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## #1 2005-09-25 02:41:47

CidroN
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### solve tan, sin & cos

I) tan v = -1

II) 2 - tan v = 0

III) 2 tan v + 5 = 0

IV) sin x = tan x

V) 2 sin x = tan x

Thanks for help!

## #2 2005-09-25 03:21:39

kylekatarn
Power Member

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### Re: solve tan, sin & cos

All the above equations can be reduced to the form T(x) = a where T(x) is a simple trig function

//the first 3 ones are easy
//since tan is periodic: tan(x) = tan(x+kπ); k∈Z

tan(x+kπ) = a
arctan(tan(x+kπ)) = arctan(a)
x+kπ = arctan(a)

wich is the same as
x = arctan(a)+kπ; k∈Z
(no, I didn't made a mistake with the signal... It doesn't matter if its +kπ or -kπ, because K=...,-2,-1,0,1,2,..... the solution doesn't change)

//the others require a bit of manipulation
a.sin(x) = tan(x)

//divide by sin(x) both members
a = tan(x)/sin(x)
a = 1/cos(x)
1/a = cos(x)

let 1/a = b

//cos is periodic. cos(x) = cos(x+2π).
b = cos(x+2π)
arccos(b) = arccos(cos(x+2π))

//Remember also that cos(±x) = cos(x)
±arccos(b) = x+2kπ
±arccos(b)-2kπ = x

x=±arccos(b)+2kπ; k∈Z

Now It's up to you! Solve your equations using the methods I gave you! Good luck!

Last edited by kylekatarn (2005-09-25 03:22:19)

## #3 2005-10-04 19:11:34

abhishek_rttc
Novice

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### Re: solve tan, sin & cos

I) tan v= -1 = tan (- π/4)
Particualr Value, v= -π/4
General Value,   v= nπ - π/4, where n=0,1,2,.....

II) tan v= 2 = tan (63.4349...)
Particualr Value, v= tan^-1(2)
General Value,   v= nπ +tan^-1(2), where n=0,1,2,.....

III) tan v= 5/2 = tan (68.1986...)
Particualr Value, v= tan^-1(5/2)
General Value,   v= nπ +tan^-1(5/2), where n=0,1,2,....

IV) sin x = tan x
sin x cos x - sinx = 0  ( as tan x = sin x/ cos x )
sin x (cos x -1) =0
either sin x = 0 to cos x = 1
in general  x = nπ, where n= 0,1,2,....

V)  2sin x = tan x
2 sin x cos x - sinx = 0  ( as tan x = sin x/ cos x )
sin x (2 cos x -1) =0
either sin x = 0 to cos x = 1/2
if sin x = 0, in general  x = nπ, where n= 0,1,2,...
if cos x = 1/2, in general x = 2nπ +/- π/3, n= 0,1,2,...

Hence either x = nπ or x = 2nπ +/- π/3, n= 0,1,2,...

Last edited by abhishek_rttc (2005-10-04 19:13:03)