Exercises on trigonometric circle with answer

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Practice trigonometric circle with this list of exercises solved step by step. Ask your questions and be prepared for your assessments.

question 1

Determine in which quadrant the angle of 2735° in the positive direction is located.

See Answer

Since each complete revolution is 360°, we divide 2735 by 360.

2735 degree sign space divided by space 360 degree sign equals space 7 multiplication sign 360 degree sign space plus space 215 degree sign

That's seven full turns plus 215º.

The 215° angle is in the third quadrant in the positive (counterclockwise) direction.

question 2

Let A be the set formed by the first six multiples of pi over 3 typographic , determine the sine of each of the arcs.

See Answer

The first six multiples are, in degrees:

$straight pi over 3 space multiplication sign space 1 space equals straight pi over 3 equals 60 degree sign straight pi over 3 space multiplication sign space 2 equals numerator 2 straight pi over denominator 3 end of fraction equals 120 degree sign straight pi over 3 space multiplication sign space 3 equals numerator 3 straight pi over denominator 3 end of fraction equals straight pi equals 180 degree sign straight pi over 3 space multiplication sign space 4 equals numerator 4 straight pi over denominator 3 end of fraction equal to 240 straight degree sign pi over 3 space multiplication sign space 5 equals numerator 5 straight pi over denominator 3 end of fraction equal to 300 sign of degree straight pi over 3 space multiplication sign space 6 space equals numerator 6 straight pi over denominator 3 end of fraction equals 2 straight pi space equals space 360 degree sign$

Let's determine the sine values per quadrant of the trigonometric circle.

1st quadrant (positive sine)

sin space 2 straight pi space equals sin space 360 degree sign equals 0

$sin straight space pi over 3 space equals sin space 60 degree sign equals numerator square root of 3 over denominator 2 end of fraction$

2nd quadrant (positive sine)

$sin space numerator 2 straight pi over denominator 3 end of fraction equals sin space 120 degree sign equals numerator square root of 3 over denominator 2 end of fraction$

3rd quadrant (negative sine)

$sin space numerator 4 straight pi over denominator 3 end of fraction equals sin space 240 degree sign equals minus numerator square root of 3 over denominator 2 end of fraction$

4th quadrant (negative sine)

$sin space numerator 5 straight pi over denominator 3 end of fraction equals sin space 300 degree sign equals minus numerator square root of 3 over denominator 2 end of fraction$

question 3

Considering the expression $numerator 1 over denominator 1 minus cos straight space x end of fraction$ , with straight x not equal straight k.2 straight pi , determine the value of x to obtain the smallest possible result.

See Answer

The smallest possible result occurs when the denominator is maximum. For this, the cos x must be as small as possible.

The smallest value of cosine is -1, and occurs when x is 180º or, straight pi .

$numerator 1 over denominator 1 minus cos straight space pi end of fraction equals numerator 1 over denominator 1 minus parenthesis left minus 1 right parenthesis end of fraction equals numerator 1 over denominator 1 plus 1 end of fraction equals bold 1 over bold 2$

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question 4

Calculate the value of the expression: $tg open parentheses numerator 4 straight pi over denominator 3 end of fraction close parentheses minus tg open parentheses numerator 5 straight pi over denominator 6 end of fraction close parentheses$ .

See Answer

$tg open parentheses numerator 4 straight pi over denominator 3 end of fraction close parentheses minus tg open parentheses numerator 5 straight pi over denominator 6 end of fraction close parentheses equal to tg open parentheses numerator 4,180 over denominator 3 end of fraction close parentheses minus tg open parentheses numerator 5,180 over denominator 6 end of fraction close parentheses equals tg space 240 space minus space tg space 150 space equal to$

The tangent is positive for the 240° angle as it is in the third quadrant. It is equivalent to the tangent of 60° in the first quadrant. Soon,

t g space 240 space equals space square root of 3

The tangent of 150° is negative as it is in the second quadrant. It is equivalent to the tangent of 30° in the first quadrant. Soon,

$tg space 150 equals minus numerator square root of 3 over denominator 3 end of fraction$

Returning the expression:

$tg space 240 space minus space tg space 150 equals square root of 3 space minus space opens parentheses minus numerator square root of 3 over denominator 3 end of fraction close parentheses equals square root of 3 space plus numerator square root of 3 over denominator 3 end of fraction equals numerator 3 square root of 3 space plus space square root of 3 over denominator 3 end of fraction equals bold numerator 4 square root of bold 3 over denominator bold 3 end of fraction$

question 5

The fundamental relationship of trigonometry is an important equation relating sine and cosine values, expressed as:

sin squared right x plus cos squared right x equals 1

Considering an arc in the 4th quadrant and the tangent of this arc equal to -0.3, determine the cosine of this same arc.

See Answer

The tangent is defined as:

$tg straight space x equals numerator sin straight space x over denominator cos straight space x end of fraction$

Isolating the sine value in this equation, we have:

sin straight space x space equals space tg straight space x space. space cos straight space x sin straight space x space equals space minus 0 comma 3. cos straight space x

Substituting in the fundamental relation:

$open parentheses minus 0 comma 3. cos straight space x close parentheses squared space plus space cos squared space x space equals space 1 0 comma 09. cos squared x space plus space cos squared space x space equals space 1 cos squared x space left parenthesis 0 comma 09 space plus space 1 right parenthesis equals 1 cos squared x space. space 1 comma 09 space equals space 1 cos squared x space equals numerator space 1 over denominator 1 comma 09 end of fraction cos space x equals space square root of numerator 1 over denominator 1 comma 09 end of fraction end of root cos space x approximately equals 0 comma 96$

question 6

(Fesp) The expression OK:

a) 5/2

b) -1

c) 9/4

d) 1.

e) 1/2

Answer explained

$numerator 5 cos 90 space minus space 4 space cos 180 over denominator 2 sin 270 space minus space 2 sin 90 end of equal fraction numerator 5.0 space minus space 4. left parenthesis minus 1 right parenthesis over denominator 2. left parenthesis minus 1 right parenthesis space minus space 2.1 end of fraction equals numerator 4 over denominator minus 2 space minus space 2 end of fraction equals numerator 4 over denominator minus 4 end of fraction equals bold minus bold 1$

question 7

(CESGRANRIO) If is an arc of the 3rd quadrant and then é:

The) $minus numerator square root of 5 over denominator 2 end of fraction$

B) minus 1

w) less space 1 medium

d) $minus numerator square root of 2 over denominator 2 end of fraction$

It is) $minus numerator square root of 3 over denominator 2 end of fraction$

Answer explained

As tg x = 1, x must be a multiple of 45º that generates a positive value. So, in the third quadrant, this angle is 225º.

In the first quadrant, cos 45º = $numerator square root of 2 over denominator 2 end of fraction$ , in the third quadrant, cos 225º = $minus numerator square root of 2 over denominator 2 end of fraction$ .

question 8

(UFR) Performing the expression has as a result

a) 0

b) 2

c) 3

d) -1

e) 1

Answer explained

$numerator sin squared space 270 space minus space cos space 180 space plus sen space space 90 over denominator tg squared space 45 end of equal fraction numerator sin space 270 space. space sin space 270 space minus space cos space 180 space plus space sin space 90 over denominator tg space 45 space. tg space 45 end of fraction equals numerator minus 1 space. space left parenthesis minus 1 right parenthesis space minus space left parenthesis minus 1 right parenthesis space plus space 1 over denominator 1 space. space 1 end of fraction equals numerator 1 space minus space left parenthesis minus 1 right parenthesis space plus space 1 over denominator 1 end of fraction equals numerator 1 space plus space 1 space plus space 1 over denominator 1 end of fraction equals a3 over 1 equals bold 3$

question 9

Knowing that x belongs to the second quadrant and that cos x = –0.80, it can be stated that

a) cosec x = –1.666...

b) tg x = –0.75

c) sec x = –1.20

d) cotg x = 0.75

e) sin x = –0.6

Answer explained

By the trigonometric circle, we obtain the fundamental relation of trigonometry:

Once we have the cosine, we can find the sine.

right squared sin x plus right cos squared x equals 1 right squared sin x equals 1 minus right cos squared x sin squared right x equals 1 minus left parenthesis minus 0 comma 80 right parenthesis squared sin to the power of 2 end of right exponential x equals 1 minus 0 comma 64sin squared straight x equals 0 comma 36sin straight space x equals the square root of 0 comma 36 end of rootsen straight space x equals 0 comma 6

The tangent is defined as:

$tg straight space x equals numerator sin straight space x over denominator cos straight space x end of fractiontg straight space x equals numerator 0 comma 6 over denominator minus 0 comma 8 end of fractionbold tg bold space bold x bold equals bold minus bold 0 bold comma bold 75$

question 10

(UEL) The value of the expression é:

The) $numerator square root of 2 space minus space 3 over denominator 2 end of fraction$

B) minus 1 half

w) 1 half

d) $numerator square root of 3 over denominator 2 end of fraction$

It is) $numerator square root of 3 over denominator 2 end of fraction$

Answer explained

Passing radian values to arcs:

$cos space open parentheses numerator 2,180 over denominator 3 end of fraction close parentheses plus space sin open parentheses numerator 3,180 over denominator 2 end of fraction close parentheses space plus space tg open parentheses numerator 5,180 over denominator 4 end of fraction close parentheses equal acos space 120 space plus space sin space 270 space plus space tg space 225 equal to$

From the trigonometric circle, we see that:

cos space 120 space equals space minus space cos space 60 space equals space minus 1 half

sin space 270 space equals space minus space sin space 90 space equals space minus 1

tg space 225 space equals space tg space 45 space equals space 1

Soon,

cos space 120 space plus space sin space 270 space plus space tg space 225 equal minus 1 half plus left parenthesis minus 1 right parenthesis plus 1 equals bold minus bold 1 over bold 2

Learn more about:

Trigonometric Table
Trigonometric Circle
Trigonometry
Trigonometric Relations

ASTH, Rafael. Exercises on trigonometric circle with answer.All Matter, [n.d.]. Available in: https://www.todamateria.com.br/exercicios-sobre-circulo-trigonometrico/. Access at:

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