Identify the Equation of a Trigonometric Graph

Example on how to identify the equation of a trigonometric graph, this corresponds to question 8 in trigonometry_2.

Example 1


Which function corresponds to the graph shown below?

  1. $y=\dfrac{1}{2}\cos(x)$
  2. $y=-\dfrac{1}{2}\cos(x)$
  3. $y=\dfrac{1}{2}\sin(4x)$
  4. $y=-\dfrac{1}{2}\sin(4x+\pi/2)$
  5. $y=-\dfrac{1}{2}\cos(4x+\pi)$

Solution


  1. Let us start by evaluating at $x=0$ the 5 functions given as possible answers.
    1. $y(0)=\dfrac{1}{2}\cos(0)=\dfrac{1}{2}$
    2. $y(0)=-\dfrac{1}{2}cos(0)=-\dfrac{1}{2}$
    3. $y(0)=\dfrac{1}{2}\sin(4(0))=\dfrac{1}{2}\sin(0)=0$
    4. $y(0)=-\dfrac{1}{2}\sin(4(0)+\pi/2)=-\dfrac{1}{2}\sin(0+\pi/2)=-\dfrac{1}{2}\sin(\pi/2)=-\dfrac{1}{2}$
    5. $y=-\dfrac{1}{2}cos(4(0)+\pi)=-\dfrac{1}{2}\cos(4(0)+\pi)=-\dfrac{1}{2}\cos \pi=-\dfrac{1}{2}(-1)=\dfrac{1}{2}$
  2. It can be seen that the answer can only be $A$ or $E$. We now compare the periods of function $y=\dfrac{1}{2}\cos(x)$ in $A$ which is $2\pi$ and the period of function $y=-\dfrac{1}{2}\cos(4x+\pi)$ in E which is $\dfrac{2\pi}{4}=\dfrac{\pi}{2}$ which corresponds to the period of the graph.

    Answer E