VCE Methods Question Thread!

[TheCommando]

These are exact values. You need to memorise them.

Ah yes

[n__n]

Hey friends, can someone confirm the following for me;

Two simultaneous equations have:
- no solutions when gradient and y intercepts don't equal
- unique (one) solutions when the gradients don't equal
- infinite solutions when both gradient and y intercept equals

Thanks for all help.

[zhen]

Hey friends, can someone confirm the following for me;

Two simultaneous equations have:
- no solutions when gradient and y intercepts don't equal
- unique (one) solutions when the gradients don't equal
- infinite solutions when both gradient and y intercept equals

Thanks for all help.

Actually there are no solutions when the gradients are equal, but there are no intersection points (parallel lines). The rest look fine, but for inifinite solutions I think it's better to think about it as they are the same line, so the gradient, y-int, x-int and any point have to be the same, giving an infinite number of solutions. Hope that helps.

[LPadlan]

Hey friends, can someone confirm the following for me;

Two simultaneous equations have:
- no solutions when gradient and y intercepts don't equal
- unique (one) solutions when the gradients don't equal
- infinite solutions when both gradient and y intercept equals

Thanks for all help.

-no solutions when lines have the same gradient but different equations. eg.(2x+y=5,2x+y=7)
The rest are correct

[TheCommando]

Hey guys why do we add 2pi since sin theta = sin(2pie+theta)
And how did they get 2(pie+theta)

[zhen]

Hey guys why do we add 2pi since sin theta = sin(2pie+theta)
And how did they get 2(pie+theta)

Basically when we add 2π, we're going around the unit circle back to the original angle, since 2π is 360°.
So sin(θ)=sin(2π+θ)=sin(4π+θ)=sin(6π+θ)

[n__n]

Actually there are no solutions when the gradients are equal, but there are no intersection points (parallel lines). The rest look fine, but for inifinite solutions I think it's better to think about it as they are the same line, so the gradient, y-int, x-int and any point have to be the same, giving an infinite number of solutions. Hope that helps.

-no solutions when lines have the same gradient but different equations. eg.(2x+y=5,2x+y=7)
The rest are correct

So essentially when the c in y=mx+c is different?

[zhen]

So essentially when the c in y=mx+c is different?

Yup. No solutions when m is the same but c is different.

[TheCommando]

Hey guys for the second solution here why is it pie-pie/six
I thought it was meant to be + since sin is positive in the second quadrant
https://postimg.org/image/g2xhv8789/

[zhen]

Hey guys for the second solution here why is it pie-pie/six
I thought it was meant to be + since sin is positive in the second quadrant
https://postimg.org/image/g2xhv8789/

π+π/6=7π/6 is in the third quadrant making sine negative. Since we want 1/2, it has to be in the first or second quadrant to make sine positive. π-π/6=5π/6 is in the second quadrant, making sin positive. It also makes an angle of π/6 with the x-axis of the unit circle. So, sin(5π/6)=1/2. If you want to better understand it look at the image attached. The y coordinate of π/6 and 5π/6 are the same. So since sin(π/6)=1/2, then sin(5π/6)=1/2.

[TheCommando]

Pretty sure the first one is wrong. -sin(π/2-θ)=-cos(θ), since sin(π/2-θ) is in the first quadrant and is positive, so the negative of it would be negative.
-cos(π/2-θ)=-sin(θ), similar to first one
sin(π/2+θ) is in the second quadrant which is positive for sine. So sin(π/2+θ)=cos(θ)
cos(π/2+θ) is in the second quadrant, which is negative for cosine, so it's -sin(θ)

This image explains why these identities work if you want to understand it,
Also someone should correct me if I got anything wrong, since I may have made a mistake, which may be why I got a different answer to you.

Hi for this image i a bit confused with the flipping of the triangle on the left so why is sin negative and cos postive

[TheCommando]

π+π/6=7π/6 is in the third quadrant making sine negative. Since we want 1/2, it has to be in the first or second quadrant to make sine positive. π-π/6=5π/6 is in the second quadrant, making sin positive. It also makes an angle of π/6 with the x-axis of the unit circle. So, sin(5π/6)=1/2. If you want to better understand it look at the image attached. The y coordinate of π/6 and 5π/6 are the same. So since sin(π/6)=1/2, then sin(5π/6)=1/2.

could you clarify this Since we want 1/2, it has to be in the first or second quadrant to make sine positie

[zhen]

Hi for this image i a bit confused with the flipping of the triangle on the left so why is sin negative and cos postive

For that it's cos(θ+π/2), which is in the second quadrant, so it's negative. Since cos(θ-π/2)=sin(θ), then
cos(θ+π/2)=-cos(θ-π/2)=-sin(θ).

[zhen]

could you clarify this Since we want 1/2, it has to be in the first or second quadrant to make sine positie

1/2 is positive. First and second quadrant is above the x-axis of the unit circle. So, sine is positive. So to get sin(x)=1/2, x has to be in the first or second quadrant.

[TheCommando]

1/2 is positive. First and second quadrant is above the x-axis of the unit circle. So, sine is positive. So to get sin(x)=1/2, x has to be in the first or second quadrant.

I finally understand now. Actually thank you so much. ❤
Also are u assuming 1/2 as the yvalue since if 1/2 is x on the second quadrant.
Or neither and half is just positive and first and second quadrant is positive as the top section of the graph is positive if your looking at it from the y-axis