Ken's specialist question thread!

[kensan]

If w=   and  z=
Find Arg(zw)

For this questions do I find the individual Args then multiply or add together? My teacher added them and got
If this is correct that means I didn't know you can just add the Args together, I thought you would multiply.

[TrueTears]

Yes just an application of de moivre's theorem http://en.wikipedia.org/wiki/Complex_number#Multiplication.2C_division_and_exponentiation_in_polar_form

[kensan]

Yes just an application of de moivre's theorem http://en.wikipedia.org/wiki/Complex_number#Multiplication.2C_division_and_exponentiation_in_polar_form

Oh wow, yep I get it, I don't know why my brain doesn't see these things lol   thanks

[kensan]

Ok so we are doing vectors now, specifically scalar product.
This question
If  |a|=|b|  simplify  (a+b)(a-b) and (a+b)^2

Not too sure what I'm looking for, do I use the scalar product to simplify or am I meant to recognize relations between vectors if the modulus is equal?

[rife168]

If  |a|=|b|  simplify  (a+b)(a-b) and (a+b)^2

Not too sure what I'm looking for, do I use the scalar product to simplify or am I meant to recognize relations between vectors if the modulus is equal?

Remember that for any vector
Note that is invalid notation, instead just use or

Attempt the questions with that knowledge and also the face that and hopefully you can get it, if not, I'll be able to help


edit: the reason that is invalid is because you can't "square" a vector, because there's no distinct "multiply" operation defined for all vectors.

[kensan]

I got the first one, which is 0

With the second one, I get which is
In the answers it has simplified but I can't see why :/

[nubs]

I got the first one, which is 0

With the second one, I get which is
In the answers it has simplified but I can't see why :/

What has it simplified 2a.b to?

[kensan]

The answer is

[John President]

The answer is

   2|a|^2 + 2a.b
= 2|a|^2 + 2|a||b| cos(theta) [Definition of the dot product]
= 2|a|^2 + 2|a|^2 cos(theta) [Since |a|=|b|]
= 2|a|^2 (1+cos(theta)) [Taking 2|a|^2 out as a common factor].

Hope that makes sense.

If you get a different answer to the recommended one in the future, always check if it can be simplified further. It is very useful to be able to recognise equivalent forms of a certain expression - especially in multiple choice questions!
     

[brightsky]

just use a.b = |a||b|cos(t) and since |a|=|b|, a.b = |a|^2cos(t). sub this in and hoorah.

[kensan]

Awesome thanks guys! I always get how things are done but unable to get there by myself.. kind of annoying haha.
One more for tonight:
Let a=xi+2j-3k
and b=2i-6j+zk
For what values of x and z are a and b perpendicular to each other?
So I've done the dot product and got 2x-12-3z=0 ,where do I get the other equation from?

[Lasercookie]

Edit: Made a big error, crossed the stuff out, this post probably doesn't help that much then.

Possibly this:

Since they're perpendicular:



as

[nina_rox]

Can someone help me with this one:

tan(2x)/1+ sec(2x)

and

For pi/2 < A < pi and 0 < B < pi/2  with sinA=t and cosB=t, cos(B+A) is equal to?

[nina_rox]

Also:
If x is an acute angle and cos(x)= 4/5, Find
cos(2x)

[kensan]

Possibly this:

Since they're perpendicular:



(as )



I'll leave you to it.

edit: fixed LaTeX

I actually tried this last night, but I got false on my CAS, probably because you have |a||b|cos(theta), but since theta is 90 degrees it ends up being zero which makes the whole thing zero, if that makes sense.

Can someone help me with this one:

tan(2x)/1+ sec(2x)

and

For pi/2 < A < pi and 0 < B < pi/2  with sinA=t and cosB=t, cos(B+A) is equal to?

For the first one, are you trying to simplify? If so I got tan(x). Try turning everything into sin and cos and simplify from there.

Also:
If x is an acute angle and cos(x)= 4/5, Find
cos(2x)

Use the double angle formula
And since you already know cos(x), you just sub in!