Find all angles in a range given specific output - Trigonometry

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Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-$\frac{1}{2}$$ .

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At first glance, you may think that this problem has infinite answers, since there would be infinitely many negative coterminal angles that could satisfy this; however, notice that the question asks only for positive values of $\theta$ . In other words, this question is simply asking for values of $\theta$ between $0^\circ$ and $360^\circ$ that satisfy this equation.

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\sin \theta=.6478$ .

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This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\sin \theta=.6478$ is positive, solutions for $\theta$ will be in Quadrants I and II because these are the quadrants where sine is positive. Use inverse sine and a calculator to find $\theta$ :

$\sin R=.6478$

$\sin^$-^1$(.6478)=40.38^\circ$

In Quadrant I, we have $R=\theta$ , so $\theta=40.38^\circ$ .

In Quadrant II, $R=180^\circ-\theta$ , so $R=180^\circ-40.38^\circ=139.62^\circ$ .

Therefore $\theta = 40.38^\circ$ and $139.62^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos \theta=-.3333$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\cos \theta=-.3333$ is negative, solutions for $\theta$ will be in Quadrants II and III because these are the quadrants where cosine is negative.

Use inverse cosine and a calculator to find $\theta$ :

$\cos R = -.3333$

$\cos^$-^1$(.3333)=70.53^\circ = R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-70.53^\circ=109.47^\circ$ .

In Quadrant III, $R=\theta-180^\circ$ , so $\theta=180^\circ+R=180^\circ+70.53^\circ=250.53^\circ$ .

Therefore $\theta=109.47^\circ$ and $250.53^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\tan\theta=-1.3456$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\tan\theta=-1.3456$ is negative, solutions for $\theta$ will be in Quadrants II and IV because these are the quadrants where tangent is negative. Use inverse tangent and a calculator to find $\theta$ :

$\tan R=-1.3456$

$\tan^$-^1$(1.3456)=53.38^\circ=R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-53.38^\circ=126.62^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-53.38^\circ=306.62^\circ$ .

Therefore $\theta=126.62^\circ$ and $306.62^\circ$ .

Find all angles $\theta$ when $\tan\theta=.5432$ .

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We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\tan R=.5432$ , so $\tan ^$-^1$(.5432)=28.51^\circ=R$ . Next, think about where tangent is positive, or reference the Function Signs column of the above table. Tangent is positive in Quadrants I and III.

In Quadrant I, $\theta=R=28.51^\circ$ .

In Quadrant III, $\theta=180^\circ+R=180^\circ+28.51^\circ=208.51^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $28.51^\circ$ and $208.51^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

Find all angles $\theta$ when $\sin \theta=-.1919$ .

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We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\sin \theta=-.1919$ so $\sin^$-^1$(.1919)=11.06^\circ = R$ . Next, think about where sine is negative, or reference the Function Signs column of the above table. Sine is negative in Quadrants III and IV.

In Quadrant III, $\theta=180^\circ+R=180^\circ+11.06^\circ=191.06^\circ$ .

In Quadrant IV, $\theta=306^\circ-R=360^\circ-11.06^\circ=348.94^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $191.06^\circ$ and $348.94^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos\theta=.8999$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\cos\theta=.8999$ is positive, solutions for $\theta$ will be in Quadrants I and IV because these are the quadrants where cosine is positive. Use inverse cosine and a calculator to find $\theta$ :

$\cos\theta=.8999$

$\cos^$-^1$=(.8999)=25.86^\circ=R$

In Quadrant I, we have $R=\theta$ , so $\theta=25.86^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-25.86^\circ=334.14^\circ$ .

Therefore $\theta=25.86^\circ$ and $334.14^\circ$ .

Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-$\frac{1}{2}$$ .

Tap to see back →

At first glance, you may think that this problem has infinite answers, since there would be infinitely many negative coterminal angles that could satisfy this; however, notice that the question asks only for positive values of $\theta$ . In other words, this question is simply asking for values of $\theta$ between $0^\circ$ and $360^\circ$ that satisfy this equation.

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\sin \theta=.6478$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\sin \theta=.6478$ is positive, solutions for $\theta$ will be in Quadrants I and II because these are the quadrants where sine is positive. Use inverse sine and a calculator to find $\theta$ :

$\sin R=.6478$

$\sin^$-^1$(.6478)=40.38^\circ$

In Quadrant I, we have $R=\theta$ , so $\theta=40.38^\circ$ .

In Quadrant II, $R=180^\circ-\theta$ , so $R=180^\circ-40.38^\circ=139.62^\circ$ .

Therefore $\theta = 40.38^\circ$ and $139.62^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos \theta=-.3333$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\cos \theta=-.3333$ is negative, solutions for $\theta$ will be in Quadrants II and III because these are the quadrants where cosine is negative.

Use inverse cosine and a calculator to find $\theta$ :

$\cos R = -.3333$

$\cos^$-^1$(.3333)=70.53^\circ = R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-70.53^\circ=109.47^\circ$ .

In Quadrant III, $R=\theta-180^\circ$ , so $\theta=180^\circ+R=180^\circ+70.53^\circ=250.53^\circ$ .

Therefore $\theta=109.47^\circ$ and $250.53^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\tan\theta=-1.3456$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\tan\theta=-1.3456$ is negative, solutions for $\theta$ will be in Quadrants II and IV because these are the quadrants where tangent is negative. Use inverse tangent and a calculator to find $\theta$ :

$\tan R=-1.3456$

$\tan^$-^1$(1.3456)=53.38^\circ=R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-53.38^\circ=126.62^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-53.38^\circ=306.62^\circ$ .

Therefore $\theta=126.62^\circ$ and $306.62^\circ$ .

Find all angles $\theta$ when $\tan\theta=.5432$ .

Tap to see back →

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\tan R=.5432$ , so $\tan ^$-^1$(.5432)=28.51^\circ=R$ . Next, think about where tangent is positive, or reference the Function Signs column of the above table. Tangent is positive in Quadrants I and III.

In Quadrant I, $\theta=R=28.51^\circ$ .

In Quadrant III, $\theta=180^\circ+R=180^\circ+28.51^\circ=208.51^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $28.51^\circ$ and $208.51^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

Find all angles $\theta$ when $\sin \theta=-.1919$ .

Tap to see back →

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\sin \theta=-.1919$ so $\sin^$-^1$(.1919)=11.06^\circ = R$ . Next, think about where sine is negative, or reference the Function Signs column of the above table. Sine is negative in Quadrants III and IV.

In Quadrant III, $\theta=180^\circ+R=180^\circ+11.06^\circ=191.06^\circ$ .

In Quadrant IV, $\theta=306^\circ-R=360^\circ-11.06^\circ=348.94^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $191.06^\circ$ and $348.94^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos\theta=.8999$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\cos\theta=.8999$ is positive, solutions for $\theta$ will be in Quadrants I and IV because these are the quadrants where cosine is positive. Use inverse cosine and a calculator to find $\theta$ :

$\cos\theta=.8999$

$\cos^$-^1$=(.8999)=25.86^\circ=R$

In Quadrant I, we have $R=\theta$ , so $\theta=25.86^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-25.86^\circ=334.14^\circ$ .

Therefore $\theta=25.86^\circ$ and $334.14^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\sin \theta=.6478$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\sin \theta=.6478$ is positive, solutions for $\theta$ will be in Quadrants I and II because these are the quadrants where sine is positive. Use inverse sine and a calculator to find $\theta$ :

$\sin R=.6478$

$\sin^$-^1$(.6478)=40.38^\circ$

In Quadrant I, we have $R=\theta$ , so $\theta=40.38^\circ$ .

In Quadrant II, $R=180^\circ-\theta$ , so $R=180^\circ-40.38^\circ=139.62^\circ$ .

Therefore $\theta = 40.38^\circ$ and $139.62^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\cos \theta=-.3333$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\cos \theta=-.3333$ is negative, solutions for $\theta$ will be in Quadrants II and III because these are the quadrants where cosine is negative.

Use inverse cosine and a calculator to find $\theta$ :

$\cos R = -.3333$

$\cos^$-^1$(.3333)=70.53^\circ = R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-70.53^\circ=109.47^\circ$ .

In Quadrant III, $R=\theta-180^\circ$ , so $\theta=180^\circ+R=180^\circ+70.53^\circ=250.53^\circ$ .

Therefore $\theta=109.47^\circ$ and $250.53^\circ$ .

Find all angles $\theta$ between $0^\circ$ and $360^\circ$ when $\tan\theta=-1.3456$ .

Tap to see back →

This problem relies on understanding reference angles and coterminal angles. A reference angle for an angle $\theta$ in standard position is the positive acute angle between the x axis and the terminal side of the angle $\theta$ . A table of reference angles for each quadrant is given below.

Since $\tan\theta=-1.3456$ is negative, solutions for $\theta$ will be in Quadrants II and IV because these are the quadrants where tangent is negative. Use inverse tangent and a calculator to find $\theta$ :

$\tan R=-1.3456$

$\tan^$-^1$(1.3456)=53.38^\circ=R$

In Quadrant II, we have $R=180^\circ-\theta$ , so $\theta=180^\circ-53.38^\circ=126.62^\circ$ .

In Quadrant IV, $R=360^\circ-\theta$ , so $R=360^\circ-53.38^\circ=306.62^\circ$ .

Therefore $\theta=126.62^\circ$ and $306.62^\circ$ .

Find all angles $\theta$ when $\tan\theta=.5432$ .

Tap to see back →

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\tan R=.5432$ , so $\tan ^$-^1$(.5432)=28.51^\circ=R$ . Next, think about where tangent is positive, or reference the Function Signs column of the above table. Tangent is positive in Quadrants I and III.

In Quadrant I, $\theta=R=28.51^\circ$ .

In Quadrant III, $\theta=180^\circ+R=180^\circ+28.51^\circ=208.51^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $28.51^\circ$ and $208.51^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=28.51^\circ+n360^\circ$ and $\theta=208.51^\circ+n360^\circ$

Find all angles $\theta$ when $\sin \theta=-.1919$ .

Tap to see back →

We can use reference angles, inverse trig, and a calculator to solve this problem. Below is a table of reference angles.

We have $\sin \theta=-.1919$ so $\sin^$-^1$(.1919)=11.06^\circ = R$ . Next, think about where sine is negative, or reference the Function Signs column of the above table. Sine is negative in Quadrants III and IV.

In Quadrant III, $\theta=180^\circ+R=180^\circ+11.06^\circ=191.06^\circ$ .

In Quadrant IV, $\theta=306^\circ-R=360^\circ-11.06^\circ=348.94^\circ$ .

If this problem asked for values of $\theta$ between $0^\circ$ and $360^\circ$ , our work would be done, but this problem does not restrict the range, so we need to give all possible values of $\theta$ by generalizing our answers. To do this, we must understand that all angles that are coterminal to $191.06^\circ$ and $348.94^\circ$ will also be solutions. Coterminal angles add or subtract multiples of $360^\circ$ . To write this generally, we write:

$\theta=191.06^\circ+n360^\circ$ and $\theta=348.94^\circ+n360^\circ$ .

Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-$\frac{1}{2}$$ .

Tap to see back →

At first glance, you may think that this problem has infinite answers, since there would be infinitely many negative coterminal angles that could satisfy this; however, notice that the question asks only for positive values of $\theta$ . In other words, this question is simply asking for values of $\theta$ between $0^\circ$ and $360^\circ$ that satisfy this equation.

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$