Right let's see if i can go through this..We start with a triangle using your given co-ordinates..then rotate it through an angle θ around the origin so that the triangle lies on the X-Y axis as asked for by the question (shown in diagram)
Now let's take a closer look at that angle. Have a close look at this second diagram.
Notice that i'm now looking only at the point X. What's special about this point is that it's the one that we're putting onto the x-axis. Now, imagine that we are making a transformation
not from X to X', but from X' to something else in the XY-plane. I've shown this as the point (5, 2).
What i'm trying to get across is that it doesnt matter what point i take to transform, the transformation itself is always equivalent. If i take any random co-ordinate w = (283746527, 2983745) say, and apply a rotation θ to it, then no matter what cos θ and sin θ will remain the same. This is for the simple reason that θ is exactly the same as the one I used earlier.
Now then, let's take my little transformation i've made from the x-axis, and use that to find tan θ, cos θ and sin θ. Have a look at this new diagram. I've performed two transformations using our angle θ. First is the one I talked about above, going from X' to the point (5, 2) (essentially 'mirroring' X in the x-axis).
Now, we need to use some helpful properties:
When we rotate a vector about the origin, the length of said vector remains the same.
If we rotate a point (r, 0) through angle θ about the origin (i'm basically talking about rotating an arbitrary point on the x-axis here), then the resulting point is given by: (r cos θ, r sin θ)
Now, this may not seem very useful, but put it all together and we have a recipe for obtaining all three quantities we want.
Note first that we are, indeed, rotating our points about the origin.
Note second, that what I've done is rotate a point
that we know on the x-axis through θ (I mean X').
Now, if you havent made the connection, here's the key:
X' is a point on the x-axis. That means that it is given by some co-ordinates (r, 0) as I said above. So it has
length r.
This new point that I've created, from what I've said above, is therefore given by (r cos θ, r sin θ)!!
We can actually work out r very easily using pythagoras' theorem. The x-component of my point is 5, the y-component of my point is 2, so
r = sqrroot(5^2 + 2^2) = sqrroot(29)
Thus we have the result that: (5, 2) = ( √29 * (cos θ), √29 * (sin θ) ).
(wish i had latex to use here
write the expression out and it'll be a lot clearer)
Now, look at my diagram. Can you use the result I've shown for you here and what you see on the diagram, along with what you know about cos and sin to find the values of
cos θ, sin θ and tan θ?
If you can't i'll take you through the rest of the method
Also this is quite a simplified version, you could use rotation matrices to make life a lot simpler for yourself, but I dont know what level i'm coming in at. Sorry if this is a tad patronising
EDIT: Not sure if the images are showing...
EDIT 2: Should have the images showing now
(Maths was never my strong point)
