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Trigonometry · Trigonometry

Trigonometric Ratios of an Angle

Before you start, make sure you know: Oriented angles and the unit circle (Secondary Year 3)
Before you start, make sure you know: Right triangles (Secondary Year 3)

The access ramp at a Metrobús station is $6$ m long and rises $0.6$ m. How steep is it? With the right triangle you already know how to answer for acute angles. Here you extend that idea to any angle (from $0^\circ$ to $360^\circ$, or more), calculate exact values without a calculator, and begin proving identities.

From the legs to the unit circle

In a right triangle with acute angle $\alpha$, the sine is the opposite leg divided by the hypotenuse, the cosine is the adjacent leg divided by the hypotenuse, and the tangent is the opposite leg divided by the adjacent leg. If you place the vertex of the angle at the origin and use the hypotenuse as the radius of a circle of radius $1$, the sine and cosine turn out to be the coordinates of the point where the side of the angle meets the circle.

Unit circle with a point P in the first quadrant; its x-coordinate is the cosine of the angle and its y-coordinate is the sine. P (cos α, sen α) cos α sen α α 1
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Calculate the exact value of the sine, cosine and tangent of the smallest angle of a right-angled triangle, knowing that its shorter sides measure $5$ cm and $12$ cm.

Calculate the exact value of the sine, cosine and tangent of the smallest angle of a right-angled triangle, knowing that its shorter sides measure $5$ cm and $12$ cm.

Solve

Degrees and radians

Besides degrees, angles are measured in radians. A full turn is $360^\circ = 2\pi$ rad, so a half turn is $180^\circ = \pi$ rad. To convert:

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Convert the angles of $135^\circ$ and $330^\circ$ to radians, and express the angles of $\frac{5\pi}{6}$ rad and $\frac{7\pi}{4}$ rad in sexagesimal degrees.

Convert the angles of $135^\circ$ and $330^\circ$ to radians, and express the angles of $\frac{5\pi}{6}$ rad and $\frac{7\pi}{4}$ rad in sexagesimal degrees.

Solve
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A grinder disc rotates at $7200$ revolutions per minute. Express this angular velocity in radians per second and calculate the exact angle (in radians) swept out by a…

A grinder disc rotates at $7200$ revolutions per minute. Express this angular velocity in radians per second and calculate the exact angle (in radians) swept out by a particle on its outer edge in $0.15$ seconds.

Solve

The special angles

The angles $30^\circ$, $45^\circ$, and $60^\circ$ have exact ratios involving square roots. They come from two triangles: the isosceles right triangle (half of a square of side $1$) and half of an equilateral triangle of side $2$.

Two triangles that give the special values: an isosceles right triangle with legs of 1 and half of an equilateral triangle of side 2, with the shorter leg 1 opposite 30 degrees and the longer leg root 3 opposite 60 degrees. 1 1 √2 45° √3 1 2 30° 60°

In the second triangle, the side of length $1$ is opposite $30^\circ$, the side of length $\sqrt3$ is opposite $60^\circ$, and the hypotenuse is $2$. The ratios in the table come from there.

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Evaluate the following expression by replacing the trigonometric ratios with their exact values, without using a calculator:…

Evaluate the following expression by replacing the trigonometric ratios with their exact values, without using a calculator:

$\sin^2(30^\circ) + \cos^2(45^\circ) - \tan(60^\circ) \times \sin(60^\circ)$

Solve
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Calculate the numerical value of the following trigonometric expression corresponding to angles located on the semi-axes:…

Calculate the numerical value of the following trigonometric expression corresponding to angles located on the semi-axes:

$3\sin(90^\circ) - 2\cos(180^\circ) + 4\sin(270^\circ) - \cos(360^\circ)$

Solve

The Pythagorean identity and the sign by quadrant

The point $P(\cos\alpha, \sin\alpha)$ is at distance $1$ from the origin. By the Pythagorean theorem:

Unit circle divided into four quadrants with the signs of the sine, cosine, and tangent in each one. Quadrant I sen + cos + tan + Quadrant II sen + cos − tan − Quadrant III sen − cos − tan + Quadrant IV sen − cos + tan −

The tangent is sine divided by cosine, so its sign comes from that division: it is positive in I and III (same signs) and negative in II and IV.

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Knowing that $\cos \alpha = \frac{4}{5}$ and that $\alpha$ is an acute angle, find the exact value of $\sin \alpha$ and $\tan \alpha$ using fundamental trigonometric…

Knowing that $\cos \alpha = \frac{4}{5}$ and that $\alpha$ is an acute angle, find the exact value of $\sin \alpha$ and $\tan \alpha$ using fundamental trigonometric identities.

Solve
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Determine which quadrant the angle $\beta$ lies in if both $\sin \beta < 0$ and $\tan \beta > 0$ hold simultaneously.

Determine which quadrant the angle $\beta$ lies in if both $\sin \beta < 0$ and $\tan \beta > 0$ hold simultaneously.

Solve
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Given that $\tan \gamma = -\frac{12}{5}$ and that $\gamma$ belongs to the second quadrant, calculate the exact value of $\sin \gamma$ and $\sec \gamma$.

Given that $\tan \gamma = -\frac{12}{5}$ and that $\gamma$ belongs to the second quadrant, calculate the exact value of $\sin \gamma$ and $\sec \gamma$.

Solve

Reciprocal ratios

The other three ratios are the multiplicative inverses of the first three.

Also, $1 + \tan^2\alpha = \sec^2\alpha$ and $1 + \cot^2\alpha = \csc^2\alpha$: they come from dividing the Pythagorean identity by $\cos^2\alpha$ or by $\sin^2\alpha$.

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In a right-angled triangle, the tangent of one of its acute angles is $\frac{3}{4}$. Determine the exact value of the secant and cosecant of this angle.

In a right-angled triangle, the tangent of one of its acute angles is $\frac{3}{4}$. Determine the exact value of the secant and cosecant of this angle.

Solve
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Calculate the exact values of $\cot \alpha$ and $\csc \alpha$ given that $\sec \alpha = 3$ and that the sine of the angle is negative.

Calculate the exact values of $\cot \alpha$ and $\csc \alpha$ given that $\sec \alpha = 3$ and that the sine of the angle is negative.

Solve

Reduction to the first quadrant

For large angles, first remove full turns; then bring the angle into the first quadrant.

Step 1. Divide by $360^\circ$ and keep the remainder.

Step 2. With the angle $\theta$ in the first quadrant, use:

Angle of 150 degrees in quadrant II and its related angle of 30 degrees in quadrant I, symmetric about the vertical axis. 30° 150° sen 150° = sen 30° cos 150° = −cos 30°
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Find the equivalent angle in the first revolution (between $0^\circ$ and $360^\circ$) for the angle of $1950^\circ$ and determine the sign of its secant and cotangent.

Find the equivalent angle in the first revolution (between $0^\circ$ and $360^\circ$) for the angle of $1950^\circ$ and determine the sign of its secant and cotangent.

Solve
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Find the exact value of the following expression by first reducing all the angles to the first quadrant:…

Find the exact value of the following expression by first reducing all the angles to the first quadrant:

$\cos(120^\circ) \times \sin(225^\circ) - \tan(330^\circ) \times \cos(210^\circ)$

Solve
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Simplify the following trigonometric expression as much as possible, assuming that $x$ is an angle in the first quadrant. Pay special attention to the signs when…

Simplify the following trigonometric expression as much as possible, assuming that $x$ is an angle in the first quadrant. Pay special attention to the signs when reducing to that quadrant:

$\frac{\sin(\pi - x) \times \cos\left(\frac{\pi}{2} + x\right) \times \tan(2\pi - x)}{\cos(\pi + x) \times \sin\left(\frac{3\pi}{2} - x\right)}$

Solve

Constructing right triangles, elevation and depression

Ratios are used to construct right triangles and to measure what cannot be measured directly, such as the height of a building or the distance to a faraway object.

Constructing a right triangle

A right triangle is determined if you know, besides the right angle, one side and one acute angle, or else two sides. You can also build it from a ratio: if $\tan\alpha = \dfrac34$, you take legs proportional to $3$ and $4$.

Angles of elevation and depression

Both are measured from the horizontal through the observer's eye. The angle of elevation is formed when you look up; the angle of depression, when you look down.

A tower and a car: from the top of the tower, the angle of depression is measured downward from the horizontal; from the car, the angle of elevation is measured upward from the ground and equals the angle of depression. depression elevation horizontal car tower
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A building casts a shadow of $24$ meters when the angle of elevation of the sun above the horizon is $53^\circ$. Write the trigonometric expression to calculate the…

A building casts a shadow of $24$ meters when the angle of elevation of the sun above the horizon is $53^\circ$. Write the trigonometric expression to calculate the height of the building and determine its value. If the angle of elevation is halved, does the length of the shadow double? Justify your answer based on the non-linearity of the tangent ratio.

Solve
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To optimize solar energy collection in winter, a rectangular panel $2$ meters long is raised by resting one of its shorter sides on flat ground, so that the opposite…

To optimize solar energy collection in winter, a rectangular panel $2$ meters long is raised by resting one of its shorter sides on flat ground, so that the opposite side is at a height of exactly $1.2$ meters. Determine the exact value of the cosine and tangent of the angle of inclination of the panel.

Solve

Geometric applications

The trick is to identify a right triangle and choose the ratio that connects the given data to the unknown.

Pendulum of 50 centimeters displaced 25 degrees from the vertical; the horizontal displacement is the opposite leg of the triangle formed by the string. 25° 50 sen 25°

Back to the beginning: the $6$ m ramp that rises $0.6$ m satisfies $\sin\alpha = \dfrac{0.6}{6} = 0.1$, so $\alpha = \sin^{-1}(0.1) \approx 5.7^\circ$.

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A pendulum of length $80$ cm moves from its vertical equilibrium position to form an angle of $18^\circ$ with the vertical. Calculate the exact horizontal distance that…

A pendulum of length $80$ cm moves from its vertical equilibrium position to form an angle of $18^\circ$ with the vertical. Calculate the exact horizontal distance that the pendulum bob has moved and the height it has risen with respect to its lowest point. Express the results in terms of the trigonometric ratios of $18^\circ$.

18° x = ? h = ? 80 cm
Solve
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Prove that the area of a regular polygon of $n$ sides inscribed in a circle of radius $R$ is given by the expression…

Prove that the area of a regular polygon of $n$ sides inscribed in a circle of radius $R$ is given by the expression $A = \frac{n \times R^2}{2} \sin\left(\frac{360^\circ}{n}\right)$. Then, apply the formula to calculate the exact area of a regular octagon inscribed in a circle of radius $10$ cm.

O 360°/n R R
Solve

Simple identities and equations

An identity holds for all angles where it is defined. To prove one, work on one side until it looks the same as the other. The tools are: rewrite everything in terms of sine and cosine, factor out a common factor, add fractions, and use $\sin^2\alpha + \cos^2\alpha = 1$.

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Simplify the following trigonometric expression as much as possible using fundamental identities, expressing the result in terms of a single trigonometric ratio:…

Simplify the following trigonometric expression as much as possible using fundamental identities, expressing the result in terms of a single trigonometric ratio:

$\frac{\sin^3 x + \sin x \times \cos^2 x}{\cos x}$

Solve
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Prove the following trigonometric identity, showing the algebraic steps and fundamental relations used. Assume that $\alpha$ is an angle for which all ratios are…

Prove the following trigonometric identity, showing the algebraic steps and fundamental relations used. Assume that $\alpha$ is an angle for which all ratios are defined:

$\frac{1 - \sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{1 - \sin \alpha} = 2\sec \alpha$

Solve
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Given the relation $5\sin \theta - 2\cos \theta = 0$, determine all possible exact values for $\sin \theta$, $\cos \theta$ and $\tan \theta$. Discuss with reasons which…

Given the relation $5\sin \theta - 2\cos \theta = 0$, determine all possible exact values for $\sin \theta$, $\cos \theta$ and $\tan \theta$. Discuss with reasons which quadrants the angle $\theta$ can lie in.

Solve
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Find all angles $\alpha \in [0^\circ, 360^\circ)$ satisfying the following trigonometric equation: $2\cos^2 \alpha - 3\sin \alpha = 0$ (Hint: express the entire equation…

Find all angles $\alpha \in [0^\circ, 360^\circ)$ satisfying the following trigonometric equation:

$2\cos^2 \alpha - 3\sin \alpha = 0$

(Hint: express the entire equation in terms of a single trigonometric ratio using the fundamental identity).

Solve
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A light ray passes from a liquid medium with refractive index $n_1 = 1.33$ to a crystal with index $n_2 = 1.52$. Snell's law states that…

A light ray passes from a liquid medium with refractive index $n_1 = 1.33$ to a crystal with index $n_2 = 1.52$. Snell's law states that $n_1 \sin \theta_1 = n_2 \sin \theta_2$, where $\theta_1$ and $\theta_2$ are the angles of incidence and refraction measured relative to the normal to the surface. If the ray is incident at an angle of $\theta_1 = \frac{\pi}{4}$ rad, determine the exact value of the cosine of the angle of refraction $\theta_2$.

Solve
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Two circular pulleys of radii $R = 15$ cm and $r = 5$ cm are connected by a taut belt crossed between them. The distance between the centers of the pulleys is $40$ cm.…

Two circular pulleys of radii $R = 15$ cm and $r = 5$ cm are connected by a taut belt crossed between them. The distance between the centers of the pulleys is $40$ cm. If $\alpha$ is the angle formed by the straight section of the belt with the imaginary line joining the centers, calculate the exact value of $\sin \alpha$, $\cos \alpha$ and the total length of the straight section of the belt between the two points of tangency.

α R r d = 40 cm
Solve
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Prove that if $\alpha$ and $\beta$ are two acute complementary angles (that is, $\alpha + \beta = \frac{\pi}{2}$ rad), then the following trigonometric relation always…

Prove that if $\alpha$ and $\beta$ are two acute complementary angles (that is, $\alpha + \beta = \frac{\pi}{2}$ rad), then the following trigonometric relation always holds:

$\frac{\tan \alpha + \tan \beta}{\sec \alpha \times \sec \beta} = 1$

Solve

Summary

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