VCE Specialist 3/4 Question Thread!

[Cort]

If someone doesn't mind giving me a hand or some videos regarding this, but:
How on earth do you figure out the symmetry/complementary properties of trigonometric functions and its reciprocals beyond the first quadrant? I found one video that explains it perfectly for the first quadrant, but when the angles become more obtuse I do not have the faintest clue.

Attached is shown. Secondly, I would also like to ask if knowing such angles off the top of your end is necessary in Specialist Mathematics. How often does it get tested in the examination? How important is it? Would it be fine to just copy it down and snag a peak at it if necessary?

Thanks, Cort.

[brightsky]

consider the unit circle. select a point in the first quadrant on the circumference of the unit circle. connect this point to the origin (i.e. centre of the unit circle). let the angle between the radius that you just drew and the positive direction of the x-axis be θ. then the point on the circumference of the unit circle would have coordinates (cosθ, sinθ). pay careful attention to the implications of this; this means that cosθ corresponds to the x-coordinate, and sinθ corresponds to the y-axis. these are the unit circle definitions of sine and cosine.

with the unit circle definitions of sine and cosine in mind, you can derive all trigonometric identities. consider the following question:

simplify sin(pi - θ).

okay, what do we do? first of all let's draw the angle pi - θ. if θ is in the first quadrant, then pi - θ should be in the second quadrant. the point on the circumference is (cos(pi - θ), sin(pi - θ)). let us focus on the y-coordinate. physically draw in this y-coordinate. what do you notice about the value of this y-coordinate, and that of the y-coordinate of the original point (cos(θ), sin(θ))? that's right, the values are exactly the same. how do we know that? congruent triangles.

so we conclude that sin(pi - θ) = sin(θ). all the symmetry properties, etc. can be proven in a similar way.

[Cort]

consider the unit circle. select a point in the first quadrant on the circumference of the unit circle. connect this point to the origin (i.e. centre of the unit circle). let the angle between the radius that you just drew and the positive direction of the x-axis be θ. then the point on the circumference of the unit circle would have coordinates (cosθ, sinθ). pay careful attention to the implications of this; this means that cosθ corresponds to the x-coordinate, and sinθ corresponds to the y-axis. these are the unit circle definitions of sine and cosine.

with the unit circle definitions of sine and cosine in mind, you can derive all trigonometric identities. consider the following question:

simplify sin(pi - θ).

okay, what do we do? first of all let's draw the angle pi - θ. if θ is in the first quadrant, then pi - θ should be in the second quadrant. the point on the circumference is (cos(pi - θ), sin(pi - θ)). let us focus on the y-coordinate. physically draw in this y-coordinate. what do you notice about the value of this y-coordinate, and that of the y-coordinate of the original point (cos(θ), sin(θ))? that's right, the values are exactly the same. how do we know that? congruent triangles.

so we conclude that sin(pi - θ) = sin(θ). all the symmetry properties, etc. can be proven in a similar way.

Sorry, if it's in the second quadrant, won't cos be negative? It is in the negative x direction after all, right? So something like:
sin(pi-o)
would be (-cos(pi-0,sin(pi-0)? Similarly, can we also 'connect' this dot even if it's in like, the 3rd quadrant? Excuse the questions, but I also wanted to clarify the physical connection, just wondering if the drawing below is correct.

Thank you, nonetheless, you cleared up my butts. If cos is negative in the 2nd quadrant, then use the coordination system (eg: -cosθ,-sinθ in 3rd quadrant) would explain why so.

Cort.

[b^3]

Sorry, if it's in the second quadrant, won't cos be negative? It is in the negative x direction after all, right? So something like:
sin(pi-o)
would be (-cos(pi-0,sin(pi-0)? Similarly, can we also 'connect' this dot even if it's in like, the 3rd quadrant? Excuse the questions, but I also wanted to clarify the physical connection, just wondering if the drawing below is correct.

Thank you, nonetheless, you cleared up my butts. If cos is negative in the 2nd quadrant, then use the coordination system (eg: -cosθ,-sinθ in 3rd quadrant) would explain why so.

Cort.

Just whipped up something that might help you visualise it a bit more. Depending on what you want to do move the slider or change the values of the angles and .
https://www.desmos.com/calculator/9hoemfupd1
Hope it helps

[Cort]

Just whipped up something that might help you visualise it a bit more. Depending on what you want to do move the slider or change the values of the angles and .
https://www.desmos.com/calculator/9hoemfupd1
Hope it helps

Thank you, both of you. Sometimes I'm beginning to think it was a bad choice for me to do spesh.

[survivor]

The solutions of the quadratic equation z^2 + pz + q = 0 are 1+i and 4+3i. Find the complex numbers p and q.

Thanks

[Stick]

The solutions of the quadratic equation z^2 + pz + q = 0 are 1+i and 4+3i. Find the complex numbers p and q.

Thanks

Since we are dealing with a quadratic equation here, we know that there are only two solutions for . They both happen to be provided for us. There's a couple of ways we could deal with this problem, but I'm basically going to take the solutions and go backwards to determine the original quadratic equation.

Look if you get stuck

Solutions: and











and

[LOLs99]

The solutions of the quadratic equation z^2 + pz + q = 0 are 1+i and 4+3i. Find the complex numbers p and q.

Thanks

Since 1+i and 4+3i are the roots. Therefore the factor would be (z-1-i) and (z-4-3i)
Expand (z-1-i)(z-4-3i).  (z-1-i)(z-4-3i)= z^2-4z-3iz-z+4+3i-iz+4i-3.
Group according to the power of z. So it becomes z^2+(-5-4i)z+(7i+1)

Finally you can just equate the coefficient: p= -5-4i and q= 7i+1

EDIT: Beaten by Stick

[Cort]

While going through the revision for essentials on circular functions, I came across this. I don't think I ever encountered this before during my time in the methods, or I should have. Nonetheless, it's new stuff for me. Is it related to the pythagorean identity at all?

Thanks,
Cort.

[nhmn0301]

While going through the revision for essentials on circular functions, I came across this. I don't think I ever encountered this before during my time in the methods, or I should have. Nonetheless, it's new stuff for me. Is it related to the pythagorean identity at all?

Thanks,
Cort.

As I rmb, it does cover in methods, but not too much. Yup, what most people do is using "Triangle method" as I believe. Imagine you have a triangle, tan(theta)= 4/3. You do know that tan = opp/adj. So here, the opposite side is 4 and adjacent side is 3. YOu can work out the hypotenuse now, using Pythagoras's theorem, it should equal 5. Then looking at this triangle, you can point out exactly what does sin equal, sin = opp/hypo, which means sin=4/5.  This is a pretty easy example but it just gives you a general idea of how does method works.
Hope this help, let me know if there're any errors!

[fadzsta1]

Hey guys, I've been doing hundreds of these questions, but for some reason I can't get my head around this one.

If and , use the double angle formulae to find:

i)
ii) , and hence
iii)

Thanks

[brightsky]

find cosA, then use the formula cosA = 1 - 2sin^2(A/2) = 2cos^2(A/2) - 1. after you've find sin(A/2) and cos(A/2), find the quotient of the two to obtain tan(A/2).

[fadzsta1]

Ah didn't look at those two formulae. Silly me. Thanks!!

[fadzsta1]

Sorry again, but regarding the same question I posted above:

= 4/5. Why is it positive? Is it because A is in the third quadrant, thus A/2 would be in the second quadrant, hence positive?

If this is true, then is -3/5 because it's not in the fourth quad. Is this correct? Do I account for the change in the angle ie. A/2 or do I still keep in mind the domain of the original angle A?

Thanks again,
PS. totally not spamming this thread, promise

[brightsky]

yes that is correct. pi < A < 3pi/2, so pi/2 < A/2 < 3pi/4. this means that A/2 is in the 2nd quadrant, in which sin is positive and cosine is negative.