?

Green Yoda · Apr 24, 2016

Btw is the rule sin(x) = 1/2*sin(x)cos(x)?
I don't seem to completely understand that rule

jathu123 · Apr 24, 2016

applying the double angle formula again to sin(4x), you will get 2sin(2x)cos(2x). You can see that you already have sin(2x)cos(2x). so if you do what you just did before and you will get 1/4 sin(4x)

edit: its just the double angle formula rearranged;
sin(2x)=2sin(x)cos(x)
1/2*sin(2x)=sin(x)cos(x)

InteGrand · Apr 24, 2016

Rathin said:
Btw is the rule sin(x) = 1/2*sin(x)cos(x)?
I don't seem to completely understand that rule

The rule is sin(A)cos(A) = (1/2)*sin(2A).

Green Yoda · Apr 24, 2016

shitz I wrote sin(a)=sin(a)cos(a)/2 in by book lol
I think I got it now

InteGrand · Apr 24, 2016

^ A quick way to check that that can't be right is to note that that would imply that cos(a)/2 = 1 whenever sin(a) is not 0. But that would require cos(a) = 2, which we know can't happen for real a, since cos(a) is between -1 and 1.

Green Yoda · Apr 24, 2016

Just to clarify 1/2*(sin2x)=sin(x)cos(x)
and sin2x=sin(x)cos(x)/2?

InteGrand · Apr 24, 2016

Rathin said:
Just to clarify 1/2*(sin2x)=sin(x)cos(x)
and sin2x=sin(x)cos(x)/2?

These two can't be simultaneously correct identities. The first one is the one that is correct. Rearrange that to obtain: sin(2x) = 2*sin(x)cos(x).

Green Yoda · Apr 24, 2016

Yup got it completely now! Had problems with teacher writing wrong formulas.

Green Yoda · Apr 24, 2016

Another question
prove:
(sin(A)+cos(A)tan(B))/(cos(A)-sin(A)tan(B)) = tan(A+B)

parad0xica · Apr 24, 2016

DIY...

It seems that the availability of help has destroyed your perseverance.

Green Yoda · Apr 24, 2016

I kept note of all the questions I had troubled with throughout the day with maths and post it here (on the prelim maths help forum) for help. Trig is one of my worst topic so I am doing alot of questions to get better at it and whilst doing these questions I come across some questions which I don't get for which I seek help for.

parad0xica · Apr 24, 2016

Rathin said:
I kept note of all the questions I had troubled with throughout the day with maths and post it here (on the prelim maths help forum) for help. Trig is one of my worst topic so I am doing alot of questions to get better at it and whilst doing these questions I come across some questions which I don't get for which I seek help for.

Have you investigated on why you don't get those questions?

Green Yoda · Apr 24, 2016

parad0xica said:
Have you investigated on why you don't get those questions?

Maybe my mathematical ability is not too good :/

Green Yoda · Apr 24, 2016

I have just realised how dumb I am..gg hsc

Green Yoda · Apr 25, 2016

Fk it I'll still ask for help
Prove:
sin(2θ)+1/cos(2θ)=cos(θ)+sin(θ)/cos(θ)-sin(θ)

Paradoxica · Apr 25, 2016

Rathin said:
Fk it I'll still ask for help
Prove:
sin(2θ)+1/cos(2θ)=cos(θ)+sin(θ)/cos(θ)-sin(θ)

Rathin, I'm sure I'm not the first to mention this but you REALLY need to use brackets.

That expression is too ambiguous for me to decipher instantaneously.

That being said...

$$\noindent$ \frac{1+\sin{2\theta}}{\cos{2\theta}} \equiv \frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}} \\ LHS \equiv \frac{\cos^2{\theta}+2\sin{\theta} \cos{\theta}+\sin^2{\theta}}{\cos^2{\theta}-\sin^2{\theta}} \equiv \frac{(\cos{\theta}+\sin{\theta})^2}{(\cot{\theta}-\sin{\theta})(\cos{\theta}+\sin{\theta})}$

And the result follows through from cancellation.

$$\noindent Identities used:$ 1 \equiv \cos^2{\theta}+\sin^2{\theta},\,\sin{2\theta} \equiv 2\sin{\theta}\cos{\theta}$

Green Yoda · Apr 25, 2016

Paradoxica said:
Rathin, I'm sure I'm not the first to mention this but you REALLY need to use brackets.

That expression is too ambiguous for me to decipher instantaneously.

That being said...

$$\noindent$ \frac{1+\sin{2\theta}}{\cos{2\theta}} \equiv \frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}} \\ LHS \equiv \frac{\cos^2{\theta}+2\sin{\theta} \cos{\theta}+\sin^2{\theta}}{\cos^2{\theta}-\sin^2{\theta}} \equiv \frac{(\cos{\theta}+\sin{\theta})^2}{(\cot{\theta}-\sin{\theta})(\cos{\theta}+\sin{\theta})}$

And the result follows through from cancellation.

$$\noindent Identities used:$ 1 \equiv \cos^2{\theta}+\sin^2{\theta},\,\sin{2\theta} \equiv 2\sin{\theta}\cos{\theta}$

I did not think of doing sin^2(θ)-cos^2(θ)
Thank you so much

Would this be better (sin(2θ)+1)/(cos(2θ))=(cos(θ)+sin(θ))/(cos(θ)-sin(θ))?
Legit question, How do i get better at further trig? I dont know why I am so bad...:/

eyeseeyou · Apr 25, 2016

Rathin said:
I did not think of doing sin^2(θ)-cos^2(θ)
Thank you so much
Would this be better (sin(2θ)+1)/(cos(2θ))=(cos(θ)+sin(θ))/(cos(θ)-sin(θ))?
Legit question, How do i get better at further trig? I dont know why I am so bad...:/

Trig is basically algebra, you need practice. Just revise and practice constantly

the first thing you do whenever you see a question is you need to interpret the question and look at what it's trying to ask you to do. Once you know that change it to a more suitable form

Prelim 2016 Maths Help Thread (7 Viewers)

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