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# 2nd DAY: SOLVE THIS INTERESTING PROBLEM...

IT'S THE SECOND DAY OF POSTING AND DISCUSSING VARIOUS MATH PROBLEMS THAT I EVENTUALLY COME ACROSS.....HAPPY SOLVING...:-)*

Note by Raja Metronetizen
4 years, 4 months ago

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since m and l are real .it means that l<1 and m<1.(as if they were greater than 1 the square of the other would be complex)

hence write l = cosx (for some x)

and m is the sin x

then divide al+bm=c by sqrta^2+b^2 (As a^2 + b^2 is not 0)

write the lhs as sin(j+x) (for some j)

hence as the sin function oscillates between -1 and 1

c oscillates between -sqrt a^2 + b^2 to sqrt a^2+b^2

i know i didnt write clearly but hope u get what im saying · 4 years, 4 months ago

yea, i got to what you are trying to say......it's great and right.....a good job...then wait for tomorrow for an another great problem......keep in touch....:--] · 4 years, 4 months ago

using cauchy-schwarz inequality,we have (a^2+b^2)(l^2+m^2)>=(al+bm)^2=c^2,which implies A · 4 years, 4 months ago

good idea......you are then showing acumen over problem solving....great application...thanx..... · 4 years, 4 months ago

The answer is a · 4 years, 4 months ago

how got to answer.....?..... :-o · 4 years, 4 months ago

It can be seen through this... most of the above portion ... an expression of the form {Asinx +(or -) Bcosx} oscillates between +(-) sqrt(a^2 + b^2) ... · 4 years, 4 months ago