
How do you graph $y = 2\cos x$?
Answer
545.7k+ views
Hint: In order to plot the given trigonometric function, we first find the time period. Then we see how many intervals there are going to be in that time period.
Since it’s a graph of cosine and as we know that $\cos {0^ \circ } = 1$, therefore the graph of cosine should start from $1$ but as we have our trigonometric function as $y = 2\cos x$, therefore all the values are multiplied by $2$.
Thus our graph starts from $2$ on the y-axis. Once we have our time period and time intervals, we can plot it on the x-axis and make our amplitude on the y-axis and get our required graph.
Complete step-by-step solution:
The given trigonometric function is: $y = 2\cos x$
In order to solve this, we first need to find the time period by using the formula: $\dfrac{{2\pi }}{b}$ , where in an expression $y = 2\cos x$, $a = 2$ and coefficient of $x = b$. Here the coefficient of $x = 1$
Therefore time period=$\dfrac{{2\pi }}{1} = 2\pi $
Now we know that $\cos {0^ \circ } = 1$ , therefore the graph should start from $1$ on the y-axis but as the function is given as $y = 2\cos x$, therefore the values are multiplied with $2$
We take suitable units distance and plot the time period on the x-axis in terms of $\pi $ and plot the y-axis with usual numbers.
According to the graph that we have, the time period is till $2\pi $ , therefore there are four time intervals such as: $0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi $ . We make our graph as shown below:
Since it’s a graph of cosine and as we know that $\cos {0^ \circ } = 1$, therefore the graph of cosine should start from $1$ but as we have our trigonometric function as $y = 2\cos x$, therefore all the values are multiplied by $2$.
Thus our graph starts from $2$ on the y-axis. Once we have our time period and time intervals, we can plot it on the x-axis and make our amplitude on the y-axis and get our required graph.
Complete step-by-step solution:
The given trigonometric function is: $y = 2\cos x$
In order to solve this, we first need to find the time period by using the formula: $\dfrac{{2\pi }}{b}$ , where in an expression $y = 2\cos x$, $a = 2$ and coefficient of $x = b$. Here the coefficient of $x = 1$
Therefore time period=$\dfrac{{2\pi }}{1} = 2\pi $
Now we know that $\cos {0^ \circ } = 1$ , therefore the graph should start from $1$ on the y-axis but as the function is given as $y = 2\cos x$, therefore the values are multiplied with $2$
We take suitable units distance and plot the time period on the x-axis in terms of $\pi $ and plot the y-axis with usual numbers.
According to the graph that we have, the time period is till $2\pi $ , therefore there are four time intervals such as: $0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi $ . We make our graph as shown below:
Note: The trigonometric functions are real functions which relate an angle of a right-angled triangle to the ratio of the other two sides. The trigonometric functions widely used are: sine, cosine and tangent and their reciprocals such as cosecant, secant and cot respectively.
If we plot the value of sine on a graph, then we will find that the values of sine are positive from $0$ to $\pi $ which corresponds to positive values of sine in the first and the second quadrant.
If we plot the value of cosine on a graph, then we will find that the values of cosine are positive from $0$ to $\dfrac{\pi }{2}$ and then negative from $\dfrac{\pi }{2}$ to $\dfrac{{3\pi }}{2}$ corresponding to the first quadrant and the second and third quadrants respectively.
If we plot the value of sine on a graph, then we will find that the values of sine are positive from $0$ to $\pi $ which corresponds to positive values of sine in the first and the second quadrant.
If we plot the value of cosine on a graph, then we will find that the values of cosine are positive from $0$ to $\dfrac{\pi }{2}$ and then negative from $\dfrac{\pi }{2}$ to $\dfrac{{3\pi }}{2}$ corresponding to the first quadrant and the second and third quadrants respectively.
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