Derive log⁡(a+b)=log⁡(a)−2log⁡(cos⁡(arctan⁡(ba)))

Margie Marx

Margie Marx

Answered

2022-01-22

Derive log(a+b)=log(a)2log(cos(arctan(ba)))

Answer & Explanation

Thomas Lynn

Thomas Lynn

Expert

2022-01-22Added 28 answers

Nothing too magical is happening - its
lovagwb

lovagwb

Expert

2022-01-23Added 50 answers

log(a+b)=log(a)2log(cos(arctan(ba)))
Start from RHS:
Let arctan(ba)=ytany=ba
Using the theorem according to pythagoras:
cosy=aa+b
The equation then reduces to:
log(a)2log(cosy)=log(a)2log(aa+b)=log(a)log(aa+b)
log(a)log(aa+b)=log(a)log(a)(log(a+b))=log(a+b)
=LHS
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

By subtracting loga, 12log(1+ba)=logcosarctanba Using the identity cosarctanx=1x2+1 logcosarctanba=log1ba+1=12log(1+ba) as desired.

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