The period of the tangent function, tan(x), is π because the tangent function has a repeating pattern every π units. This is due to the nature of the tangent function, which has vertical asymptotes at intervals of π. As x increases by π, the tangent function repeats its values, resulting in a period of π for the function.
Well, honey, the period of the tangent function is actually pi, not equals pi. It's because the tangent function repeats every pi units, just like how your annoying neighbor repeats the same old stories every time you run into them. So, when you see tan x with a period of pi, just remember it's like Groundhog Day but with trig functions.
If you look at the graph of the tangent function, you will see it has a vertical asymptote at x = pi/2. This is because the tangent approaches infinity as t approaches pi/2. It approaches negative infinity as t approaches -pi/2.
The distance between - pi/2 and pi/2 is pi which is the period.
Another way to see this is tanx is sinx divided by cosx.
We cannot have cosx equal to zero or we would be dividing by zero.
So when does cosx=0? It happens when x=pi/2 or -pi/2. It also equal 0 as
multiples, both positive and negative ones, or pi/2 such as 3pi/2, 5pi/2 etc.
First: note 3 things about cot and tan, and note the given statement:cot = 1/tantan is cyclic with a period of π, that is tan(nπ + x) = tan(x)tan is an odd function, that is tan(-x) = -tan(x)tan(π/4) = 1Now apply them to the problem:cot(π - π/4) = 1/tan(π - π/4)= 1/tan(-π/4)= 1/-tan(π/4)= 1/-1 = -1Thus:cot(π - π/4) = -1.
tan(3x)=1 3x= PI/4 x=PI/12 is the smallest positive number
2
y = -5tan(x) can also be written y/(-5) = tan(x). In other words, the -5 just changes the y values and the orientation of the graph (it flips tan(x) over the y axis and stretches the graph up and down). So -5tan(x), like tan(x), has a period of pi. This is because tan is the y value divided by the x value at any given point on the unit circle. At 0 degrees, x is at one and y is at zero, so tan0o = 0. As we travel counterclockwise around the unit circle, tan is next equal to zero when x is equal to zero. This occurs halfway around the circle at 180o, or (in radians) pi.
To solve for tan x degree 90 you do a few things. First, if x equals 90, then this equals 1.5597 radian or 89.36 degrees. This is the easiest way to solve tan x degree 90.
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
First: note 3 things about cot and tan, and note the given statement:cot = 1/tantan is cyclic with a period of π, that is tan(nπ + x) = tan(x)tan is an odd function, that is tan(-x) = -tan(x)tan(π/4) = 1Now apply them to the problem:cot(π - π/4) = 1/tan(π - π/4)= 1/tan(-π/4)= 1/-tan(π/4)= 1/-1 = -1Thus:cot(π - π/4) = -1.
tan(3x)=1 3x= PI/4 x=PI/12 is the smallest positive number
The same as the period of y = sin x. This period is equal to (2 x pi).
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
2 pi
2
It is 2*pi radians.
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
5/8 = tan(x) x = tan-1(5/8) = tan-1(0.625) = 0.558599 + k*pi radians or 32.00538 + k*180 degrees where k is an integer
tan (A-B) + tan (B-C) + tan (C-A)=0 tan (A-B) + tan (B-C) - tan (A-C)=0 tan (A-B) + tan (B-C) = tan (A-C) (A-B) + (B-C) = A-C So we can solve tan (A-B) + tan (B-C) = tan (A-C) by first solving tan x + tan y = tan (x+y) and then substituting x = A-B and y = B-C. tan (x+y) = (tan x + tan y)/(1 - tan x tan y) So tan x + tan y = (tan x + tan y)/(1 - tan x tan y) (tan x + tan y)tan x tan y = 0 So, tan x = 0 or tan y = 0 or tan x = - tan y tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C) tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B) A, B and C are all angles of a triangle, so are all in the range (0, pi). So A-B and B-C are in the range (- pi, pi). At this point I sketched a graph of y = tan x (- pi < x < pi) By inspection I can see that: A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi A = B or B = C or A = C or A = C +/- pi But A and C are both in the range (0, pi) so A = C +/- pi has no solution So A = B or B = C or A = C A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).
y = -5tan(x) can also be written y/(-5) = tan(x). In other words, the -5 just changes the y values and the orientation of the graph (it flips tan(x) over the y axis and stretches the graph up and down). So -5tan(x), like tan(x), has a period of pi. This is because tan is the y value divided by the x value at any given point on the unit circle. At 0 degrees, x is at one and y is at zero, so tan0o = 0. As we travel counterclockwise around the unit circle, tan is next equal to zero when x is equal to zero. This occurs halfway around the circle at 180o, or (in radians) pi.