VCE Specialist 3/4 Question Thread!

[Sayf44]

If tan(a)=1/12, tan(b)=2/5, tan(c)=1/3, where a,b and c are acute angles, show that a+b+c= pi/4.

Someone please explain

[Conic]

Using trig identities we have:





Because we know that , and , it follows that , and , as a,b and c are acute.

Adding these gives . a,b and c are all greater than 0, as they are acute, hence we have .

So we know that , and that , so , as required.

[Sayf44]

Using trig identities we have:





Because we know that , and , it follows that , and , as a,b and c are acute.

Adding these gives . a,b and c are all greater than 0, as they are acute, hence we have .

So we know that , and that , so , as required.

Thank you SO MUCH

[Sayf44]

one more question

Let cosec(a)=b where a € [pi/2, pi]
Find, in terms of a, two values of x in the range of [0,2pi] which satisfy the equation cosec(x) = -b

I love this website. Heaps of thanks to those who answer

[lzxnl]

one more question

Let cosec(a)=b where a € [pi/2, pi]
Find, in terms of a, two values of x in the range of [0,2pi] which satisfy the equation cosec(x) = -b

I love this website. Heaps of thanks to those who answer

Remember how cosec behaves like sine in terms of its symmetry properties? Cosec(pi-x) = cosec(x) and cosec(2pi-x) = -cosec(x)
Using what you know about sin x, cosec(pi+a) and cosec(2pi-a) should both give -b
Note how pi+a, 2pi-a still leave you in the range [0,2pi] for a E [pi/2,pi]

[Sayf44]

Remember how cosec behaves like sine in terms of its symmetry properties? Cosec(pi-x) = cosec(x) and cosec(2pi-x) = -cosec(x)
Using what you know about sin x, cosec(pi+a) and cosec(2pi-a) should both give -b
Note how pi+a, 2pi-a still leave you in the range [0,2pi] for a E [pi/2,pi]

ohhh truue! Thanks!!

[Eugenet17]

does the formula for a rotation about the y-axis only apply if the region is bounded by the y-axis?

Also, does the formula for regions not bounded by the x-axis also apply to regions not bounded by y-axis?

[Sayf44]

Sorry eug I don't know the answer to that question.

If someone can please answer my question after eug's

Consider the function with the rule :
f(x)= pi/2 + arccos(2x)

Let g(x)= 2cosec(f(x))

a. Show that g(x)= 1/x         < This was simple but the next one..
 
b. State the domain and range for g(x)

[keltingmeith]

Euge - you're being very wordy there. Try drawing what you're trying to do. Remember - what you're trying to do when you apply your formula (V = pi * integral(x^2 dy) or integral(y^2 dx)) is find the area between a curve and the axis on the end of the integral, knowing that there's a bunch of maths you don't understand taking place that means you're finding something called a "volume of the solid of revolution". Drawing things out should hopefully answer your question for you - because you're not expected to understand the maths here, you're just expected to apply it.

Say - very interesting - applications of methods questions, here. So, for this compound function to exist, the number that comes out of f(x) needs to be able to go into g(x) - that is, the range of f(x) needs to be a subset of g(x). ran_f = [pi/2, 3pi/2] which is contained in g(x)

So, true!

A result of this means that the domain of g(x) is all the x values of f(x) (because all the x values you put in g(x) is subsequently all the x values you put in f(x)), so dom_g(x) you'd find as you normally would:

-1 <= 2x <= 1
-1/2 <= x <= 1/2

For the range, you know that g(x) = 1/x, and that the dom_g = [-1/2, 1/2], so graph that and see what comes out.

[Sayf44]

Euge - you're being very wordy there. Try drawing what you're trying to do. Remember - what you're trying to do when you apply your formula (V = pi * integral(x^2 dy) or integral(y^2 dx)) is find the area between a curve and the axis on the end of the integral, knowing that there's a bunch of maths you don't understand taking place that means you're finding something called a "volume of the solid of revolution". Drawing things out should hopefully answer your question for you - because you're not expected to understand the maths here, you're just expected to apply it.

Say - very interesting - applications of methods questions, here. So, for this compound function to exist, the number that comes out of f(x) needs to be able to go into g(x) - that is, the range of f(x) needs to be a subset of g(x). ran_f = [pi/2, 3pi/2] which is contained in g(x)

So, true!

A result of this means that the domain of g(x) is all the x values of f(x) (because all the x values you put in g(x) is subsequently all the x values you put in f(x)), so dom_g(x) you'd find as you normally would:

-1 <= 2x <= 1
-1/2 <= x <= 1/2

For the range, you know that g(x) = 1/x, and that the dom_g = [-1/2, 1/2], so graph that and see what comes out.

Thanks! you made it really simple, didnt think of methods and thank you so much i understand the range part totally now . But one more thing, i graphed the function on the calculator and y=0 is an asymptote, and so x= 0 is not included. Is there a rule here? or is it that we should know x=0 would not be included in 1/x?

[keltingmeith]

Honestly, I'd just be thinking of the graph of 1/x, which you should know has an asymptote at x=0 and y=0. Once again, methods stuff. They can assume you know that stuff, which textbooks especially like to do.

[liamh96]

Hi guys,
how do you integrate (2x+1)/(x+3) ?

Thanks in advance
Liam.22

[b^3]

Expand it out first using long division or this little trick. This turns it into a form that is easier to work with.

What I've done above is to try and get the factor that we have in the denominator, in the numerator. Since we have the coefficient of in the denominator being 1, we take the factor out of the top to reform this, then by adding and subtracting the same constant we can obtain this factor without altering the actual meaning of the expression. Pulling the leftover terms out then allows us to simplify the fraction.

Then you can integrate as you would normally do

[liamh96]

Thanks b^3, this makes sense now.

[Sayf44]

Given that the domain of sin(x) is restricted [-pi/2 , pi/2], find the implied domain and range of f(x) = sin[2cos^-1(x)]

Someone please help with this, thanks. I know it involves composite functions, but the domain of sin(x) is also restricted, and I'm really confused.