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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.
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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 the function y equals 2sin x?
2 pi
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 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.
