Question

How do you graph $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$?

Answer 1

Here, we are given the cosine function. We know that cosine is the periodic function. We will first use the general form equation of the cosine function to find certain values such as the amplitude, period, phase shift and vertical shift of the function.

Formula used:
$y = A\cos \left( {Bx + C} \right) + D$,where, $A$ is the vertical stretch factor and its absolute value gives the amplitude, $B$ is used to find the period $P = \dfrac{{2\pi }}{B}$, $- \dfrac{C}{B}$ is the phase shift and $D$ is the vertical shift.

Complete step by step solution:
We will first use the general equation of the cosine function and compare it with the given one to find the amplitude, period, phase shift and vertical shift of the function.
The general form of cosine equation is: $y = A\cos \left( {Bx + C} \right) + D$.
If we compare it with the given function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ , we get $A = 1$ and $B = 1$, $C = 45^\circ$ and $D = - 1$.
Thus, we can say that,
The amplitude of the function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ is $A = 1$.
The period of the function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ is $P = \dfrac{{2\pi }}{B} = \dfrac{{2\pi }}{1} = 2\pi = 360^\circ$ .
Phase shift of the function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ is $- \dfrac{C}{B} = - \dfrac{{45}}{1} = - 45^\circ$. This value is negative and thus the phase shift is to the left side.
Vertical shift of the function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ is $D = - 1$. This value is negative and thus the vertical shift is in downward direction.
By using this information, the graph of the given function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ can be obtained as:

Note:
The important thing to note here is that the shape of the given function $y = \cos \left( {x^\circ + 45^\circ } \right) - 1$ is the same as that of the simple cosine function $y = \cos x$. However, here $45^\circ$is added to the angle $x^\circ$. This will result in the phase shift of the graph to $45^\circ$units to the left. Also, 1 is subtracted from the cosine of the given angle. This will shift the graph one unit in the downward direction.