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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.
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What is cot of pi - pi over 4 given that tan of pi over 4 equals 1?
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.
What is the smallest positive number for which tan of 3x equals 1?
tan(3x)=1 3x= PI/4 x=PI/12 is the smallest positive number
What is the period of y equals 3 sin pi x?
2
What is the period of y equals -5 tan x?
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.
What is tan x degree 90?
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.
What is the period of y equals tan2x?
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
What is cot of pi - pi over 4 given that tan of pi over 4 equals 1?
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.
What is the smallest positive number for which tan of 3x equals 1?
tan(3x)=1 3x= PI/4 x=PI/12 is the smallest positive number
What is the period of y equals 2sinx?
The same as the period of y = sin x. This period is equal to (2 x pi).
1 over tan x equals what?
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
What is the period of y equals 3 sin pi x?
2
What is the period of y equals cosecant x?
It is 2*pi radians.
What is the period of the function y equals 2sin x?
2 pi
What is the answer to cot squared x - tan squared x equals 0?
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.
Solve 5 over 8 equals Tan x?
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
If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?
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).
What is the period of y equals -5 tan x?
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.
