Trigonometric Functions and Graphs - Math

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Question

Simplify the function below:

Answer

We need to use the following formulas:

a)

and

b)

We can simplify as follows:

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Question

Which of the following is not in the range of the function $y = $\sin{x}$$ ?

Answer

The range of the function $y= \sin x$ is all numbers between and (the sine wave never goes above or below this).

Of the choices given, $\frac{3}{2}$ is greater than and thus not in this range.

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Question

What is the period of

$\tan \left ( $\frac{x}{\pi }$ \right )$ ?

Answer

The period for $\tan(x)$ is $\pi$ . However, if a number is multiplied by , you divide the period $\pi$ by what is being multiplied by . Here, $(1/\pi )$ is being multiplied by . $\pi/(1/\pi )$ equals .

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Question

Which of the given functions has the greatest amplitude?

Answer

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

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Question

What is the amplitude of $y=3\sin(2x+5)$ ?

Answer

The amplitude of a wave function like $y=3\sin(2x+5)$ is always going to be the coefficient of the function. In this case, that is .

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Question

What is the local maximum of $\sin(2x)$ between $-\pi$ and $\pi$ ?

Answer

The fastest way to solve this problem is to graph it and observe the answer. However, the other option is to think of this equation in terms of period.

When the coefficient of the variable increases, the frequency increases and the period decreases by that rate.

Since our equation is $\sin(2x)$ , our period will be $\frac{1}{2}$ the normal period of a $\sin(x)$ wave. Since only the period is changing, the amplitude is not. Therefore the amplitude (the highest and lowest points) of $\sin(2x)$ will be the same as that of $\sin(x)$ . The amplitude of a sine wave is , so the amplitude of $\sin(2x)$ will also be .

Therefore, our maximum will be .

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Question

Simplify the following trionometric function:

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.

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:

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 $\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$ .

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 $\frac{sin\ x}{sin\ x+cos\ 2x}$ .

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\ $304^{\circ}$}$ .

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}$$ , what is the value of $(sin\ x+cos\ $x)^2$$ ?

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 .

Answer

We know that .

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Question

Simplify:

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 $x)^{2}$ =$

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

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

Simplify the trigonometric expression.

$\frac{\tan(\theta)\csc(\theta)}{\sec(\theta)}$

Answer

Using basic trigonometric identities, we can simplify the problem to

$$\frac{\frac{\sin(\theta)}{\cos(\theta)}$$\frac{1}{\sin(\theta)}$}{$\frac{1}{\cos(\theta)}$}$ .

We can cancel the sine in the numerator and the one over cosine cancels on top and bottom, leaving us with 1.

$$\frac{\frac{\sin(\theta)}{\cos(\theta)}$$\frac{1}{\sin(\theta)}$}{$\frac{1}{\cos(\theta)}$}=\frac{\frac{1}{\cos(\theta)}$}{$\frac{1}{\cos(\theta)}$}=\frac{1}{\cos(\theta)}$\cdot $\frac{\cos(\theta)}{1}$=1$

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Question

Give the value of :

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

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}$$

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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