Trigonometric Functions and Graphs - Trigonometry

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Question

Let be a function defined as follows:

$f(x)=-\sin(x+\pi)+4$

The 4 in the function above affects what attribute of the graph of ?

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Answer

The period of the function is indicated by the coefficient in front of ; here the period is unchanged.

The amplitude of the function is given by the coefficient in front of the $\sin$ ; here the amplitude is -1.

The phase shift is given by the value being added or subtracted inside the $\sin$ function; here the shift is $\pi$ units to the right.

The only unexamined attribute of the graph is the vertical shift, so 4 is the vertical shift of the graph. A vertical shift of 4 means that the entire graph of the function will be moved up four units (in the positive y-direction).

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Question

Let be a function defined as follows:

$f(x)=3\cos(x+\pi)-3$

What is the vertical shift in this function?

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Answer

The period of the function is indicated by the coefficient in front of ; here the period is unchanged.

The amplitude of the function is given by the coefficient in front of the $\sin$ ; here the amplitude is 3.

The phase shift is given by the value being added or subtracted inside the cosine function; here the shift is $\pi$ units to the right.

The only unexamined attribute of the graph is the vertical shift, so -3 is the vertical shift of the graph. A vertical shift of -3 means that the entire graph of the function will be moved down three units (in the negative y-direction).

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Question

This graph shows a translated cosine function. Which of the following could be the equation of this graph?

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Answer

The correct answer is $f(x)=\cos x+3$ . There are no sign changes with vertical shifts; in other words, when the function includes , it directly translates to moving up three units. If you thought the answer was $f(x)=\cos x+4$ , you may have spotted the y-intercept at and jumped to this answer. However, recall that the y-intercept of a regular $y=\cos x$ function is at the point . Beginning at and ending at corresponds to a vertical shift of 3 units.

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Question

The graph below shows a translated sine function. Which of the following functions could be shown by this graph?

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Answer

A normal $y=\sin x$ graph has its y-intercept at . This graph has its y-intercept at . Therefore, the graph was shifted down three units. Therefore the function of this graph is $y=\sin x-3$ .

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Question

Consider the function $4sec(5(x+\frac{\pi}{3}$)) +10$ . What is the vertical shift of this function?

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Answer

The general form for the secant transformation equation is . represents the phase shift of the function. When considering $4sec(5(x+\frac{\pi}{3}$)) +10$ we see that , so our vertical shift is and we would shift this function units up from the original secant function’s graph.

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Question

Which of the following is the graph of $\sec x$ with a vertical shift of ?

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Answer

The graph of $\sec x$ with a vertical shift of is shown below. This can also be expressed as $y=\sec x-3$ .

Here is a graph that shows both $y=\sec x$ and $y=\sec x-3$ , so that you can see the "before" and "after." The original function is in blue and the translated function is in purple.

The graphs of the incorrect answer choices are $y=\sec x$ (no vertical shift applied), $y=\sec x+3$ (shifted upwards instead of downwards), $y=-3\sec x+3$ (amplitude modified, and shifted upwards instead of downwards), and $y=3\sec x-3$ (shifted downwards 3 units, but this is not the correct original graph of simply $y=\sec x$ since the amplitude was modified.)

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Question

Which of the following graphs shows one of the original six trigonometric functions with a vertical shift of applied?

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Answer

We are looking for an answer choice that has one of the six trigonometric functions, as well as that function shifted up 3 units. The only answer choice that displays that is this graph of $y=\tan x$ (purple) and $y=\tan x+3$ (blue).

The incorrect answers depict $y=\sin x$ and $y=\sin x-3$ , $y=\cos x$ and $y=\cos x+\pi$ , and $y=\sec x$ and $y=\sec (x-\pi)$ .

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Question

Simplify the following trionometric function:

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Answer

To solve the problem, you need to know the following information:

Replace the trigonometric functions with these values:

$$\frac{4(\frac{3}{4}$)-(1)}{($\frac{3\sqrt{2}$}{2})}$

$$\frac{2}{(\frac{3\sqrt{2}$}{2})}$

$$\frac{4}{3\sqrt{2}$} = $\frac{4\cdot \sqrt{2}$}{3$\sqrt{2}$\cdot $\sqrt{2}$} = $\frac{4\sqrt{2}$}{6} = $\frac{2\sqrt{2}$}{3}$

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Question

Change a angle to radians.

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Answer

In order to change an angle into radians, you must multiply the angle by $$\frac{\pi $}{180^{\circ}$$}$ .

Therefore, to solve:

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Question

Simplify the following trigonometric function in fraction form:

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Answer

To determine the value of the expression, you must know the following trigonometric values:

$sin $(30^{\circ}$)=\frac{1}{2}$$

Replacing these values, we get:

$$\frac{2}{4}$-$\frac{1}{2}$=\frac{1}{2}$-$\frac{1}{2}$=0$

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Question

If $cos\ x=\frac{1}{3}$$ and $sin\ x\neq 0$ , give the value of $\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$ .

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Answer

Based on the double angle formula we have, $sin\ 2x=2(sin\ x)(cos\ x)$ .

$\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$=\frac{sin\ x+2(sin\ x)(cos\ x)}{sin\ x-2(sin\ x)(cos\ x)}$

$=\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}$=\frac{1+2cos\ x}{1-2cos\ x}$$

$=\frac{1+2(\frac{1}{3}$)}{1-2($\frac{1}{3}$)}=\frac{\frac{3+2}{3}$}{$\frac{3-2}{3}$}=5$

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Question

If $x=\frac{\pi}{3}$$ , give the value of $\frac{sin\ x}{sin\ x+cos\ 2x}$ .

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Answer

$x=\frac{\pi}{3}$\Rightarrow sin\ x=sin($\frac{\pi}{3}$)=\frac{\sqrt{3}$}{2}$

Now we can write:

$x=\frac{\pi}{3}$\Rightarrow 2x=2($\frac{\pi}{3}$)=\frac{2\pi}{3}$\Rightarrow cos\ 2x=cos($\frac{2\pi}{3}$)$

$=cos(\pi-$\frac{\pi}{3}$)$

$=-cos ($\frac{\pi}{3}$)$

$=-$\frac{1}{2}$$

Now we can substitute the values:

$$\frac{sin\ x}{sin\ x+cos\ 2x}$=\frac{\frac{\sqrt{3}$}{2}}{$\frac{\sqrt{3}$}{2}-$\frac{1}{2}$}=\frac{\sqrt{3}$}{$\sqrt{3}$-1}=\frac{\sqrt{3}$}{$\sqrt{3}$-1}\times $\frac{{\sqrt{3}$+1}}{{$\sqrt{3}$+1}}=\frac{3+\sqrt{3}$}{2}$

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Question

If $cot\ $34^{\circ}$=1.5$ , give the value of $$\frac{2sin\ $326^{\circ}$$+3sin\ $56^{\circ}$}{cos\ $304^{\circ}$}$ .

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Answer

Now we can simplify the expression as follows:

$$\frac{2sin\ $326^{\circ}$$+3sin\ $56^{\circ}$}{cos\ $304^{\circ}$}=\frac{-2sin\ $34^{\circ}$$+3cos\ $34^{\circ}$}{sin\ $34^{\circ}$}$

$=-2+3cot\ $34^{\circ}$=-2+3\times 1.5=-2+4.5=2.5$

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Question

If $sin\ x\times cos\ x=\frac{1}{2}$$ , what is the value of $(sin\ x+cos\ $x)^2$$ ?

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Answer

$(sin\ x+cos\ $x)^2$$=sin^2x$$+cos^2x$+2sin\x cos\ x$

$=1+2(sin\ x)(cos\ x)=1+2($\frac{1}{2}$)=1+1=2$

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Question

If , give the value of .

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Answer

We know that .

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Question

Simplify:

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Answer

We know that .

Then we can write:

$\Rightarrow $A=\frac{(cos^2x$$-sin^2x$$)}{cos^2x$$}$+tan^2x$$

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Question

$tan (x)sin(x) + sec(x)(cos $x)^{2}$ =$

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Answer

$tan (x)sin(x) + sec(x)(cos $x)^{2}$ =$

$$\frac{sin (x)}{cos (x)}$* sin(x) + $\frac{1}{cos (x)}$*(cos $x)^{2}$ =$

$$\frac{(sin $x)^{2}$$ + (cos $x)^{2}$}{cos (x)} = $\frac{1}{cos (x)}$ = sec (x)$

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Question

Simplify the following expression:

$A=cos($\frac{3\pi}{2}$+x)+sin (\pi+x)+cos($\frac{\pi}{2}$+x)+sin(\pi-x)$

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Answer

We need to use the following identities:

$cos($\frac{3\pi}{2}$+x)=sin\ x$

$sin (\pi+x)=-sin\ x$

$cos($\frac{\pi}{2}$+x)=-sin\ x$

$sin(\pi-x)=sin\ x$

Use these to simplify the expression as follows:

$A=cos($\frac{3\pi}{2}$+x)+sin (\pi+x)+cos($\frac{\pi}{2}$+x)+sin(\pi-x)$

$\Rightarrow A=sin\ x-sin\ x-sin\ x+sin\ x = 0$

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Question

Give the value of :

$A=\frac{2sin(\frac{\pi}{2}$)+4cos($\frac{\pi}{2}$)}{4sin($\frac{\pi}{2}$)-cos($\frac{\pi}{2}$)}$

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Answer

$sin($\frac{\pi}{2}$)=1$

$cos($\frac{\pi}{2}$)=0$

Plug these values in:

$A=\frac{2sin(\frac{\pi}{2}$)+4cos($\frac{\pi}{2}$)}{4sin($\frac{\pi}{2}$)-cos($\frac{\pi}{2}$)}=\frac{2\times 1+4\times 0}{4\times 1-0}$=\frac{2}{4}$=\frac{1}{2}$=0.5$

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Question

If $tan\ x = $\frac{1}{4}$$ , solve for

$D=\frac{sin\ x}{sin\ x-cos\ x}$+\frac{sin\ x+cos\ x}{cos\ x}$$

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Answer

$tan\ x=\frac{1}{4}$\Rightarrow $\frac{sin\ x}{cos\ x}$=\frac{1}{4}$\Rightarrow cos\ x=4sin\ x$

Substitute $cos\ x=4sin \ x$ into the expression:

$D=\frac{sin\ x}{sin\ x-cos\ x}$+\frac{sin\ x+cos\ x}{cos\ x}$=\frac{sin\ x}{sin\ x-4sin\ x}$+\frac{sin\ x+4sin\ x}{4sin\ x}$$

$\Rightarrow D=\frac{sin\ x}{-3sin\ x}$+\frac{5sin\ x}{4sin\ x}$$

$\Rightarrow D=-$\frac{1}{3}$+\frac{5}{4}$=\frac{-4+15}{12}$=\frac{11}{12}$$

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