I know how lớn solve the trig equations I just get stuck putting the 2 at the over or not. I understand pi is 180 & 2pi is 360

For example

2θ=pi/6

θ=5pi/12 +kpi & θ=11pi/12 +kpi

but for another question θ= pi/6 và θ= 5pi/6 but has 2kpi in front of it


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Trig functions are periodic, which basically means they end up repeating values infinitely at a set interval.

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Like sin(x) & cos(x) both have periods of 2pi, which means that if we showroom a multiple of 2pi we'd get the same number as if we haven't. So:

sin(x + k2pi) = sin(x) và cos(x + k2pi) = cos(x)

So if a specific angle works for sin, then that angle + 2kpi will also work.

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Now tan(x) is periodic with period only pi, so we only need multiples of pi (not 2pi lượt thích with sin & cos). So

tan(x + k*pi) = tan(x)

So if so angle works for tan, that angle plus key performance indicator will work as well.

I guess this is a long winded way of saying that it's problem dependent if you need to use 2kpi vs kpi based on if you used cos, sin (and sec, csc) vs chảy (and cot)


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level 2
Op · 4 yr. Ago
College

I am still kinda confused. So will Cos & Sin always be 2pi? Because the first one started out as sec2(2θ)=4/3 & the second was 2sin2(θ)-3sin(θ)+1=0


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level 1

· 4 yr. Ago
kpi is used for tan. If you can imagine a unit circle in your head, you'll realise that the tan values move up by 1 radian. Sin & cos work differently, they move up by 2 radians

you can see this if you look at both of their graphs

https://d2jmvrsizmvf4x.cloudfront.net/iCwAMYstT42Sb7rj9RqN_tannyfannyx.png https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/Sin.svg/500px-Sin.svg.png

the values for rã repeat every radian, và likewise 2 radians for sin và cos


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level 2
Op · 4 yr. Ago
College

I am still kinda confused. So will Cos & Sin always be 2pi? Because the first one started out as sec2(2θ)=4/3 & the second was 2sin2(θ)-3sin(θ)+1=0


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