Some help :)

[balkis]

Hey guys,

Hope you're all studying away and getting ready for tomorrow's exam! I just need a hand with a few things that i must clear up before i hit the paper tomorrow. Any help would be appreciated!

The first question i want to ask is regarding The General Solution of trigonometric functions. I know how to determine the general solution with questions such as Sin(2x)= 1/2...however, when questions like this come up where the value on the right side of the equal sign are negative (e.g sin(x)= -1/2) i always seem to stuff them up. My issue with this is, when i form the solution, would i use -Pi/6? If someone could clarify this with Sin, Cos and Tan, that would be helpful!

Also another thing is questions that i've come across in the multiple choice of both 2008 and 2009 VCAA Mathematical Methods (CAS) Exam 2. These type of questions like the ones attached (sorry i don't know how to show them in the post) always seem to get me.

Any help would be appreciated! Hopefully this all made sense! And good luck to all

Balkis

[BubbleWrapMan]

You'd need another angle, so you'd have -5π/6 + 2kπ and -π/6 + 2kπ. For sin and cos you generally need 2 different base angles. For tan you only need one, and remember the period of tan is half that of sin and cos, so for example tan(x) = 1 has solution π/4 + kπ

For Q15

Pr({1,2}) = 2/6 = 1/3
Pr({2,4,6}) = 3/6 = 1/2

(Note that Pr({1,2}), for example, is the probability that the number is a 1 or a 2)

Intersection of {1,2} and {2,4,6} is {2}

Pr({2}) = 1/6 = Pr({1,2}) x Pr({2,4,6}), so they are independent

Try that method out for Q17

[kensan]

Edit: Beaten 
Ok so with the probability ones, you know it has to be independent.
Therefore 
So basically you have to go though them all and see which satisfies this rule.

Q17
Option A has and    (I got these by seeing how many options there were and dividing by the total which is 12)



Then we have a look at which numbers are similar in the two sets, in option A these are; 1 ,7
This gives 2 out of a possible 12 which is
  So it's option A.
If you go through with the other options, you'll see that  meaning they are not independent.

[balkis]

Edit: Beaten 
Ok so with the probability ones, you know it has to be independent.
Therefore 
So basically you have to go though them all and see which satisfies this rule.

Q17
Option A has and    (I got these by seeing how many options there were and dividing by the total which is 12)



Then we have a look at which numbers are similar in the two sets, in option A these are; 1 ,7
This gives 2 out of a possible 12 which is
  So it's option A.
If you go through with the other options, you'll see that  meaning they are not independent.

THANK YOU THE HELPED ALOT! If only i could repay you haha

You'd need another angle, so you'd have -5π/6 + 2kπ and -π/6 + 2kπ. For sin and cos you generally need 2 different base angles. For tan you only need one, and remember the period of tan is half that of sin and cos, so for example tan(x) = 1 has solution π/4 + kπ

For Q15

Pr({1,2}) = 2/6 = 1/3
Pr({2,4,6}) = 3/6 = 1/2

(Note that Pr({1,2}), for example, is the probability that the number is a 1 or a 2)

Intersection of {1,2} and {2,4,6} is {2}

Pr({2}) = 1/6 = Pr({1,2}) x Pr({2,4,6}), so they are independent

Try that method out for Q17

Thank you to you too! I understand that there should be two for sin and cos, and one for tan, but when you inverse the -1/2, you get -Pi/6, i was just curious, did you plug that in to formula and then simplify to get one with -pi/6 and the second -5*pi/6? (I hope i'm making sense! And thanks again!

[BubbleWrapMan]

I didn't get it from the formula, I just looked for another angle that wasn't equivalent to the first one. For example if I picked 11π/6 it wouldn't have worked since it's just -π/6 + 2π. But -5π/6 isn't a multiple of 2π from -π/6. I could have also used 7π/6, for instance.

[balkis]

I didn't get it from the formula, I just looked for another angle that wasn't equivalent to the first one. For example if I picked 11π/6 it wouldn't have worked since it's just -π/6 + 2π. But -5π/6 isn't a multiple of 2π from -π/6. I could have also used 7π/6, for instance.

Ahh i clicked Thank you so much!!

[vcestudent94]

I didn't get it from the formula, I just looked for another angle that wasn't equivalent to the first one. For example if I picked 11π/6 it wouldn't have worked since it's just -π/6 + 2π. But -5π/6 isn't a multiple of 2π from -π/6. I could have also used 7π/6, for instance.

Hey. For this method, does it matter which how you express your base angles? For example, in the instance above, can you use 7pi/6 instead of -5pi/6. Is there a common notation like using the base angles which are closest to 0? Thanks

[BubbleWrapMan]

Nah doesn't matter, as long as you have two base angles that aren't 2kπ apart from each other. Like I said, I could have used 7π/6 instead of -5π/6.