The rule sin(A)cos(A) = (1/2)sin(2A), which is what you used to get up to what you were up to.
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Prelim 2016 Maths Help Thread (2 Viewers)
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Green Yoda
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I still don't understand, sorry.
Do I do 1/2*1/2*cos(x)sin(x)cos(2x)?
Green Yoda
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Btw is the rule sin(x) = 1/2*sin(x)cos(x)?
I don't seem to completely understand that rule
I don't seem to completely understand that rule
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)
edit: its just the double angle formula rearranged;
sin(2x)=2sin(x)cos(x)
1/2*sin(2x)=sin(x)cos(x)
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The rule is sin(A)cos(A) = (1/2)*sin(2A).
Green Yoda
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shitz I wrote sin(a)=sin(a)cos(a)/2 in by book lol
I think I got it now
I think I got it now
^ 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
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Just to clarify 1/2*(sin2x)=sin(x)cos(x)
and sin2x=sin(x)cos(x)/2?
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
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Yup got it completely now! Had problems with teacher writing wrong formulas.
Green Yoda
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Another question
prove:
(sin(A)+cos(A)tan(B))/(cos(A)-sin(A)tan(B)) = tan(A+B)
prove:
(sin(A)+cos(A)tan(B))/(cos(A)-sin(A)tan(B)) = tan(A+B)
parad0xica
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DIY...
It seems that the availability of help has destroyed your perseverance.
It seems that the availability of help has destroyed your perseverance.
Green Yoda
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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
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Have you investigated on why you don't get those questions?
Green Yoda
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Maybe my mathematical ability is not too good :/
Green Yoda
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I have just realised how dumb I am..gg hsc
Green Yoda
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Fk it I'll still ask for help
Prove:
sin(2θ)+1/cos(2θ)=cos(θ)+sin(θ)/cos(θ)-sin(θ)
Prove:
sin(2θ)+1/cos(2θ)=cos(θ)+sin(θ)/cos(θ)-sin(θ)
Paradoxica
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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...
And the result follows through from cancellation.
Green Yoda
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I did not think of doing sin^2(θ)-cos^2(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...
And the result follows through from cancellation.
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
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Trig is basically algebra, you need practice. Just revise and practice constantlyI 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...:/
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
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