Exercises on trigonometric functions with answers

A periodic function repeats itself along the x-axis. In the graph below we have the representation of a function of the type straight f left parenthesis straight x right parenthesis equals straight A space. space sin space left parenthesis straight omega. straight x right parenthesis. Product A. straight omega é:

Answer key explained

The amplitude is the magnitude of the measurement between the equilibrium line (y = 0), and a crest (highest point) or valley (lowest point).

Thus, A = 2.

The period is the length in x of a complete wave, which on the graph is straight pi.

The coefficient of x can be obtained from the relationship:

straight omega equals numerator 2 straight pi over straight denominator T end of fractionright omega equals numerator 2 straight pi over straight denominator pi end of fractionright omega equals 2

The product between A and straight omega é:

straight to space. straight space omega space equals space 2 space. space 2 space equals space 4

The real function defined by straight f left parenthesis straight x right parenthesis equals straight A. sin left parenthesis straight omega. straight x right parenthesis has period 3straight pi and image [-5,5]. The function law is

Answer key explained

In the trigonometric function sin x or cos x, the parameters A and w modify their characteristics.

Determination of A

A is the amplitude and changes the image of the function, that is, the maximum and minimum points that the function will reach.

In the sinx and cos x functions, the range is [-1, 1]. Parameter A is an image amplifier or compressor, as we multiply the result of the function by it.

Since the image is [-5, 5], A must be 5, because: -1. 5 = -5 and 1. 5 = 5.

Determination of omega bold

straight omegais multiplying x, therefore, it modifies the function on the x axis. It compresses or stretches the function in an inversely proportional way. This means that it changes the period.

If it is greater than 1 it compresses, if it is less than 1 it stretches.

When multiplying by 1, the period is always 2pi, when multiplying by straight omega, the period became 3straight pi. Writing the proportion and solving the rule of three:

2 straight pi space. space 1 space equals space 3 straight pi space. straight space omeganumerator 2 straight pi over denominator 3 straight pi end of fraction equals straight omega2 over 3 equals straight omega

The function is:

f (x) = 5.sin (2/3.x)

A comet with an elliptical orbit passes close to Earth at regular intervals described by the function straight c left parenthesis straight t right parenthesis equal to sin open parentheses 2 over 3 straight t close parentheses where t represents the interval between their appearances in tens of years. Suppose the last appearance of the comet was recorded in 1982. This comet will pass by Earth again in

Answer key explained

We need to determine the period, time for a complete cycle. This is the time in tens of years for the comet to complete its orbit and return to Earth.

The period can be determined by the relationship:

straight omega equals numerator 2 straight pi over straight denominator T end of fraction

Explaining T:

straight T equals numerator 2 straight pi over straight denominator omega end of fraction

The value straight omega is the coefficient of t, that is, the number that multiplies t, which in the function given by the problem is 2 over 3.

Considering straight pi equals 3 comma 1 and substituting the values ​​in the formula, we have:

straight T equals numerator 2.3 comma 1 over denominator start style show 2 over 3 end of style end of fraction equals numerator 6 comma 2 over denominator start style show 2 over 3 end style end of fraction equal to 6 comma 2.3 over 2 equal to numerator 18 comma 6 over denominator 2 end of fraction equal to 9 comma 3

9.3 tens is equal to 93 years.

As the last appearance occurred in 1982, we have:

1982 + 93 = 2075

Conclusion

The comet will pass again in 2075.

(Enem 2021) A spring is released from the stretched position as shown in the figure. The figure on the right represents the graph of the position P (in cm) of mass m as a function of time t (in seconds) in a Cartesian coordinate system. This periodic movement is described by an expression of the type P(t) = ± A cos (ωt) or P(t) = ± A sin (ωt), where A >0 is the maximum displacement amplitude and ω is the frequency, which is related to the period T by the formula ω = 2π/T.

Consider the absence of any dissipative forces.

The algebraic expression that represents the positions P(t) of mass m, over time, on the graph, is

Answer key explained

Analyzing the initial instant t = 0, we see that the position is -3. We will test this ordered pair (0, -3) in the two function options provided in the statement.

For straight P left parenthesis straight t right parenthesis equal to plus or minus sin space left parenthesis ωt right parenthesis

straight P left parenthesis straight t right parenthesis equal to plus or minus A. sin space left parenthesis ωt right parenthesisstraight P left parenthesis 0 right parenthesis equal to plus or minus A. sin space left parenthesis straight omega.0 right parenthesisstraight P left parenthesis 0 right parenthesis equal to plus or minus A. sin space left parenthesis 0 right parenthesis

We have that sine of 0 is 0. This information is obtained from the trigonometric circle.

Thus, we would have:

straight P left parenthesis 0 right parenthesis equal to plus or minus A. sin space left parenthesis 0 right parenthesisstraight P left parenthesis 0 right parenthesis equal to plus or minus A. space 0straight P left parenthesis 0 right parenthesis equals 0

This information is false, because at time 0 the position is -3. That is, P(0) = -3. Thus, we discard the options with the sine function.

Testing for the cosine function:

straight P left parenthesis straight t right parenthesis equal to more or less straight A. cos left parenthesis straight omega. straight t right parenthesisrect P left parenthesis 0 right parenthesis equal to more or less straight A. cos left parenthesis straight omega.0 right parenthesis straight P left parenthesis 0 right parenthesis equal to more or less straight A. cos left parenthesis 0 right parenthesis

Once again, we know from the trig circle that the cosine of 0 is 1.

straight P left parenthesis 0 right parenthesis equal to more or less straight A. cos left parenthesis 0 right parenthesisstraight P left parenthesis 0 right parenthesis equals more or less straight A.1straight P left parenthesis 0 right parenthesis equals more or less straight A

From the graph, we saw that the position at time 0 is -3, therefore, A = -3.

Combining this information, we have:

straight P left parenthesis straight t right parenthesis equals negative 3. cos left parenthesis straight omega. straight t right parenthesis

The period T is removed from the graph, it is the length between two peaks or two valleys, where T = straight pi.

The expression for frequency is provided by the statement, being:

straight omega equals numerator 2 straight pi over straight denominator T end of fractionright omega equals numerator 2 straight pi over straight denominator pi end of fractionright omega equals 2

The final answer is:

start style math size 18px straight P left parenthesis straight t right parenthesis equals minus 3. cos space left parenthesis 2 straight t right parenthesis end of style

(Enem 2018) In 2014, the largest Ferris wheel in the world, the High Roller, was opened in Las Vegas. The figure represents a sketch of this Ferris wheel, in which point A represents one of its chairs:

From the indicated position, where the OA segment is parallel to the ground plane, the High Roller is rotated counterclockwise, around point O. Let t be the angle determined by the segment OA in relation to its initial position, and f be the function that describes the height of point A, in relation to the ground, as a function of t.

Answer key explained

For t = 0 the position is 88.

cos(0) = 1

sin(0) = 0

Substituting these values, in option a, we have:

straight f left parenthesis 0 right parenthesis equals 80 sin left parenthesis 0 right parenthesis plus 88straight f left parenthesis 0 right parenthesis equals 80.0 space plus space 88straight f left parenthesis 0 right parenthesis equal to 88
Answer key explained

The maximum value occurs when the value of the denominator is the smallest possible.

straight f straight left parenthesis x right parenthesis equal to numerator 1 over denominator 2 plus cos straight left parenthesis x right parenthesis end of fraction

The term 2 + cos (x) should be as small as possible. Thus, we must think about the smallest possible value that cos (x) can assume.

The cos (x) function varies between -1 and 1. Substituting the smallest value into the equation:

straight f left parenthesis straight x right parenthesis equal to numerator 1 over denominator 2 plus cos left parenthesis 0 right parenthesis end of fractionrecto f left parenthesis straight x parenthesis right equals numerator 1 over denominator 2 plus left parenthesis minus 1 right parenthesis end of fractionright f straight left parenthesis x right parenthesis equals numerator 1 over denominator 2 space minus 1 end of fractionstraight f left parenthesis straight x right parenthesis equal to 1 over 1bold f bold left parenthesis bold x bold right parenthesis bold equal in bold 1

(UECE 2021) In the plane, with the usual Cartesian coordinate system, the intersection of the graphs of real functions of real variable f (x)=sin (x) and g (x)=cos (x) are, for each integer k, the points P(xk, yk). Then the possible values ​​for yk are

Answer key explained

We want to determine the intersection values ​​of the sine and cosine functions which, as they are periodic, will repeat themselves.

The values ​​of sine and cosine are the same for angles of 45° and 315°. With the help of a table of notable angles, for 45°, the sine and cosine values ​​of 45° are numerator square root of 2 over denominator 2 end of fraction.

For 315° these values ​​are symmetric, that is, minus numerator square root of 2 over denominator 2 end of fraction.

The correct option is the letter a: numerator square root of 2 over denominator 2 end of fraction spaceIt is minus numerator square root of 2 over denominator 2 end of fraction.

ASTH, Rafael. Exercises on trigonometric functions with answers.All Matter, [n.d.]. Available in: https://www.todamateria.com.br/exercicios-sobre-funcoes-trigonometricas/. Access at:

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