Question

Select the functions that have identical graphs.

Answer 1

Answer:

c. 1 and 3

Step-by-step explanation:

To quickly solve this problem, we can use a graphing tool or a calculator to plot each equation.

Please see the attached image below, to find more information about the graph s

The equations are:

1) y = sin (3x + π/6)

2) y = cos (3x - π/6)

3) y = cos (3x - π/3)

Looking at the graphs, we can see that the identical ones

are equations one and three

Correct option:

c. 1 and 3

Answer 2

Answer:

The correct option is:

             option: c     c.   1 and 3

Step-by-step explanation:

The first trignometric function is given by:

     $y=\sin (3x+\dfrac{\pi}{6})$

and also we know that:

$\sin \theta=\cos(\dfrac{\pi}{2}-\theta)$

This means that:

$\sin (3x+\dfrac{\pi}{6})=\cos (\dfrac{\pi}{2}-(3x+\dfrac{\pi}{6})\\\\\\\sin (3x+\dfrac{\pi}{6})=\cos (\dfrac{\pi}{2}-3x-\dfrac{\pi}{6})\\\\\\\sin (3x+\dfrac{\pi}{6})=\cos (\dfrac{\pi}{2}-\dfrac{\pi}{6}-3x)\\\\\\\sin (3x+\dfrac{\pi}{6})=\cos (\dfrac{2\pi}{6}-3x)\\\\\\\sin (3x+\dfrac{\pi}{6})=\cos (\dfrac{\pi}{3}-3x)\\\\\\\sin (3x+\dfrac{\pi}{6})=cos (-(3x-\dfrac{\pi}{3}))$

As we know that:

$\cos (-\theta)=cos(\theta)$

Hence, we have:

$\sin (3x+\dfrac{\pi}{6})=\cos (3x-\dfrac{\pi}{3})$

Also,  by the graph we may see that the graph of 1 and 2 function do not match.

Hence, they are not equivalent.