Help getting the intuition when applying sin and cos to vector problems needed.

[austere needle]

I understand how to work this problem, you need to convert the vectors v and w to component form using sin and cos. We use cos for the x coordinates and sin for the y coordinates. But I don't quite get the problem per se. In the solutions provided by Khan Academy, they just use cos with the 100 and 290 right away, but I don't understand how. In my mind I would need to drop a line perpendicular to the x axis, and work out the problems using sin and cos in the artificial right triangle I've just created... any help would be appreciated. Thanks in advance :D!!

Thanks zfn!!

[brave coyote]

well, i guess you just memorize that a vector is in the form (rcos(theta), rsin(theta))

[austere needle]

"r" in like cos^-1?

[brave coyote]

no as in the length of the vector

= r

you can draw the right triangle as you described if you want

cos(theta) = adj/hyp, adj = hyp*cos(theta) = r*cos(theta)
and same with the sin

[austere needle]

I see....! So its a new function then. And we just need to memorise that as a fact?

[brave coyote]

well its not really a new function

it's just the radius (aka the length) multiplied by the cosine of the angle

[austere needle]

AHHH!! So "r" is short for radius. I thought it were like a new function or somethin.

[brave coyote]

nah

[austere needle]

xD!!

[brave coyote]

confused because acos

?

[austere needle]

Thank you!! Do you have any resources that help me get why it is the way it is? (its fine if you don't, though).

[brave coyote]

here

[austere needle]

arccos.. yea...!

Sure... give me a sec

[brave coyote]

yep

[austere needle]

What I did originally was this. I already F is 90 degrees, G is 360 - 290 = 70, so E is 20.

And then I use sin and cos on E to find out the length of GF and FE, which is the coordinates for that point E itself.

[brave coyote]

i think finding angle E is not necessary but that work

s

[austere needle]

So this approach is also valid?

But how does this approach connects with just using sin(290) and cos(290) right away?

[brave coyote]

sin(290) and cos(290) are both based on the main angle they give you
you can also do
sin(290) and sin(90 - 290) (where 90 - 290 is the other angle) which gets you the same pair of numbers

as cos(x) = sin(90 - x)

[austere needle]

This is what I was wondering about... because the angle 290 doesn't exist in the triangle which I just sketched. So how can I use and sin and cos on 290, right?

[brave coyote]

well the angle 290 exists wdym

angles are measured counterclockwise from the +x

[austere needle]

xD! I meant it doesn't in the triangle I just sketched, right?

[brave coyote]

it does

[austere needle]

Can you please show me how?

[brave coyote]

if you mean that, like, lengths shouldnt be negative so how is cos and sin producing negative numbers

if that's unintuitive to you, then you can definitely turn everything into an acute angle

and add negative signs depending on the quadrant

[austere needle]

Oh no I'm not meaning that.

[brave coyote]

oh

[austere needle]

In my mind, sin(290) = opp/hyp... but I don't really understand how the opp and hyp of sin(290) applies to the triangle I just sketched.

I know its a little confusing xD

[brave coyote]

because the hypotenuse is the one labeled w

and the opposite is EF?

[austere needle]

I think we are getting somewhere. cos(290) = cos(70), right?!!

[brave coyote]

mhm

[austere needle]

So when we are calculating cos(290), we are calculating cos of G!!

And G is a part of the triangle!!!

[brave coyote]

well when you calculate sin(290) it makes a negative number

[austere needle]

We do have to take into account the quadrants and thing, and just doing sin(290) and cos(290) elimates the need for that. But intuition wise we are just calculating the sines and cosines of G (taking into account the quadrant ofc).

[brave coyote]

pretty much, yes

[austere needle]

Thanks for helping me walk through this!!

Thanks!!

[brave coyote]

yw

[austere needle]

+close