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Trigonometry : Trigonometric Operations

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Example Question #26 : Trigonometry

Evaluate: sec(60)-csc(45)

Possible Answers:

$2-\frac{\sqrt2}{2}$

$2\sqrt2+2$

$2+\sqrt2$

$2+\frac{\sqrt2}{2}$

$2-\sqrt2$

Correct answer:

$2-\sqrt2$

Explanation:

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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Example Question #1 : Sec, Csc, Ctan

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

Possible Answers:

$\sec C=\frac{AB}{AC}$

$\sec C=\frac{CB}{AC}$

$\sec C=\frac{AC}{CB}$

None of the other answers.

$\sec C=\frac{AC}{AB}$

Correct answer:

$\sec C=\frac{CB}{AC}$

Explanation:

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

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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Example Question #1 : Sec, Csc, Ctan

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

Possible Answers:

$\frac{28}{45}$

$\frac{45}{28}$

$\frac{45}{53}$

$\frac{53}{28}$

Correct answer:

$\frac{45}{28}$

Explanation:

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 .

y^2+28^2=53^2

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

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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Example Question #1 : Arcsin, Arccos, Arctan

Trig_id

What is $\theta$ if o=10 and a=8 ?

Possible Answers:

$0.90^{\circ}$

$51.34^{\circ}$

$89.60^{\circ}$

$53.14^{\circ}$

$55.14^{\circ}$

Correct answer:

$51.34^{\circ}$

Explanation:

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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Example Question #561 : Trigonometry

Soh_cah_toa

For the above triangle, what is $\theta$ if o = 20 and a = 14 ?

Possible Answers:

$35.0^{\circ}$

$55.0^{\circ}$

$60.0^{\circ}$

$44.4^{\circ}$

$45.6^{\circ}$

Correct answer:

$55.0^{\circ}$

Explanation:

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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Example Question #562 : Trigonometry

Soh_cah_toa

For the above triangle, what is $\theta$ if o = 17 and h = 25 ?

Possible Answers:

$42.8^{\circ}$

$55.8^{\circ}$

$34.2^{\circ}$

$47.2^{\circ}$

$11.8^{\circ}$

Correct answer:

$42.8^{\circ}$

Explanation:

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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Example Question #4 : Arcsin, Arccos, Arctan

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

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

Possible Answers:

$60^\circ$

$30^\circ$

$360^\circ$

$1^\circ$

$2^\circ$

Correct answer:

$60^\circ$

Explanation:

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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Example Question #5 : Arcsin, Arccos, Arctan

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

Possible Answers:

$\textup{The answer cannot be determined.}$

Correct answer:

Explanation:

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:

D: [-1,1]

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

Inverse cosine:

D: [-1,1]

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

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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Example Question #1 : Arcsin, Arccos, Arctan

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

what value(s) does take?

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

Possible Answers:

No real solution.

$\frac{\pi}{2}, -\frac{\pi}{2}$

$0, \pi$

$\frac{\pi}{2}$

Correct answer:

$0, \pi$

Explanation:

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

sec(x) = 1 and sec(x) = -1

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

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Example Question #6 : Arcsin, Arccos, Arctan

Calculate Arcsec(2) .

Possible Answers:

$\frac{5\pi}{3}$

$\frac{\pi}{3}$ and $\frac{5\pi}{3}$

$\frac{3\pi}{2}$

$\frac{\pi}{3}$

$\frac{\pi}{3} and \frac{3\pi}{2}$

Correct answer:

$\frac{\pi}{3}$

Explanation:

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 0rad and $\pi rad$ . Since $cos(\frac{\pi}3)=\frac{1}{2}$ , and since $\frac{\pi}{3}$ lies between 0rad and $\pi rad$ ,

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

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Trigonometry : Trigonometric Operations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #26 : Trigonometry

Example Question #1 : Sec, Csc, Ctan

Example Question #1 : Sec, Csc, Ctan

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #561 : Trigonometry

Example Question #562 : Trigonometry

Example Question #4 : Arcsin, Arccos, Arctan

Example Question #5 : Arcsin, Arccos, Arctan

Example Question #1 : Arcsin, Arccos, Arctan

Example Question #6 : Arcsin, Arccos, Arctan

All Trigonometry Resources

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