Trigonometric Operations - Trigonometry

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Question

What is $\theta$ if and ?

Answer

In order to find $\theta$ we need to utilize the given information in the problem. We are given the opposite and adjacent sides. We can then, by definition, find the $\tan$ of $\theta$ and its measure in degrees by utilizing the $\arctan$ function.

$\tan=\frac{opposite}{adjacent}$

$\tan=\frac{10}{8}$

$\tan=1.25$

Now to find the measure of the angle using the $\arctan$ function.

$\theta=\arctan1.25$

$\rightarrow 51.34^{\circ}$

If you calculated the angle's measure to be $0.90^{\circ}}$ then your calculator was set to radians and needs to be set on degrees.

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Question

Which of the following is the degree equivalent of the inverse trigonometric function

$\arccos x=\frac{1}{2}$ ?

Answer

The $\arccos$ is the reversal of the cosine function. That means that if $\cos \theta=y$ , then $\arccos y=\theta$ .

Therefore,

$\theta=\arccos\left(\frac{1}{2}\right)=60^\circ$

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Question

For the above triangle, what is $\theta$ if and ?

Answer

We need to use a trigonometric function to find $\theta$ . We are given the opposite and adjacent sides, so we can use the $\tan$ and $\arctan$ functions.

$\tan(\theta) = \frac{\textup{opposite}}{\textup{adjacent}}$

$\tan(\theta) = \frac{20}{14}$

$\theta = \arctan\left ( \frac{20}{14} \right )$

$\theta = 55.0^{\circ}$

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Question

For the above triangle, what is $\theta$ if and ?

Answer

We need to use a trigonometric function to find $\theta$ . We are given the opposite and hypotenuse sides, so we can use the $\sin$ and $\arcsin$ functions.

$\sin(\theta) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin(\theta) = \frac{17}{25}$

$\theta = \arcsin\left ( \frac{17}{25} \right )$

$\theta = 42.8^{\circ}$

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Question

Assuming the angle in degrees, determine the value of $sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)$ .

Answer

To evaluate $sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)$ , it is necessary to know the existing domain and range for these inverse functions.

Inverse sine:

$R \textup{ (in radians)}: [-\frac{\pi}{2},\frac{\pi}{2}]$

Inverse cosine:

$R\textup{ (in radians)}: [0,\pi]$

Inverse tangent:

$D: \textup{All real numbers.}$

$R\textup{ (in radians)}: (-\frac{\pi}{2},\frac{\pi}{2})$

Evaluate each term. The final answers must return an angle.

$sin^{-1}(1)=\frac{\pi}{2} \textup{ radians}=90 \textup{ degrees}$

$cos^{-1}(1)=0 \textup{ degrees}$

$tan^{-1}(1)=\frac{\pi}{4} \textup{ radians}= 45 \textup{ degrees}$

$sin^{-1}(1)-cos^{-1}(1)-tan^{-1}(1)=90-0-45=45 \textup{ degrees}$

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Question

If

$cos(arcsin(sec \ x)) = 0{}$ ,

what value(s) does take?

Assume that $0\leq x< 2\pi$

Answer

If $cos(arcsin(sec \ x)) = 0{}$ , then we can apply the cosine inverse to both sides:

$cos^{-1}(cos(arcsin(sec \ x))) = cos^{-1}(0)$

$arcsin(sec \ x)) = \frac{\pi }{2} \ or -\frac{\pi }{2}$

Since cosine and cosine inverse undo each other; we can then apply sine and secant inverse functions to obtain the solution.

${}sin(arcsin(sec \ x))) = sin\left(\frac{\pi }{2} \right)$ and $sin(arcsin(sec \ x))) = sin\left(-\frac{\pi }{2}\right) {}$

and

$x = 0, \pi$ are the two solutions.

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Question

Calculate .

Answer

The arcsecant function takes a trigonometric ratio on the unit circle as its input and results in an angle measure as its output. The given function can therefore be rewritten as

$Arcsec2=Arccos\frac{1}{2}$

and is the angle measure $\Theta$ which, when applied to the cosine function $cos\Theta$ , results in $\frac{1}{2}$ . Notice that the arcsecant function as expressed in the statement of the problem is capitalized; hence, we are looking for the "principal" angle measure, or the one which lies between and $\pi rad$ . Since $cos(\frac{\pi}3)=\frac{1}{2}$ , and since $\frac{\pi}{3}$ lies between and $\pi rad$ ,

$Arcsec2=\frac{\pi}{3}$ .

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Question

Calculate .

Answer

The domain on the argument for is

$-\frac{\pi}{2}<x<\frac{\pi}{2}$ .

The range of the function is not defined at $-\frac{\pi}{2}$ or $\frac{\pi}{2}$ , and so the domain of its inverse, , does not include those values. Hence, we must find the angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ for which .

Since $\frac{sinx}{cosx}=tanx$ , the equation can be rewritten as

$\frac{sinx}{cosx}=1$ ,

or

for some x between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ .

Now, when $x=\frac{\pi}{4}$ , since $sin(\frac{\pi}{4})=cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ .

Therefore,

$arctan1=\frac{\pi}{4}$ .

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the secant of $\angle A$ ?

Answer

The value of the secant of an angle is the value of the hypotenuse over the adjacent.

Therefore:

$sec \angle A = \frac{hypotenuse}{adjacent} = \frac{25}{24}$

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Question

Which of the following is the equivalent to $\frac{1}{\csc\theta}$ ?

Answer

Since $\csc\theta=\frac{1}{\sin\theta}$ :

$\frac{1}{\csc\theta}=\frac{1}{\frac{1}{\sin\theta}}=1*\frac{\sin\theta}{1}=\sin\theta$

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Question

For the above triangle, what is $\sec (\theta)$ if , and ?

Answer

Secant is the reciprocal of cosine.

$\sec \left ( \theta\right ) = \frac{1}{\cos \left ( \theta\right )}$

It's formula is:

$\sec(\theta) = \frac{\textup{hypotenuse}}{\textup{adjacent}}$

Substituting the values from the problem we get,

$\sec(\theta) = \frac{17}{15} = 1.13$

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Question

For the above triangle, what is $\cot (\theta)$ if , and ?

Answer

Cotangent is the reciprocal of tangent.

$\cot \left ( \theta\right ) = \frac{1}{\tan \left ( \theta\right )}$

It's formula is:

$\cot(\theta) = \frac{\textup{adjacent}}{\textup{opposite}}$

Substituting the values from the problem we get,

$\cot(\theta) = \frac{24}{18} = 1.33$

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Question

Evaluate:

Answer

Evaluate each term separately.

$sec(60)= \frac{1}{cos(60)}=\frac{1}{0.5}= 2$

$csc(45)= \frac{1}{sin(45)}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}=\sqrt2$

$sec(60)-csc(45)=2-\frac{2\sqrt2}{2}=2-\sqrt2$

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Question

Determine the value of $cot(\frac{\pi}{4})$ .

Answer

Rewrite $cot(\frac{\pi}{4})$ in terms of sine and cosine.

$cot(\frac{\pi}{4})= \frac{cos(\frac{\pi}{4})}{sin(\frac{\pi}{4})}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=1$

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Question

Pick the ratio of side lengths that would give sec C.

Answer

$\sec\theta=\frac{1}{\cos\theta }$

Find the ratio of Cosine and take the reciprocal.

$\cos\theta=\frac{AC}{CB}\rightarrow \sec\theta=\frac{1}{\frac{AC}{CB}}=\mathbf{\frac{CB}{AC}}$

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Question

If $\sin x=\frac{28}{53}$ , $\cot x=$

Answer

The sine of an angle in a right triangle (that is not the right angle) can be found by dividing the length of the side opposite the angle by the length of the hypotenuse of the triangle.

From this, the length of the side opposite the angle is proportional to 28, and the length of the hypotenuse is proportional to 53.

Without loss of generality, we'll assume that the sides are actually of length 28 and 53, respectively.

We'll use the Pythagorean theorem to determine the length of the adjacent side, which we'll refer to as .

$\begin{align} y&=\sqrt{53^2-28^2}\ &=\sqrt{2809-784}\ &=\sqrt{2025}\ &=45 \end{align}$

The cotangent of an angle in a right triangle (that is not the right angle) is can be found by dividing the length of the adjacent side by the length of the opposite side.

$\cot x=\frac{45}{28}$

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Question

If cos x = 0.2 $\small \cosx$ and sin x = 0.4, what is the value of tan x?

Answer

$\small \tan{x}=\frac{\sin{x}}{\cos{x}}$

$\small \small \tan{x}=\frac{0.4}{0.2}$

$\small \tan{x}=0.035$

$\small x=\tan^{-1}0.035$

$\small x=2$

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Question

Find the value of the trigonometric function in fraction form for triangle .

What is the cosine of $\angle B$ ?

Answer

The cosine of an angle is the value of the adjacent side over the hypotenuse.

Therefore:

$cos \angle B = \frac{adjacent}{hypotenuse} = \frac{7}{25}$

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Question

Which of the following describes the ratio of sine?

Answer

Sine is by definition of sides in a right triangle is opposite side over hypotenuse.

To remember this, use SOH CAH TOA.

SOH: Sine=Opposite/Hypotenuse

CAH: Cosine=Adjacent/Hypotenuse

TOA: Tangent=Opposite/Adjacent

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Question

What is the value of ?

Answer

Solve each term separately.

$cos(30)= \frac{\sqrt3}{2}$

$cos(60)=\frac{1}{2}$

Add both terms.

$\frac{\sqrt3}{2}+\frac{1}{2}= \frac{\sqrt3+1}{2}$

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