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Investigating trigonometric functions

Investigating trigonometric functions

The steps of an investigation

Investigating a trigonometric function goes through the same steps as any investigation, and starts by asking whether the function is periodic:

  • The period: if f\displaystyle f is periodic, investigate it on an interval of one period, for example 0≤x≤2π\displaystyle 0\leq x\leq2\pi, and repeat the graph. If not – on the interval the question gives.
  • The domain: tan⁡x\displaystyle \tan x, a denominator and a root leave out points or intervals.
  • The derivative: solve f′(x)=0\displaystyle {f'(x)}={0} in the interval – a trigonometric equation, whose solutions in the interval are a finite number of points.
  • A table of signs of f′\displaystyle f': the extremum points and where the function rises and falls.
  • The second derivative: the sign of f′′\displaystyle f'' gives the intervals of concavity, and a point where it changes is an inflection point.
  • A sketch: the points you found, then repeated by the period.
Example: x+2sin⁡x\displaystyle x+2\sin x

The example below investigates f(x)=x+2sin⁡x\displaystyle {f(x)}={x+2\sin x} on the interval 0≤x≤2π\displaystyle 0\leq x\leq2\pi. The function is not periodic, because of the term x\displaystyle x.

  • f′(x)=1+2cos⁡x\displaystyle {f'(x)}={1+2\cos x}. f′(x)=0\displaystyle {f'(x)}={0} where cos⁡x=−12\displaystyle {\cos x}={-\frac{1}{2}}, and in the interval: x=2π3\displaystyle {x}={\frac{2\pi}{3}} or x=4π3\displaystyle {x}={\frac{4\pi}{3}}.
x\displaystyle x0<x<2π3\displaystyle 0<x<\frac{2\pi}{3}2π3\displaystyle \frac{2\pi}{3}2π3<x<4π3\displaystyle \frac{2\pi}{3}<x<\frac{4\pi}{3}4π3\displaystyle \frac{4\pi}{3}4π3<x<2π\displaystyle \frac{4\pi}{3}<x<2\pi
f′(x)\displaystyle f'(x)+0−0+
f(x)\displaystyle f(x)risesmaximumfallsminimumrises
  • f(2π3)=2π3+2⋅32=2π3+3≈3.83\displaystyle {f\left(\frac{2\pi}{3}\right)}={\frac{2\pi}{3}+2\cdot\frac{\sqrt{3}}{2}}={\frac{2\pi}{3}+\sqrt{3}\approx3.83}; f(4π3)=4π3−3≈2.46\displaystyle {f\left(\frac{4\pi}{3}\right)}={\frac{4\pi}{3}-\sqrt{3}\approx2.46}.
  • f′′(x)=−2sin⁡x\displaystyle {f''(x)}={-2\sin x}: negative where 0<x<π\displaystyle 0<x<\pi – concave down; positive where π<x<2π\displaystyle \pi<x<2\pi – concave up.
  • At x=π\displaystyle {x}={\pi} the concavity changes: an inflection point (π,π)\displaystyle (\pi,\pi).
  • At the ends of the interval: f(0)=0\displaystyle {f(0)}={0} – a minimum at an end, f(2π)=2π\displaystyle {f(2\pi)}={2\pi} – a maximum at an end.
f(x)=x+2sin⁡x\displaystyle f(x)=x+2\sin x
y=x\displaystyle y=x
M=(2π3,2π3+3)\displaystyle M=(\frac{2\pi}{3},\frac{2\pi}{3}+\sqrt{3})
N=(4π3,4π3−3)\displaystyle N=(\frac{4\pi}{3},\frac{4\pi}{3}-\sqrt{3})
I=(π,π)\displaystyle I=(\pi,\pi)

The graph winds around the line y=x\displaystyle {y}={x} (dashed): 2sin⁡x\displaystyle 2\sin x adds up to 2 to it and takes up to 2 from it. It crosses the line where sin⁡x=0\displaystyle {\sin x}={0} – at the inflection point and at the ends of the interval.

Example: sin⁡x\displaystyle \sqrt{\sin x}

The example below investigates f(x)=sin⁡x\displaystyle {f(x)}={\sqrt{\sin x}} – the root of a trigonometric function:

  • The domain: sin⁡x≥0\displaystyle \sin x\geq0, that is 0≤x≤π\displaystyle 0\leq x\leq\pi, and in every such period: 2πk≤x≤π+2πk\displaystyle 2\pi k\leq x\leq\pi+2\pi k. Between these intervals the function has no values.
  • The period is 2π\displaystyle 2\pi, so it is enough to investigate the interval 0≤x≤π\displaystyle 0\leq x\leq\pi.
  • f′(x)=cos⁡x2sin⁡x\displaystyle {f'(x)}={\frac{\cos x}{2\sqrt{\sin x}}}, and inside the interval f′(x)=0\displaystyle {f'(x)}={0} where cos⁡x=0\displaystyle {\cos x}={0}, that is x=π2\displaystyle {x}={\frac{\pi}{2}}.
  • f′\displaystyle f' is positive where 0<x<π2\displaystyle 0<x<\frac{\pi}{2} and negative where π2<x<π\displaystyle \frac{\pi}{2}<x<\pi: a maximum (π2,1)\displaystyle \left(\frac{\pi}{2},1\right).
f(x)=sin⁡x\displaystyle f(x)=\sqrt{\sin x}
  • At the ends of the interval, (0,0)\displaystyle (0,0) and (π,0)\displaystyle (\pi,0), the function reaches its least value: minimum points at the ends of the domain.
  • There the denominator of f′\displaystyle f' is 0 and the derivative is not defined: the graph leaves the ends with a vertical tangent, like x\displaystyle \sqrt{x} at 0.
The family y=sin⁡x+ax\displaystyle {y}={\sin x+ax}

All the functions of the family f(x)=sin⁡x+ax\displaystyle {f(x)}={\sin x+ax} have the same inflection points, but only some of them have extremum points:

  • f′′(x)=−sin⁡x\displaystyle {f''(x)}={-\sin x}, with no a\displaystyle a: the concavity changes at every x=πk\displaystyle {x}={\pi k}, for every value of a\displaystyle a.
  • f′(x)=cos⁡x+a\displaystyle {f'(x)}={\cos x+a}, and f′(x)=0\displaystyle {f'(x)}={0} where cos⁡x=−a\displaystyle {\cos x}={-a}. This equation has a solution only when −1≤a≤1\displaystyle -1\leq a\leq1.
  • When −1<a<1\displaystyle -1<a<1 the derivative changes sign, and there are extremum points; when a>1\displaystyle a>1 or a<−1\displaystyle a<-1 – there are none.

On the board: a=0.5\displaystyle {a}={0.5}, a=1\displaystyle {a}={1} and a=2\displaystyle {a}={2}.

f(x)=sin⁡x+0.5x\displaystyle f(x)=\sin x+0.5x
g(x)=sin⁡x+x\displaystyle g(x)=\sin x+x
h(x)=sin⁡x+2x\displaystyle h(x)=\sin x+2x
  • a=0.5\displaystyle {a}={0.5}: the graph rises and falls, with extremum points.
  • a=1\displaystyle {a}={1}: f′(x)=cos⁡x+1≥0\displaystyle {f'(x)}={\cos x+1\geq0}, and the function rises. At the points x=π+2πk\displaystyle {x}={\pi+2\pi k} the derivative is 0 but does not change sign – these are horizontal inflection points.
  • a=2\displaystyle {a}={2}: f′(x)≥1\displaystyle f'(x)\geq1, and the function rises with no extremum points; the inflection points stayed at the same x\displaystyle x.
The graph of x2sin⁡x\displaystyle x^2\sin x

You can sketch the graph of f(x)=x2sin⁡x\displaystyle {f(x)}={x^2\sin x} even before differentiating, from the properties of its two factors:

  • −1≤sin⁡x≤1\displaystyle -1\leq\sin x\leq1, so −x2≤f(x)≤x2\displaystyle -x^2\leq f(x)\leq x^2: the graph is caught between the parabolas y=x2\displaystyle {y}={x^2} and y=−x2\displaystyle {y}={-x^2} – its “envelope”.
  • f(x)=x2\displaystyle {f(x)}={x^2} where sin⁡x=1\displaystyle {\sin x}={1}, and f(x)=−x2\displaystyle {f(x)}={-x^2} where sin⁡x=−1\displaystyle {\sin x}={-1}: there the graph touches the envelope.
  • The zeros are the zeros of sin⁡x\displaystyle \sin x, x=πk\displaystyle {x}={\pi k}.
  • f(−x)=(−x)2sin⁡(−x)=−f(x)\displaystyle {f(-x)}={(-x)^2\sin(-x)}={-f(x)}: the function is odd.
f(x)=x2sin⁡x\displaystyle f(x)=x^{2}\sin x
y=x2\displaystyle y=x^{2}
y=−x2\displaystyle y=-x^{2}

The function is not periodic: the factor sin⁡x\displaystyle \sin x repeats itself, but the factor x2\displaystyle x^2 makes the waves bigger the further you go from 0. In sin⁡x+x2\displaystyle \sin x+x^2, by contrast, sin⁡x\displaystyle \sin x adds waves of a fixed height to the parabola.

Exercise 1

Investigate the function f(x)=sin⁡x+cos⁡x\displaystyle {f(x)}={\sin x+\cos x} on the interval 0≤x≤2π\displaystyle 0\leq x\leq2\pi.

a

Find f′(x)\displaystyle f'(x).

f(x)=sin⁡x+cos⁡x\displaystyle f(x)=\sin x+\cos x

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b

    Solve the equation f′(x)=0\displaystyle {f'(x)}={0} in the interval.

      c

      Find f′′(x)\displaystyle f''(x).

      f(x)=sin⁡x+cos⁡x\displaystyle f(x)=\sin x+\cos x

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        d

        Solve the equation f′′(x)=0\displaystyle {f''(x)}={0} in the interval.

          e

          Mark on the graph the extremum points and the inflection points on the board.

          f(x)=sin⁡x+cos⁡x\displaystyle f(x)=\sin x+\cos x

          f

          Write where f\displaystyle f rises and falls and the intervals of concavity of f\displaystyle f in the interval.
          What is the period of f\displaystyle f?

          Exercise 2

          Investigate the function f(x)=sin⁡2x\displaystyle {f(x)}={\sin^2x} on the interval 0≤x≤π\displaystyle 0\leq x\leq\pi – one period of it.

          a

          Find f′(x)\displaystyle f'(x), and write it with sin⁡2x\displaystyle \sin2x.

          f(x)=sin⁡2x\displaystyle f(x)=\sin^{2}x

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

            / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

            b

            Solve the equation f′(x)=0\displaystyle {f'(x)}={0} in the interval.

              c

              Find f′′(x)\displaystyle f''(x).

              f(x)=sin⁡2x\displaystyle f(x)=\sin^{2}x

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                d

                Solve the equation f′′(x)=0\displaystyle {f''(x)}={0} in the interval.

                  e

                  Mark on the graph the extremum points and the inflection points on the board.

                  f(x)=sin⁡2x\displaystyle f(x)=\sin^{2}x

                  Exercise 3

                  Investigate the function f(x)=sin⁡x2+cos⁡x\displaystyle {f(x)}={\frac{\sin x}{2+\cos x}} on the interval 0≤x≤2π\displaystyle 0\leq x\leq2\pi.

                  a

                  Find f′(x)\displaystyle f'(x).

                  f(x)=sin⁡x2+cos⁡x\displaystyle f(x)=\frac{\sin x}{2+\cos x}

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                    b

                    Solve the equation f′(x)=0\displaystyle {f'(x)}={0} in the interval.

                      c

                      Work out f(2π3)\displaystyle f\left(\frac{2\pi}{3}\right).

                      f(x)=sin⁡x2+cos⁡x\displaystyle f(x)=\frac{\sin x}{2+\cos x}

                        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                        d

                        Mark the extremum points on the graph.

                        f(x)=sin⁡x2+cos⁡x\displaystyle f(x)=\frac{\sin x}{2+\cos x}

                        e

                        Why is the function defined for every x\displaystyle x?
                        Show that f\displaystyle f is odd, and explain how this fits the extremum points you found.

                        Exercise 4

                        Investigate the function f(x)=2sin⁡x−x\displaystyle {f(x)}={2\sin x-x} on the interval 0≤x≤2π\displaystyle 0\leq x\leq2\pi.

                        a

                        Find f′(x)\displaystyle f'(x).

                        f(x)=2sin⁡x−x\displaystyle f(x)=2\sin x-x

                          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                          / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                          b

                          Solve the equation f′(x)=0\displaystyle {f'(x)}={0} in the interval.

                            c

                            Find f′′(x)\displaystyle f''(x).

                            f(x)=2sin⁡x−x\displaystyle f(x)=2\sin x-x

                              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                              d

                              Mark on the graph the extremum points and the inflection points on the board.

                              f(x)=2sin⁡x−x\displaystyle f(x)=2\sin x-x

                              e

                              Write the kind of each extremum point, and the greatest and the least value of f\displaystyle f on the interval.
                              In the family y=sin⁡x+ax\displaystyle {y}={\sin x+ax} of the example, 12f(x)\displaystyle \frac{1}{2}f(x) belongs to which a\displaystyle a? How does this explain why f\displaystyle f has extremum points?

                              a\displaystyle a

                                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power