Maths Methods 3/4 Help Thread 2011

[nacho]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

Since sin is positive in the first quad, then cos must be negative according to the equation. Sin and cos can only produce the same value at the same angle at pi/4 (hope you get what I mean )
So b must be pi/4
Which means -sin(pi/4) = negative square root 2/2
So a = 5pi/4
Therefore a = b + pi

in regards to that, is that formal/mathematical enough for VCAA?
I've always wondered how to set out these types of questions too, despite knowing how to obtain the answer! annoying.

[Andiio]

The answer is -b-pi/2!

[luffy]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

Sorry, doing my own exam prep so the following explanation will be dodgy/rough.



Draw the graphs of and and you will find a basic vertical translation of pi. Hence,



Draw a right-angled triangle. Let one angle = b, opposite side length = x and the hypotenuse = 1. You will find that the inverse cosine of this length will simply equal . Hence,







This contradicts with one of the previous answers. So, sorry if I am wrong.

EDIT: Also, I just noticed this is the methods board xD. Hence, to answer Andiio's question, this is not going to be the "best" way to do the question. There will be an easier way, but it was the first one that came to my mind.

[Andiio]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

Sorry, doing my own exam prep so the following explanation will be dodgy/rough.



Draw the graphs of and and you will find a basic vertical translation of pi. Hence,



Draw a right-angled triangle. Let one angle = b, opposite side length = x and the hypotenuse = 1. You will find that the inverse cosine of this length will simply equal . Hence,







This contradicts with one of the previous answers. So, sorry if I am wrong.

EDIT: Also, I just noticed this is the methods board xD. Hence, to answer Andiio's question, this is not going to be the "best" way to do the question. There will be an easier way, but it was the first one that came to my mind.

Interesting method! :O Just curious though, did you already know the general shape of the cos^-1(sin(b)) graph?

[luffy]

Interesting method! :O Just curious though, did you already know the general shape of the cos^-1(sin(b)) graph?

I didn't know the general shape. In fact, I didn't sketch it. I sketched and and found a relationship between them. This relationship would hence hold true for cos^-1(-sin(b)) if that makes sense.

[luken93]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

I'm gonna take a punt and say you need to use properties of trig functions for this.

Firstly,

and the domain will now be

Also,









How does that sound?

[b^3]

Yeh that looks good luken. I went to give to give it a stab that way and couldn't get it to work. I don't know why but I always use the wrong sign when you do sin(pi/2 +x) e.t.c

It's a neater method too.

[nacho]

isn't it
sin(pi/2 - x) = cos(x) ?

[b^3]

isn't it
sin(pi/2 - x) = cos(x) ?

Hey maybe I did have it right. Yeh that sounds right, otherwise you are saying cos is +ve in the second quadrant.

[Andiio]

Reckon a question like that is likely in Exam 1?

[Greatness]

Reckon a question like that is likely in Exam 1?

That's what i was thinking... But it would kind of dog out 75% of the state hahah

[luken93]

isn't it
sin(pi/2 - x) = cos(x) ?

yep, because
sin(pi/2 - x) = sin(-(x - pi/2)) which is a reflection in the y axis and then a pi/2 shift right, which then makes it cos(x)
sin(pi/2 + x) is a pi/2 shift left, which also equals cos(x)

btw, I didn't know it off by heart, I had to look it up, but I reckon it's the only reasonable method to doing it within the methods course...

Reckon a question like that is likely in Exam 1?

I hope not.

[nacho]

Whn specifying domains for things invilvig infinity,
for example R+ , but we want to use the other notation:
(0, infinity) do we have to place a '+' sign in front of infinity?

[Panicmode]

Whn specifying domains for things invilvig infinity,
for example R+ , but we want to use the other notation:
(0, infinity) do we have to place a '+' sign in front of infinity?

Nope. However, if you wanted to represent R- using interval notation it would be (- infinity , 0)

(0, infinity) is a perfectly accepted set notation of R+

[nacho]

What is acceptable working out for probability questions were your using the calculators pdf and cdf functions?

Does this get me marks?
Let X = number of wins
X ~ Bi(20, 0.574)

Pr(X=4) = binompdf(20,0.574,4)
?