VCE Specialist 3/4 Question Thread!

[brightsky]

hint: convert to polar form and then use de moivre's theorem. you may need to make use of some trig identities.

[bovawatkins]

Can anyone help with this trig question....

Given that tan(2x)=((4*sqrt(2))/7), where xE[0,pi/4), find the exact value of sin(x)

thanks

[rhinwarr]

How would you convert a term with tan(x) into polar form?

What is the required solution?

To convert to polar form: (not sure if it is right)


Polar form:

That to the power of 5 is:

[rhinwarr]

Given that tan(2x)=((4*sqrt(2))/7), where xE[0,pi/4), find the exact value of sin(x)

Using a triangle you can determine the value of cos(2x):
opp=4sqrt2, hyp=9, adj=7

Using double angle formula:





(must be positive because of the restricted domain)

[bovawatkins]

Using a triangle you can determine the value of cos(2x):
opp=4sqrt2, hyp=9, adj=7

Using double angle formula:





(must be positive because of the restricted domain)

wow thanks thats a lot easier than i thought.....

[Robert123]

Having problems with this partial fractions question. Any help is welcome

Given the function f(x)=A/(x^2-(a+b)+ab), state the equations if the asymptotes and find the turning point, stating the domain.

[alchemy]

Having problems with this partial fractions question. Any help is welcome

Given the function f(x)=A/(x^2-(a+b)+ab), state the equations if the asymptotes and find the turning point, stating the domain.

I think I may have a method. What are the answers? I am able to get the turning point as (0,A/(-(a+b)+ab)). Asymptote as 0. Domain R. Is that correct? Sorry my method isn't very rigorous, so I'm might be wrong. Let me know of the provided answers, and I'll come up with a proper method (:

[Nato]

What would be the most efficient way to sketch the following:

over the interval or even any kinds of trig graphs in general!!  ?

[lzxnl]

What would be the most efficient way to sketch the following:

over the interval or even any kinds of trig graphs in general!!  ?

1. Find asymptotes
2. Find endpoints
3. Find y intercepts (no x intercepts here)
4. Find turning points
5. Remember general shape
6. You're done

That's how I'd draw almost any graph, although some graphs don't have general shapes that are easy to use.

[alchemy]

Having problems with this partial fractions question. Any help is welcome

Given the function f(x)=A/(x^2-(a+b)+ab), state the equations if the asymptotes and find the turning point, stating the domain.

Just clarifying, this is to be solved for all real numbers right?

[dovibaker]

Need a little help with finding the double derivative of 3tan(x-4). Can you please show the working out

[alchemy]

Need a little help with finding the double derivative of 3tan(x-4). Can you please show the working out

The double derivative is exactly what it sounds like. It's the "derivative of the derivative". In other words, it tells us how much the rate of change is changing!
The derivative of 3tan(x-4) is 3sec^2(4-x), by applying the chain rule d/dx[tan(g(x))] = (g'(x)sec^2)*g(x).
The derivative of 3sec^2(4-x) is going to be the double derivative. Applying the chain rule of d/dx[sec(g(x))]=g'(x)sec(g(x))tan(g(x)), we get (-6sec^2)(4-x)tan(4-x)).
Finally, we can neatly write this as (d^2)/(dx^2) (3tan(x-4)) = -6tan(4-x)(sec^2)(4-x).

[dovibaker]

The double derivative is exactly what it sounds like. It's the "derivative of the derivative". In other words, it tells us how much the rate of change is changing!
The derivative of 3tan(x-4) is 3sec^2(4-x), by applying the chain rule d/dx[tan(g(x))] = (g'(x)sec^2)*g(x).
The derivative of 3sec^2(4-x) is going to be the double derivative. Applying the chain rule of d/dx[sec(g(x))]=g'(x)sec(g(x))tan(g(x)), we get (-6sec^2)(4-x)tan(4-x)).
Finally, we can neatly write this as (d^2)/(dx^2) (3tan(x-4)) = -6tan(4-x)(sec^2)(4-x).

Thanks so much calculus is still new to me and a little confusing at times!

[lzxnl]

Having problems with this partial fractions question. Any help is welcome

Given the function f(x)=A/(x^2-(a+b)+ab), state the equations if the asymptotes and find the turning point, stating the domain.

It's not a partial fractions question.
Asymptotes => denominator zero
x^2 -(a+b) + ab = 0
x^2 = a+b-ab
etc

Turning point of f(x) must be turning point of 1/f(x) and vice versa
So turning point of x^2 - (a+b) + ab is just at x=0, so (0, -(a+b) + ab)

[alchemy]

Thanks so much calculus is still new to me and a little confusing at times!

VTextbook has some good introductory videos on Calculus if you're interested.