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This statement (A.B=2AB(cos^2a+sin^2b) ) analysed says just that AB has the same sign as A.B because (cos squared plus sin squared) is positive. If A and B are vectors (which i suspect they are) then you should clarify what AB means in the second statement. This statement here A.B=|A||B|cos(theta) is indeed the definition of dot product but what AB without any dot or cross signifies i cannot understand. If we are talking about just numbers (in which case i cannot understand y there is a dot on the left hand side of both statements) then there are two cases for AB or A.B (again if numbers not any difference). AB being negative, considering the second statement, there are two possibilites. Either A, B have different signs and also θ is equal to π or A,B have the same sign and thus θ=0. Similarly AB being positive gives the same two possibilities. Both possibilities do not contradict the original statement A.B=2AB(cos^2a+sin^2b) which again says that A.B has the same sign as AB (whatever this AB means). There is the case that you have misstyped the statements (rather unlikely) and in this case i would think that A.B=2AB(cos^2a+sin^2b) is actually A.B=2|A||B|(cos^2a+sin^2b) where a, b, are the angles of the A, B vectors to a given (unknown) third vector and thus the second statement seems to be A.B=|A||B|cos(theta)=|A||B| where θ is the angle between vectors A, B. In that case A and B are pointing at the same direction or if A.B=|A||B|cos(theta)=-|A||B| are colinear but oposite direction which somehow must derive from the first statement A.B=2|A||B|(cos^2a+sin^2b) (not difficult i think substituting cos squared and sin squared by their double angle equals). Will be happy to assist more if clarified what is the case....
Thank you for replying. To clarify; this is the correct assumption; A.B=2AB(cos^2a+sin^2b) is actually A.B=2|A||B|(cos^2a+sin^2b) where a, b, are the angles of the A, B vectors to a given (unknown) third vector and thus the second statement seems to be A.B=|A||B|cos(theta)=|A||B| where θ is the angle between vectors A, B
So now that it's clear we have to show that A.B=|A||B|cos(theta)=|A||B| or cosθ = 1 Given that θ is the angle between A, B and \[0\leθ\leπ\] then θ=0. Whatever the case θ=a-b or θ=b-a , a must be equal to b. If that is the case then A.B=2|A||B|(cos^2a+sin^2b) turns into A.B=2|A||B|(cos^2a+sin^2a) = 2|A||B| which contradicts with A.B=|A||B| that you are trying to prove. Now if A.B=2|A||B|(cos^2a+sin^2a) = |A||B|cosθ , θ=a-b or θ=b-a is equal to 2|A||B|(cos^2a+sin^2a) = |A||B|cosθ or 2(cos^2a+sin^2a) = cosθ, again θ=a-b or θ=b-a. This equation has solutions acording to WolframAlpha but i don't think it's what you are looking for. Probably there is something wrong with the A.B=2|A||B|(cos^2a+sin^2b) equation. Where did this derive from if i may ask?
Thank you for an excellent response. The actual original question is as follows. Derive AdotB = A*B*cos(theta) = |A|*|B|*cos(theta) from AdotB = AxBx + AyBy +AzBz where x,y,z are components.