# What kind of mathematics/pattern is this?

So I was doing my math work today, sine cosine functions, i was trying to find zeros for different functions

in this instance i was working with y=cos2x, i knew what the graph looked like even before i put pen to paper but i wanted to know exactly where the zeros were, i took an educated guess that each side would have 4 zeros(4 negative 4 positive)

so i found my first zero using trial and error which was 45

THEN it happened i sort of found this pattern

the equation i sort of made up was

cos(2 * what number would =0)

the first number i got was 45, then either i got lucky and discovered this pattern or i remember looking at one of the graphs noticing the zeros were always the same distance apart from each other(They were increasing by the same amount each time)

so i did 45+45= not a zero

but when i did

45+45+45 = Zero (135)

45+45+45+45+45=zero(225)

but 45+45+45+45= not a zero

so it would skip a 45 and the next 45 would be a zero

in other words 45(1) = zero 45(2)=not a zero 45(3)=zero 45(4)=not a zero 45(5)=zero and it would go on and on

my question is what have i stumbled upon here? Is this a special kind of mathematics? They all seem to be odd numbers for zeros. This will REALLY come in handy for future problems of similar types, i will no longer find zeros using the exhausting method of trial and error, just use patterns like this to find the zeros and save a lot of time and energy

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tiny-tim
Homework Helper
hi supernova1203!

(have a degree: ° )
in other words 45(1) = zero 45(2)=not a zero 45(3)=zero 45(4)=not a zero 45(5)=zero and it would go on and on

my question is what have i stumbled upon here? Is this a special kind of mathematics? They all seem to be odd numbers for zeros. This will REALLY come in handy for future problems of similar types, i will no longer find zeros using the exhausting method of trial and error, just use patterns like this to find the zeros and save a lot of time and energy
you've found the formula cos((n + 1/2)π) = 0

(similar to the formula sin(nπ) = 0)

btw, you can "prove" it by expanding to cos(nπ)cos(π/2) - sin(nπ)sin(π/2) = cos(nπ)0 - 0sin(π/2), = 0

whoa.....i had no idea i did that @[email protected] that calculus or something(I have a knack for accidently stumbling onto problems that are um.. a little ahead of me, i was recently told in my pre calculus class i solved something they do on adv calculus, i figured out stuff about limits, and infinity, and most importantly i learned how to graph manually reciprocal functions etc ..i was so proud lol my teacher said they would never give a problem like that on exams because its too difficult and would take too long to solve...yet i was able to finish it eventually(I got the problem online)

tiny-tim
Homework Helper
whoa.....i had no idea i did that @[email protected] that calculus or something
no, it's just trig

the first equation is simply translating the obvious into an equation, and the second equation is one of the well-known trigonometric identities , cos(A+B) = cosA cosB - sinAsinB

SteamKing
Staff Emeritus