In relation to trig functions and the Unit circle

[Chuquis]

first of all I would like to know how to simplify the following cos (-t)cos t - sin (-t)sin t? I have tried but i can't seem to get it. I get stuck.

 

The other thing I am wondering about is why does cos(a-b) = cos (b-a) and sin(a-b) = sin(b-a)? I am not really sure how to put it in words but I think it is because they are both on the positive side of the x-axis but not both on the positive of the y-axis. I'm I on the right track or not?

[abvegto]

i guess the answer is 1....cos(-t) = cos t and sin(-t) = - sin t

[Crimson Sunbird]

The other thing I am wondering about is why does cos(a-b) = cos (b-a) and sin(a-b) = sin(b-a)?

The second identity should be [latex]\sin(a-b)=-\sin(b-a)[/latex]. Let [latex]t=b-a[/latex]; then the two identities are as abvegto has pointed out: [latex]\cos(-t)=\cos t[/latex] and [latex]\sin(-t)=-\sin t[/latex]. Cosine is an even function and sine is an odd function.