what can you do to the graph of y = cos(x) to obtain the graph of y = sin(x)? ​

Answers

The true statements include

- This is the pattern for a sine function.

- The amplitude is 1

- The vertical shift is 0

- An equation for this graph is y = sin(x)

The only untrue statement is

- The graph follows the zero-max-zero-min-zero pattern.

Step-by-step explanation:

Observing the graph of the function presented, one can examine the options and pick the true options from them all.

Statement 1

The graph follows the zero-max-zero-min-zero pattern.

From the graph, it is obvious that this is a false claim as the graph doesn't follow a zero-max-zero-min-zero pattern, rather, its values range from 0 to 1 to 0 to -1, then it goes back to zero to repeat the cycle all over.

Statement 2

This is the pattern for a sine function.

This is true. A sine wave follows the exact pattern that this plot follows; its values range from 0 to 1 to 0 to -1, then it goes back to zero to repeat the cycle all over.

Statement 3

The amplitude is 1

This is true. The amplitude of a motion of a body refers to the maximum displacement of that body from the equilibrium position. From this plot, it is evident that that amplitude is 1 for the vertical motion of this barnacle.

Statement 4

The vertical shift is 0

The vertical shift of the plot refers to how far the equilibrium position of the motion is from the origin line (x-axis line). And for this motion and plot, the equilibrium position for the vertical motion, is placed on the x-axis, indicating a vertical shift of 0.

Statement 5

An equation for this graph is y = sin(x)

This is true. As described under statement 2 and as evident from the corresponding values of x and y from the plot; with x in radians, the values of y can be obtained by taking the sin of the value of x, indicating that this plot is a plot of y = sin x

Hope this Helps!!!

The resulting graph is y = 5\cdot \sin x.

Step-by-step explanation:

The resulting function is of the form:

y = A\cdot \sin x + k

Where:

x - Independent variable, dimensionless.

y - Dependent variable, dimensionless.

A - Amplitude, dimensionless.

k - Midpoint value, dimensionless.

The sine function is bounded, between -1 and 1, and must be multiplied by a stretch factor. That is: A  0. According to the graph, the function is bounded between 5 (y_{max}) and -5 (y_{min}), and the midpoint value (k) is 0. The amplitude is determined by the following calculation:

A = \frac{y_{max}-y_{min}}{2}

If y_{min} = -5 and y_{max} = 5, then:

A = 5

The resulting graph is y = 5\cdot \sin x.

the origin

Step-by-step explanation:

The effect of 4 on the graph of y = sin(x) + 4 is that it shifts the graph by 4 units up.

The effect of 4 on the graph of y = sin(4x) is that its period would be 1/4th of 2π, i.e it completes one oscillation every π/2 units.

Step-by-step explanation:

The graph of y = sin(x) is a wave that keeps oscillating between -1 and 1, and repeats its shape every 2π units. The maximum value of the graph is 1 unit with a range of [-1, 1]

The graph of y = sin(x) is referred to as a periodic function because it repeats itself infinitely.

The effect of 4 on the graph of y = sin(x) + 4 is that it shifts the graph by 4 units up.

The effect of 4 on the graph of y = sin(4x) is that its period would be 1/4th of 2π, i.e it completes one oscillation every π/2 units.

The answers are right

Step-by-step explanation:

Option 1

Step-by-step explanation:

Correct. It is only interested in the portion of the graph confined to this interval

B. y=sin(x) and y=-1/2

Step-by-step explanation:

We have been given the following trigonometric inequality;

2sin(x)=>-1

The above inequality can be re-written as;

sin(x)=>-1/2

after dividing both sides by 2.

We can then formulate two separate equations, one containing the expression on the right hand side and the other containing the expression on the left hand side;

From the left hand side we form the following equation;

y = sin(x)

From the right hand side we form the following equation;

y = -1/2

Therefore, the above two equations can be graphed to help solve the given trigonometric inequality

x=\dfrac{\pi}{4},-\dfrac{\pi}{4},\dfrac{3\pi}{4},-\dfrac{\pi}{4}

Step-by-step explanation:

Function: y=  sin(x) cos(x)

To find the value of x where tangent line is horizontal on the interval [−π, π]

Slope = y'

y=\sin x\cos x

derivative of y

y'=\cos^2x-\sin^2x

For horizontal tangent line, slope must be 0

\cos^2x-\sin^2x=0

\tan^2x=1

\tan x=\pm 1

x=\dfrac{\pi}{4},-\dfrac{\pi}{4},\dfrac{3\pi}{4},-\dfrac{\pi}{4}

Horizontal tangents are,

y=\dfrac{1}{2}\ and \ y=-\dfrac{1}{2}

\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}
\end{array}
\\\\
-------------------\\\\

\bf \bullet \textit{ stretches or shrinks}\\
\left. \qquad   \right. \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
\left. \qquad   \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
\left. \qquad   \right. \textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\

\bf \bullet \textit{vertical shift by }{{  D}}\\
\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\
\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

with that template in mind, let's see

\bf \begin{array}{llll}
y=&1.5sin&(4x)\\
&\ \uparrow &\ \uparrow \\
&A&B
\end{array}\\\\
-------------------------------\\\\
A=1.5\quad \textit{shrinks by 1.5, or an extra 0.5 from the parent \underline{sin(x)}}
\\\\\\
B=4\qquad period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{4}\implies \cfrac{\pi }{2}


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