VCE Methods Question Thread!

[TheCommando]

For exact values how do i know where say 3pie over 4 is
How do i kmow where yheese things are on the unit circle

[zhen]

For exact values how do i know where say 3pie over 4 is
How do i kmow where yheese things are on the unit circle

For the unit circle, the first quadrant is 0 to π/2. Second quadrant is π/2 to π. Third quadrant is π to 3π/2. Fourth quadrant is 3π/2 to 2π. Since 3π/4 is between π/2 to π, it's in the second quadrant. You could also think about it in degrees. First quadrant is 0° to 90°, second is 90° to 180°, third quad is 180° to 270° and fourth quadrant is 270° to 360°. So 3π/4 would be 135°, which is in the second quadrant. Hope that helps.

[LPadlan]

For f(x)=3x^2 + x -2, find:
{x:f(x)=0}

[zhen]

For f(x)=3x^2 + x -2, find:
{x:f(x)=0}

Someone should correct me if I'm wrong, but I'm pretty sure that's just asking for the x values when f(x)=0. So, you just solve 3x^2+x-2=0
By factorising you get (x+1)(3x-2)=0
So, x=-1 and x=2/3

[Shadowxo]

Someone should correct me if I'm wrong, but I'm pretty sure that's just asking for the x values when f(x)=0. So, you just solve 3x^2+x-2=0
By factorising you get (x+1)(3x-2)=0
So, x=-1 and x=2/3

You're right. The question is asking "find X such that f(X)=0" (":" means such that). It's just finding the X intercepts, where f(X)=0.

[scanz_]

Hey guys I know this must be an easy question for those who are doing 3/4 but I can't get my head around this:

Given (8x-11) and (x-3) are factors of
Fully factorise P(x)

PS: Usually when I'm given only 1 factor I just do the long division method, but for this question, the worked solutions from the solution manual looks like this:





This next step and onward is where I lose track of what just happened:





Any help is appreciated!

[LPadlan]

f: (-oo,3] -->R ,f(x)= 3-x and g: R--->R, g(x) = x^2-1

Define a restriction g* of g such that fog* is defined and find fog*

[zhen]

Hey guys I know this must be an easy question for those who are doing 3/4 but I can't get my head around this:

Given (8x-11) and (x-3) are factors of
Fully factorise P(x)

PS: Usually when I'm given only 1 factor I just do the long division method, but for this question, the worked solutions from the solution manual looks like this:





This next step and onward is where I lose track of what just happened:





Any help is appreciated!

Basically, you try to find when (ax)(8x^2)=-8x^3 and when 33b=99, since when you expand it out, the only values that have no x are 33 and b and the only values that multiply to give an x^3 are ax and 8x^2. That gives a=-1 and b=3, so ax+b=-x+3.

[TheCommando]

For the unit circle, the first quadrant is 0 to π/2. Second quadrant is π/2 to π. Third quadrant is π to 3π/2. Fourth quadrant is 3π/2 to 2π. Since 3π/4 is between π/2 to π, it's in the second quadrant. You could also think about it in degrees. First quadrant is 0° to 90°, second is 90° to 180°, third quad is 180° to 270° and fourth quadrant is 270° to 360°. So 3π/4 would be 135°, which is in the second quadrant. Hope that helps.

Yeah but why is 3pie over 4 in the secnd quadrant
I can now understand it and i think about it by if its 3pie over 3 that would mean its greater than pie, however since the denminator is bigger than the numerator this means the value is lesser than pie which which means if the denominator is greater than the numerator it cant go past pie. And and if the denominator in this case is 6 or greater than six it will be in the first quadrant

[TheCommando]

Dumb question i know but how do i siove these? The questions on the book arefp different to the questions ive been taught
http://imgur.com/a/daNc2

[scanz_]

Basically, you try to find when (ax)(8x^2)=-8x^3 and when 33b=99, since when you expand it out, the only values that have no x are 33 and b and the only values that multiply to give an x^3 are ax and 8x^2. That gives a=-1 and b=3, so ax+b=-x+3.

Thanks Zhen 

[zhen]

Yeah but why is 3pie over 4 in the secnd quadrant
I can now understand it and i think about it by if its 3pie over 3 that would mean its greater than pie, however since the denminator is bigger than the numerator this means the value is lesser than pie which which means if the denominator is greater than the numerator it cant go past pie. And and if the denominator in this case is 6 or greater than six it will be in the first quadrant

The easiest way to think about it is that the second quadrant is from π/2 to π. π/2=2π/4 and π=4π/4, so 3π/4 is between π/2 and π, as 2<3<4. So it's in the second quadrant. So the easiest way is to just get them all on the same denominator.

[zhen]

Thanks Zhen 

No problems. Happy to help. 

[zhen]

Dumb question i know but how do i siove these? The questions on the book arefp different to the questions ive been taught
http://imgur.com/a/daNc2

Like my teacher always says, no maths question is dumb. I'll just start you off with the first few.
a) There's an identity that says cos(-x)=cos(x) and sin(-x)=-sin(x). So, cos(-a)=cos(a)=0.6
b) sin(π/2+a)=cos(a), which is another trig identity. Basically sin(π/2-a)=sin(π/2+a)=cos(a) because they are the same triangle on the unit circle. For these questions just be careful of the quadrants and signs. So, the answer is sin(π/2+a)=cos(a)=0.6
c) tan(-θ)=-tan(θ)=-0.7
d) cos(π/2-x)=sin(x)=0.3
I may have made a mistake so someone else should check, but from this, you should be able to do the rest.

[TheCommando]

Like my teacher always says, no maths question is dumb. I'll just start you off with the first few.
a) There's an identity that says cos(-x)=cos(x) and sin(-x)=-sin(x). So, cos(-a)=cos(a)=0.6
b) sin(π/2+a)=cos(a), which is another trig identity. Basically sin(π/2-a)=sin(π/2+a)=cos(a) because they are the same triangle on the unit circle. For these questions just be careful of the quadrants and signs. So, the answer is sin(π/2+a)=cos(a)=0.6
c) tan(-θ)=-tan(θ)=-0.7
d) cos(π/2-x)=sin(x)=0.3
I may have made a mistake so someone else should check, but from this, you should be able to do the rest.

Thank you! But do u know why exactly cos (-x) = cos(x) and elepaborate on what you mean by identity