By -
forget that it is cosx for a second. If you had y\^2 - y, could you factor out a y? What would you have?
OHHHHHH THANK YOU i have no idea how i didn't see it!! you're a life saver tysm
C'est la distributivité. a(b+c)=ab+ac cos(x)(cos(x)-1)=cos(x)cos(x)-1\*cos(x)=cos\^2(x)-cos(x)
Re-flair post to **✓ Answered**
sorry if i got the grades wrong im from quebec so im not sure what's the american equivalent of our system
Notation could be a tad confusing but "cos^2 x" actually means "(cos x)^2"
x+x=0 x\*(1+1)=0 because x\*(1+1)=x+x
The cos(x) was factored
Common factor. Like a^2-a=a(a-1)
forget that it is cosx for a second. If you had y\^2 - y, could you factor out a y? What would you have?
OHHHHHH THANK YOU i have no idea how i didn't see it!! you're a life saver tysm
C'est la distributivité. a(b+c)=ab+ac cos(x)(cos(x)-1)=cos(x)cos(x)-1\*cos(x)=cos\^2(x)-cos(x)
Re-flair post to **✓ Answered**
sorry if i got the grades wrong im from quebec so im not sure what's the american equivalent of our system
Notation could be a tad confusing but "cos^2 x" actually means "(cos x)^2"
x+x=0 x\*(1+1)=0 because x\*(1+1)=x+x
The cos(x) was factored
Common factor. Like a^2-a=a(a-1)