Z zamri New member Joined Feb 23, 2012 Messages 6 Mar 7, 2012 #1 Hi, i seem to forgot how to simplify e^i*theta + e^-i*theta, the answer should be 2cos(theta). Can someone help me understand it to the steps. thanks
Hi, i seem to forgot how to simplify e^i*theta + e^-i*theta, the answer should be 2cos(theta). Can someone help me understand it to the steps. thanks
D Deleted member 4993 Guest Mar 7, 2012 #2 zamri said: Hi, i seem to forgot how to simplify e^i*theta + e^-i*theta, the answer should be 2cos(theta). Can someone help me understand it to the steps. thanks Click to expand... eiΘ = cos(Θ) + i*sin(Θ) Now continue....
zamri said: Hi, i seem to forgot how to simplify e^i*theta + e^-i*theta, the answer should be 2cos(theta). Can someone help me understand it to the steps. thanks Click to expand... eiΘ = cos(Θ) + i*sin(Θ) Now continue....
Z zamri New member Joined Feb 23, 2012 Messages 6 Mar 7, 2012 #3 yap, i know that part...only when added up ..how to sum it up to get the final answer....thanks
O Oaky New member Joined Feb 11, 2012 Messages 19 Mar 7, 2012 #4 \(\displaystyle e^i^\theta = cos (\theta) + isin(\theta)\) \(\displaystyle e^-^i^\theta = cos (\theta) - isin(\theta)\) And adding the two gives...?
\(\displaystyle e^i^\theta = cos (\theta) + isin(\theta)\) \(\displaystyle e^-^i^\theta = cos (\theta) - isin(\theta)\) And adding the two gives...?
