The trigonometric functions of real variable correspond to any angle (expressed in radians), a trigonometric ratio, which can be sine, cosine, tangent, cotangent, secant and cosecant.
In this way we have the six trigonometric functions: sine, cosine, tangent, cosecant, secant and cotangent..
The trigonometric functions for angles between 0 and 2π are defined with the help of the unit circumference, with radius 1 and whose center coincides with that of the origin of the Cartesian coordinate system: the point (0,0).
We can locate any point P of coordinates (x, y) on this circumference.
The segment that joins the origin with P, together with the respective segments that join the projections of P on the coordinate axes, make up a right triangle, whose trigonometric ratios are known as the ratios between the sides of the triangle. A) Yes:
And now the reasons that are the inverse of the previous ones:
In the unit circle the hypotenuse of any triangle is equal to 1 and the legs are worth x and y, so:
sin θ = y
cos θ = x
In this way, the sine and cosine functions always acquire values between -1 and 1, while the rest:
tg θ = y / x
cosec θ = 1 / y
sec θ = 1 / x
They are not defined when x or Y worth 0.
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As we will see below, trigonometric functions are characterized by being periodic. Therefore, they are not bijective, except in a restricted domain..
Starting on the trigonometric circle at point P (1,0), the angle is 0 radians. Then the radius rotates counterclockwise and the sin x function gradually grows until it reaches π / 2 radians (90º), equivalent to approximately 1,571 radians..
There it reaches the value y = 1 and then decreases until it reaches zero in π radians (180 °). Later it decreases even more, since the value becomes negative until it reaches −1 when the angle is 3π / 2 radians (270 °).
Finally, it increases again until it returns to zero in 360 °, where everything begins again. This makes y = sin x a periodic function of period 2π, therefore the sine function is not bijective.
Also, the graph is symmetric with respect to the point (0,0), therefore the function is odd.
Then the graph of y = sin x:
The section in red is the first period. Negative angles are also considered, since the radius of the trigonometric circle can rotate clockwise.
Domain of sin x = All reals.
Range or path of sin x = [-1,1]
At the point P (1,0) the cosine function is equal to 1 and from there it decreases, reaching 0 when the angle is π / 2. It continues decreasing and takes negative values, until it reaches -1 at the angle π.
Then it begins to gradually increase until it reaches 0 in 3π / 2 and returns to 1 when the radius has made one complete revolution. From there the cycle repeats itself, since cos x is periodic and is also even (symmetric around the vertical axis).
The form of the cosine function is the same as that of the sine function, except that they are displaced π / 2 with respect to each other..
Domain of cos x = All reals.
Cos x range or travel = [-1,1]
The functions tg x, ctg x, sec x and cosec x are discontinuous, since they are quotients between sine and cosine, or the inverse. Since these are worth 0 in some angles, when they appear in the denominator they make the function discontinuous.
And since sine and cosine are periodic functions, the functions tg x, ctg x, sec x, cosec x are also periodic..
For the tangent function, the discontinuity values are: ± π / 2, ± 3π / 2, ± 5π / 2… There the function takes very large or very small values. In general, this happens for all multiples of π of the form (2n + 1) π / 2, both positive and negative, with n = 0, 1, 2 ...
Therefore:
Tg x domain: D = x ∈ R / x ≠ (2n + 1) π / 2; n ∈ Z
Tg x range or travel: All real.
Note that the function f (x) = tg x repeats between - π / 2 and + π / 2, therefore its period is π. In addition, it is symmetric with respect to the origin.
For this function the discontinuity values occur at 0, ± π, ± 2π…, that is, the integer multiples of π.
Like the tangent function, the cotangent function is periodic of period π. For her it is true that:
Ctg x domain: D = x ∈ R / x ≠ n π; n ∈ Z
Ctg x range or travel: All real.
The sec x function has points of discontinuity at ± π / 2, ± 3π / 2, ± 5π / 2…, where cos x = 0. It is also periodic with period π and it is also observed from the graph that the function never takes values in the interval (-1,1)
Domain of sec x: D = x ∈ R / x ≠ (2n + 1) π / 2; n ∈ Z
Sec x range or travel: All reals except (-1,1)
It is similar to the secant function, although it is shifted to the right, therefore the points of discontinuity are 0, ± π, ± 2π and all integer multiples of π. It is also periodic.
Cosec Domain X: D = x ∈ R / x ≠ n π; n ∈ Z
Harvest range or path x: All reals except (-1,1)
A 6-foot-tall man casts a shadow S whose length is given by:
S (t) = 6 │cot (π.t / 12) │
With S in feet and t the number of hours since 6 AM. How tall is the shadow at 8 AM, 12 PM, 2 PM, and 5:45 PM?
We must evaluate the function for each of the given values, note that it must take the absolute value, since the length of the shadow is positive:
-At 8 AM, 2 hours have elapsed from 6 AM, therefore t = 2 and S (t) is:
S (2) = 6 │cot (π.2 / 12) │ft = 6 │cot (π / 6) │ft = 10.39 feet.
-When it is 12 N, t = 6 hours have elapsed, therefore:
S (6) = 6 │cot (π.6 / 12) │ft = 6 │cot (π / 2) │ft = 0 feet. (At that time the Sun falls vertically on the person's head).
-At 2 PM t = 8 hours passed:
S (8) = 6 │cot (π.8 / 12) │ft = 6 │cot (2π / 3) │ft = 3.46 feet.
-When it is 5:45 PM, 11.75 hours have already passed since 6 AM, so:
S (11.75) = 6 │cot (π x 11.75 / 12) │feet = 91.54 feet. At this hour the shadows are getting longer.
Can the reader calculate the time when the person's shadow equals his height??
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