hyunah's Specialist Questions

[hyunah]

how do i know that tan(pi/2 - x) = cot (x) ?
how would i prove these complementary angles. i'm struggling to remember them... any tips please?

[Rectophobia]

You can reason it by saying that sin(pi/2 - x) = cos(x) and cos(pi/2 - x) = sin(x) as they are complimentary.

It would then follow that tan(pi/2 - x) = [sin(pi/2 - x)/cos(pi/2 - x)] = cos(x)/sin(x) = cot(x).

Other than this rough proof, the only one I can come up with involves expressing sine and cosine in terms of their exponential form. I've attached this proof, but it isn't needed for VCE.

P.S. Sorry for the shitty effects on the camera. The image wasn't perfect so I had to distort it in order for all the lettering to be visible.

P.S.S. I just realised That you could do it in the following way:

tan(pi/2 - x)

= sin(pi/2 - x)/cos(pi/2 - x)

= [sin(pi/2)cos(x) - cos(pi/2)sin(x)]/[cos(pi/2)cos(x) + sin(pi/2)sin(x)]           (by compound angle formula)

= cos(x)/sin(x)           (after simplification)

=cot(x)

[hyunah]

thanks XD

[hyunah]

how do i diff sin^-1(x)

isn't it just
u =1 u' = 0
v = sin(x) v' = cos(x)
dy/dx = (v'u - u'v)/v^2
?

[keltingmeith]

Not even close. sin^-1 is inverse sine, not 1/sin. Consult your textbook and/or formula sheet on how to diff it.

[hyunah]

oh cause normal when i find a function eg (1/x)^-1 isn't it x?
i guess its a different story with sin?

[keltingmeith]

Yep - circular functions are just stupid like that. Their reciprocal are given entirely different names, which you should be aware of of you're doing spec - sec (for cos), cosec (for sin) and cot (for tan).

[hyunah]

yes thank you

[hyunah]

Find the particular solution. Of f’(x) = sqroot(1-y^2) if f(0) = ˝
I got y = sin(x+pi/6)
Is the largest domain R
The second part of the question (whats the largest domain) is worth 3 marks… am I missing something?
Cause with the question after it
It says to find the general solution of f’(x) = 2/(7-4y)
Which I got x = 7/2y – y^2 +c
And part b worth also 3 marks in this question is to find the largest domain
I had to go y = ____________
Then have whatever in the sqroot >0 and solve for x to get the largest domain being x<(16c+49)/16
I get why that is worth 3 marks, but in the other one… just writing R because the domain of sin(x+pi/6) is R seems like im missing something?
Am I right?

[hyunah]

Find the particular solution. Of f’(x) = sqroot(1-y^2) if f(0) = ˝
I got y = sin(x+pi/6)
Is the largest domain R
The second part of the question (whats the largest domain) is worth 3 marks… am I missing something?
Cause with the question after it
It says to find the general solution of f’(x) = 2/(7-4y)
Which I got x = 7/2y – y^2 +c
And part b worth also 3 marks in this question is to find the largest domain
I had to go y = ____________
Then have whatever in the sqroot >0 and solve for x to get the largest domain being x<(16c+49)/16
I get why that is worth 3 marks, but in the other one… just writing R because the domain of sin(x+pi/6) is R seems like im missing something?
Am I right?

[hyunah]

can someone please help with question 13c)
I thought I could just do vectorGP . vectorGQ but that doesn't equal 0?

[keltingmeith]

can someone please help with question 13c)
I thought I could just do vectorGP . vectorGQ but that doesn't equal 0?

You're completely right there - maybe double check you haven't made an arithmetic error?

[Conic]

As EulerFan said, it is most likely an arithmetic error that is causing your problem. For the vectors a and b you should have:

Spoiler

You can find the vectors GP and GQ directly using the previous part and then find the dot product:

Spoiler

You can also expand the expression with GP and GQ in terms of the vectors a and b and then substitute in the values:

Spoiler

                 

                 

                 





                     

                     

[hyunah]

thank you conic

is that based on the assumption that p is equal to OP which is equal to a and that q is equal to OQ which is equal to b?

cause the solution had something complex like this which i ddin't get....

[Conic]

is that based on the assumption that p is equal to OP which is equal to a and that q is equal to OQ which is equal to b?

cause the solution had something complex like this which i ddin't get....

Yes, they said  "Denote vectors OP and OQ by a and b respectively", so as and , you can get the vectors a and b.

The solutions are using assumed geometry knowledge from the study design.



They are basically showing that G, P and Q lie on a circle around M, and that PQ is the diameter of a circle. As in the image above, this means that PGQ is a right angle. This approach is correct, but it is a lot easier to use dot products.