VCE Specialist 3/4 Question Thread!

[peanut]

I don't get how the worked example goes from one last to the next. Could someone explain?

[MightyBeh]

I don't get how the worked example goes from one last to the next. Could someone explain?

First line to second:
|a + bi| = \(\sqrt{a^2+b^2}\); here (x-1) is the 'real' term and 'y' is the imaginary.
Second line to third:
Squaring both sides to change it to a recognisable form.

[peanut]

Is sin^-1(sin(x)) = x for all x?
What about sin(sin^-1(x)) = x ?
Similarly, does this apply to cos and tan as well? I assume it does, but my calculator does not produce that result.

[Syndicate]

Is sin^-1(sin(x)) = x for all x?
What about sin(sin^-1(x)) = x ?
Similarly, does this apply to cos and tan as well? I assume it does, but my calculator does not produce that result.

This does apply to cos and tan as well.




Your calculator will not yield a result, if x is out of its domain.

[tysh]

Is sin^-1(sin(x)) = x for all x?
What about sin(sin^-1(x)) = x ?
Similarly, does this apply to cos and tan as well? I assume it does, but my calculator does not produce that result.

Also worth noting that since sin(x) cycles through between ranges of -1 to 1, a calculator can still spit out a result for arcsin(sin(x)) in the case that x is outside -pi/2 to pi/2 (range of arcsin). However, as Syndicate said, if this is the case then the result will not satisfy arcsin(sin(x)) = x (instead, it will satisfy arcsin(sin(pi/2 + x) = pi/2 - x as long as you plus or minus 2pi to x until -pi<x<pi). This is similar for arccos(cos(x)) and arctan(tan(x)) except for the differences in their respective ranges (i.e. arccos(cos(x))=x is not satisfied outside 0<x<pi, and for arctan(tan(x))=x, outside -pi/2<x<pi/2.

As for sin(arcsin(x)), and cos(arccos(x)), you can only input values between -1 and 1 anyway as this is the domain for arcsin(x). For tan(arctan(x)) = x this is always true since unlike arctan, tan has no restricted range.

[deStudent]

I need help understanding these 2 theorems http://m.imgur.com/a/PCj7b

Image 1: so pretty much the angles created by the lines drawn from 2 certain spots on the arc will always equal each other if the lines meet at the edge of the circle? Is this correct?

Image 2: I don't quite understand what they mean when they say "alternate segment". The angle theta in the triangle isn't in the alternate segment?

Cheers!

[RuiAce]

I need help understanding these 2 theorems http://m.imgur.com/a/PCj7b

Image 1: so pretty much the angles created by the lines drawn from 2 certain spots on the arc will always equal each other if the lines meet at the edge of the circle? Is this correct?

Image 2: I don't quite understand what they mean when they say "alternate segment". The angle theta in the triangle isn't in the alternate segment?

Cheers!

1: Yeah. An alternate name for that theorem is "angles standing on the same arc are equal". Another is "angles subtended by same arc to circumference are equal"

2: Ignore the tangent briefly and focus on the chord AB. To the left, the chord AB draws out the minor segment AB. To the right, the chord AB draws out the major segment AB.

Now put the tangent, and the theta on the left back in. Notice that the theta is to the left of AB. Hence, theta is more or less in the minor segment (although a bit of it is outside the circle).

Now look at the other theta. Clearly, that theta is in the major segment, not the minor segment. Hence, it is indeed the alternate segment (if you're not in the minor segment you're in the major segment).

[deStudent]

Thanks RuiAce, it make sense now. I just kept thinking the alternate segment was only that bit above the 2nd theta (furthest away from the tangent) 

[peanut]

What is the second derivative of sin^-1(x)? The first is (1-x^2)^(-1/2) and using the chain rule the second is x(1-x^2)^(-3/2)?

Following on from this, is x(1-x^2)^(-3/2) = x / sqroot((1-x^2)^3)? My calculator doesn't seem to think so.

[RuiAce]

What is the second derivative of sin^-1(x)? The first is (1-x^2)^(-1/2) and using the chain rule the second is x(1-x^2)^(-3/2)?

Following on from this, is x(1-x^2)^(-3/2) = x / sqroot((1-x^2)^3)? My calculator doesn't seem to think so.

Why ever not?

[Syndicate]

Ah I see, ThnxDid you stop mid sentence? If you draw a triangle of tan^-1(-2/3) doesnt that leave you with theta being in either quadrant 2 or 4? but the ans only gives 1 soln

I had to delete my prior statement (since it was erroneous - really don't know why I said there is a possibility of having two angles  ).

There would only be one solution to this question, as \(tan^{-1} \) is restricted to the range \( (-\frac{\pi}{2}, \frac{\pi}{2} ) \). Although \( tan^{-1}(x) \space dom = \mathbb{R} \), you cannot get multiple angles (like tan(x)) due to it's range.



In the image above, you will be able to see that tan^-1(-2/3) only has one negative solution.

[deStudent]

Could someone clarify something for this question http://m.imgur.com/N7kLunf

I got y=140 which was correct but I tried a couple other methods beforehand which failed, not sure why?

1: I tried using the sum of opposite angles for a cyclic quadrilateral with the angle x. This gave me y=100. The answer used this method but with the 40 degree angle. Idk why using angle X would give me the wrong answer?

2: I split the 40 degree angle in half by drawing a line from that point to y/2. So it's an isosceles triangle. Therefore (180 - 20)/2 = 80, so y = 160. This is wrong. But I used this method to get the correct answer but by splitting angle x in half. (180 - 40)/2, y = 140.

[Syndicate]

Could someone clarify something for this question http://m.imgur.com/N7kLunf

I got y=140 which was correct but I tried a couple other methods beforehand which failed, not sure why?

1: I tried using the sum of opposite angles for a cyclic quadrilateral with the angle x. This gave me y=100. The answer used this method but with the 40 degree angle. Idk why using angle X would give me the wrong answer?

2: I split the 40 degree angle in half by drawing a line from that point to y/2. So it's an isosceles triangle. Therefore (180 - 20)/2 = 80, so y = 160. This is wrong. But I used this method to get the correct answer but by splitting angle x in half. (180 - 40)/2, y = 140.

In an inscribed quadrilateral, two opposite angles on the circumference of the circle add upto 180 degrees. What this means is that y + 40 = 180.

Method 1: you have used the angle in the centre of the circle, which is wrong. You would only use the angle in the centre of the circle when you have a quadrilateral created by two tangents (this angle will be outside of the circle) and two radius lengths.

Method 2: how do you know that two isosceles triangles are created? This is certainly wrong. You may have got the last one right by luck.

Anyways I will be posting up my Circle geometry notes soon (tomorrow maybe?). They helped me quite a lot throughout the year.

[deStudent]

In an inscribed quadrilateral, two opposite angles on the circumference of the circle add upto 180 degrees. What this means is that y + 40 = 180.

Method 1: you have used the angle in the centre of the circle, which is wrong. You would only use the angle in the centre of the circle when you have a quadrilateral created by two tangents (this angle will be outside of the circle) and two radius lengths.

Method 2: how do you know that two isosceles triangles are created? This is certainly wrong. You may have got the last one right by luck.

Anyways I will be posting up my Circle geometry notes soon (tomorrow maybe?). They helped me quite a lot throughout the year.

Thanks buddy.

For your explanation of method one, I'm still a little confused on how the quadrilateral will be formed? Do you have a picture you can illustrate it with.

For method 2, 2 lengths are just the diameter.

[Syndicate]

Thanks buddy.

For your explanation of method one, I'm still a little confused on how the quadrilateral will be formed? Do you have a picture you can illustrate it with.

For method 2, 2 lengths are just the diameter.

The quadrilateral formed with the dotted lines (from the centre of the circle) and the tangents PA and PB.

2: lengths are only diameters when they are going through the centre of the circle.