VCE Specialist 3/4 Question Thread!

[pi]

Use the double angle formula for tan2x, and the fact that tan(pi/4)=1, to find the exact value of tan(pi/8).
Having a bit of trouble with this one; how would I go about doing this?

Hint: let 2x = pi/4 and then rearrange and algebraically have a go at it

[zvezda]

Use the double angle formula for tan2x, and the fact that tan(pi/4)=1, to find the exact value of tan(pi/8).
Having a bit of trouble with this one; how would I go about doing this?

Hint: let 2x = pi/4 and then rearrange and algebraically have a go at it

that's sneaky. Thanks for that

[zvezda]

Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

[pi]

Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

Hint: try rearranging inside the sin and cos

Spoiler

sin(theta) + i cos(theta)
= cos(pi/2 - theta) + i sin(pi/2 - theta)    <-- standard trig rearranging*
= cis(pi/2 - theta)
QED



*Proof:
sin(a-b) = sin(a)*cos(b) - sin(b)*cos(a)
Let a =  pi/2
sin(pi/2 - b) = sin(pi/2)*cos(b) - sin(b)* cos(pi/2)
sin(pi/2 - b) = 1*cos(b) - sin(b)*0
sin(pi/2 - b) = cos(b)

[zvezda]

Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

Hint: try rearranging inside the sin and cos

Spoiler

sin(theta) + i cos(theta)
= cos(pi/2 - theta) + i sin(pi/2 - theta)    <-- standard trig rearranging*
= cis(pi/2 - theta)
QED



*Proof:
sin(a-b) = sin(a)*cos(b) - sin(b)*cos(a)
Let a =  pi/2
sin(pi/2 - b) = sin(pi/2)*cos(b) - sin(b)* cos(pi/2)
sin(pi/2 - b) = 1*cos(b) - sin(b)*0
sin(pi/2 - b) = cos(b)

had to look at the spoiler lol. You had (pi/2 - theta) instead of (theta - pi/2) in your working though? Am I missing something? Or did you just misread? :p

[pi]

had to look at the spoiler lol. You had (pi/2 - theta) instead of (theta - pi/2) in your working though? Am I missing something? Or did you just misread? :p

You're right, I did + instead of -, if you do - you just "switch" what's inside cis due to cos(-x)=cos(x) and sin(-x)=-sin(x) Sorry about that haha

[Jenny_2108]

Another question:
show that sin(theta) - icos(theta) = cis(theta-pi/2)

[polar]

just another way of looking at the question:

[zvezda]

thanks for the help guys

[#1procrastinator]

Show that there is no intersection point of the line y=x+5 and the ellipse x^2+y^2/4=1

Is it a valid method to square the linear equation, equate the two and use the discrimnant to show that the resulting quadratic has no solutions? The book subs the linear equation into the equation of the ellipse and shows that the squared term cannot be equal to the negative term.

[pi]

You can't just square one equation afaik.

[Truck]

Show that there is no intersection point of the line y=x+5 and the ellipse x^2+y^2/4=1

Is it a valid method to square the linear equation, equate the two and use the discrimnant to show that the resulting quadratic has no solutions? The book subs the linear equation into the equation of the ellipse and shows that the squared term cannot be equal to the negative term.

If you square each side you'd have to multiply it so that it's y^2=(x+5)^2, not y^2=x^2 + 5^2

In future these questions will get harder - you'll have 2 simultaneous equations in terms of 3 variables, i.e. (easy start) y=kx + 2 and y=3x+k, and the question will be "find the values of k for there to be no solutions, infinite solutions or 1 unique solution".

[paulsterio]

If you square each side you're not actually multiplying each side by the same number - i.e. you're multiplying the LHS by "y" and the R.H.S. by (x+5), so although y=x+5, y^2=/=(x+5)^2.

What? y and x+5 are the same number, however.

[Truck]

If you square each side you're not actually multiplying each side by the same number - i.e. you're multiplying the LHS by "y" and the R.H.S. by (x+5), so although y=x+5, y^2=/=(x+5)^2.

What? y and x+5 are the same number, however.

Derp, I think I/he meant to say y^2=x^2+5^2 (so you times each individual item by their values).

Ignore what I said before haha, y^2=(x+5)^2, but y^2=/=x^2 + 5^2. 2 weeks out of school and my brain is already done lol

[Homer]

How would i find the asymptote for the hyperbola with equation