Trigonometry - Find all angles in a range given specific output

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

$\theta=60^\circ$

$\theta=120^\circ$

$\theta=240^\circ$

$\theta=60^\circ$ and $240^\circ$

$\theta=120^\circ$ and $240^\circ$

Explanation

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.

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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:

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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 $\theta$ when $\sin \theta=-.1919$ .

$\theta=191.06^\circ$ and $\theta=348.94^\circ$

$\theta=11.06^\circ$ and $\theta=191.06^\circ$

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

$\theta=11.06^\circ+n360^\circ$ and $\theta=191.06^\circ+n360^\circ$

Explanation

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

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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 $\cos \theta=-.3333$ .

$\theta=70.53^\circ$

$\theta=190.47^\circ$

$\theta=70.53^\circ$ and $190.47^\circ$

$\theta=70.53^\circ$ and $250.53^\circ$

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

Explanation

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.

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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 $\cos\theta=.8999$ .

$\theta=25.86^\circ$

$\theta=154.14^\circ$

$\theta=334.14^\circ$

$\theta=25.86^\circ$ and $154.14^\circ$

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

Explanation

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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 $\sin \theta=.6478$ .

$\theta = 40.38^\circ$ and $130.38^\circ$

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

$\theta = 40.38^\circ$ and $319.62^\circ$

$\theta=$ $130.38^\circ$ and $139.62^\circ$

$\theta=$ $139.62^\circ$ and $319.62^\circ$

Explanation

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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 $\tan\theta=-1.3456$ .

$\theta=53.38^\circ$

$\theta=126.62^\circ$

$\theta=53.38^\circ$ and $126.62^\circ$

$\theta=53.38^\circ$ and $306.62^\circ$

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

Explanation

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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 $\tan\theta=.5432$ .

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

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

$\theta=28.51^\circ+n360^\circ$ and $\theta=151.49^\circ+n360^\circ$

$\theta=28.51^\circ$ and $\theta=151.49^\circ$

Explanation

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

Screen shot 2020 07 30 at 11.05.57 am

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 in a range given specific output

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