i am so confused in my math class right now. i have no idea what to do or how to start. it just doesnt make sense. i am going to go in for tutoring but the assignment is due tommorrow morning and i need to get a good score otherwise my grade will go way down. so if anyone can help me i will be SO grateful.
i need to find the smallest positive number for which:
1. sin (2x)-sin x=0
One way is Newton's method.
Let 1 be your initial estimate.
\(\displaystyle 1-\frac{sin(2(1))-sin(1)}{2cos(2(1))-cos(1)}=1.04941471712\)
\(\displaystyle 1.04941471712=\frac{sin(2(1.04941471712))-sin(1.04941471712)}{2cos(2(1.04941471712))-cos(1.04941471712)}=1.40720178129\)
\(\displaystyle 1.40720178129-\frac{sin(2(1.40720178129))-sin(1.40720178129)}{2cos(2(1.40720178129))-cos(1.40720178129)}=1.0471975512={\pi}/3\)
2. sin (x+1)= sin x
Use the addition formula:
\(\displaystyle sin(x+1)=sin(x)cos(1)+cos(x)sin(1)\)
\(\displaystyle sin(x)cos(1)+cos(x)sin(1)=sin(x)\)
\(\displaystyle sin(x)cos(1)+cos(x)sin(1)-sin(x)=0\)
Factor:
\(\displaystyle sin(x)(cos(1)-1)+cos(x)sin(1)=0\)
\(\displaystyle (cos(1)-1)sin(x)=-sin(1)cos(x)\)
\(\displaystyle \frac{(cos(1)-1)}{-sin(1)}=\frac{cos(x)}{sin(x)}\)
\(\displaystyle tan(1/2)=cot(x)\)
\(\displaystyle x=cot^{-1}(tan(1/2))=\frac{({\pi}-1)}{2}\)
There, Dan and I gave you two answers. Can you struggle through the other two?.
3. sin x+ cos (2x)=0
4. sin x-cos (2x)=0
(in radians..)
-elizabeth
