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1.3 : Fonctions trigonométriques

  • Page ID
    197995
    • Edwin “Jed” Herman & Gilbert Strang
    • OpenStax
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    Objectifs d'apprentissage
    • Convertissez les mesures d'angle en degrés et en radians.
    • Reconnaître les définitions triangulaires et circulaires des fonctions trigonométriques de base.
    • Écrivez les identités trigonométriques de base.
    • Identifier les graphes et les périodes des fonctions trigonométriques.
    • Décrivez le décalage d'un graphe en sinus ou en cosinus par rapport à l'équation de la fonction.

    Les fonctions trigonométriques sont utilisées pour modéliser de nombreux phénomènes, notamment les ondes sonores, les vibrations des cordes, le courant électrique alternatif et le mouvement des pendules. En fait, presque tous les mouvements répétitifs ou cycliques peuvent être modélisés par une combinaison de fonctions trigonométriques. Dans cette section, nous définissons les six fonctions trigonométriques de base et examinons certaines des principales identités impliquant ces fonctions.

    Mesure du rayonnement

    Pour utiliser les fonctions trigonométriques, il faut d'abord comprendre comment mesurer les angles. Bien que nous puissions utiliser à la fois des radians et des degrés, les radians constituent une mesure plus naturelle car ils sont directement liés au cercle unitaire, un cercle de rayon 1. La mesure en radian d'un angle est définie comme suit. Étant donné un angle\(θ\),\(s\) soit la longueur de l'arc correspondant sur le cercle unitaire (Figure\(\PageIndex{1}\)). Nous disons que l'angle correspondant à l'arc de longueur 1 a une mesure de radian de 1.

    Image d'un cercle. Au centre exact du cercle se trouve un point. À partir de ce point, un segment de ligne s'étend horizontalement vers la droite jusqu'à un point situé sur le bord du cercle et un autre segment de ligne s'étend en diagonale vers le haut et vers la droite jusqu'à un autre point du bord du cercle. Ces segments de ligne ont une longueur d'une unité. Le segment incurvé situé sur le bord du cercle qui relie les deux points situés à l'extrémité des segments de ligne est étiqueté « s ». À l'intérieur du cercle, une flèche pointe du segment de ligne horizontale vers le segment de ligne diagonale. Cette flèche porte l'étiquette « theta = s radians ».
    Figure\(\PageIndex{1}\) : La mesure en radian d'un angle\(θ\) est la longueur\(s\) de l'arc associé sur le cercle unitaire.

    Puisqu'un angle de\(360°\) correspond à la circonférence d'un cercle, ou d'un arc de longueur\(2π\), nous concluons qu'un angle dont la mesure en degrés\(360°\) est de a une mesure en radian de\(2π\). De même, nous voyons que cela\(180°\) équivaut à\(\pi\) des radians. Le tableau\(\PageIndex{1}\) montre la relation entre les valeurs communes en degrés et en radians.

    Tableau\(\PageIndex{1}\) : Angles communs exprimés en degrés et en radians
    Diplômes Radians Diplômes Radians
    0 0 120 \(2π/3\)
    30 \(π/6\) 135 \(3π/4\)
    45 \(π/4\) 150 \(5π/6\)
    60 \(π/3\) 180 \(π\)
    90 \(π/2\)    
    Conversion entre radians et degrés
    1. Express\(225°\) using radians.
    2. Express\(5π/3\) rad using degrees.

    Solution

    Utilisez le fait que cela\(180\)° équivaut à\(\pi\) des radians comme facteur de conversion (Tableau\(\PageIndex{1}\)) :

    \[1=\dfrac{π \,\mathrm{rad}}{180°}=\dfrac{180°}{π \,\mathrm{rad}}. \nonumber \]

    1. \(225°=225°⋅\left(\dfrac{π}{180°}\right)=\left(\dfrac{5π}{4}\right)\) rad
    2. \(\dfrac{5π}{3}\) rad = \(\dfrac{5π}{3}\)\(\dfrac{180°}{π}\)=\(300\)°
    Exercice\(\PageIndex{1}\)
    1. \(210°\)Exprimez en radians.
    2. Express\(11π/6\) rad using degrees.
    Allusion

    \(π\)radians est égal à 180°

    Réponse
    1. \(7π/6\)
    2. 330°

    Les six fonctions trigonométriques de base

    Les fonctions trigonométriques nous permettent d'utiliser des mesures d'angle, en radians ou en degrés, pour trouver les coordonnées d'un point sur n'importe quel cercle, et pas seulement sur un cercle unitaire, ou pour trouver un angle donné par un point sur un cercle. Ils définissent également la relation entre les côtés et les angles d'un triangle.

    Pour définir les fonctions trigonométriques, considérez d'abord le cercle unitaire centré à l'origine et un point\(P=(x,y)\) sur le cercle unitaire. \(θ\)Soit un angle avec un côté initial situé le long de l'\(x\)axe positif et un côté terminal qui est le segment de ligne\(OP\). Un angle dans cette position est considéré comme étant en position standard (Figure\(\PageIndex{2}\)). Nous pouvons ensuite définir les valeurs des six fonctions trigonométriques pour\(θ\) en termes de coordonnées\(x\) et\(y\).

    Image d'un graphique. Un cercle est tracé sur le graphique, avec le centre du cercle à l'origine, là où se trouve un point. À partir de ce point, un segment de ligne s'étend horizontalement le long de l'axe x vers la droite jusqu'à un point situé sur le bord du cercle. Un autre segment de ligne s'étend en diagonale vers le haut et vers la droite jusqu'à un autre point du bord du cercle. Ce point est étiqueté « P = (x, y) ». Ces segments de ligne ont une longueur d'une unité. À partir du point « P », il y a une ligne verticale pointillée qui s'étend vers le bas jusqu'à atteindre l'axe x et donc le segment de ligne horizontale. À l'intérieur du cercle, une flèche pointe du segment de ligne horizontale vers le segment de ligne diagonale. Cette flèche porte l'étiquette « thêta ».
    Figure\(\PageIndex{2}\) : L'angle\(θ\) est en position standard. Les valeurs des fonctions trigonométriques pour\(θ\) sont définies en termes de coordonnées\(x\) et\(y\).
    Définition : Fonctions trigonométriques

    \(P=(x,y)\)Soit un point du cercle unitaire centré à l'origine\(O\). \(θ\)Soit un angle avec un côté initial le long de l'\(x\)axe positif et un côté terminal donné par le segment de ligne\(OP\). Les fonctions trigonométriques sont alors définies comme

    \(\sin θ=y\) \(\csc θ=\dfrac{1}{y}\)
    \(\cos θ=x\) \(\sec θ=\dfrac{1}{x}\)
    \(\tan θ=\dfrac{y}{x}\) \(\cot θ=\dfrac{x}{y}\)

    Si\(x=0, \sec θ\) et\(\tan θ\) ne sont pas définis. Si\(y=0\), alors\(\cot θ\) et\(\csc θ\) ne sont pas définis.

    Nous pouvons voir que pour un point\(P=(x,y)\) sur un cercle de rayon\(r\) avec un angle correspondant\(θ\), les coordonnées\(x\) et\(y\) satisfont

    \[\begin{align} \cos θ &=\dfrac{x}{r} \\ x &=r\cos θ \end{align} \nonumber \]

    et

    \[\begin{align} \sin θ &=\dfrac{y}{r} \\ y &=r\sin θ. \end{align} \nonumber \]

    Les valeurs des autres fonctions trigonométriques peuvent être exprimées en termes de\(x,y\), et\(r\) (Figure\(\PageIndex{3}\)).

    CNX_Calc_Figure_01_03_003.jpg
    Figure\(\PageIndex{3}\) : Pour un point situé\(P=(x,y)\) sur un cercle de rayon\(r\), les coordonnées\(x\) et\(y\) satisfont à\(x=r\cos θ\) et\(y=r\sin θ\).

    Le tableau\(\PageIndex{2}\) montre les valeurs du sinus et du cosinus aux principaux angles du premier quadrant. À partir de ce tableau, nous pouvons déterminer les valeurs du sinus et du cosinus aux angles correspondants dans les autres quadrants. Les valeurs des autres fonctions trigonométriques sont facilement calculées à partir des valeurs de\(\sin θ\) and \(\cos θ.\)

    Tableau\(\PageIndex{2}\) : Valeurs des\(\cos θ\) angles principaux\(\sin θ\) et des angles principaux\(θ\) dans le premier quadrant
    \(θ\)θ \(\sin θ\) \(\cos θ\)
    \ (θ \) θ » style="text-align:center ; « >0 \ (\ sin θ \) » style="text-align:center ; « >0 \ (\ cos θ \) » style="text-align:center ; « >1
    \ (θ \) θ » style="text-align:center ; « >\(\dfrac{π}{6}\) \ (\ sin θ \) » style="text-align:center ; « >\(\dfrac{1}{2}\) \ (\ cos θ \) » style="text-align:center ; « >\(\dfrac{\sqrt{3}}{2}\)
    \ (θ \) θ » style="text-align:center ; « >\(\dfrac{π}{4}\) \ (\ sin θ \) » style="text-align:center ; « >\(\dfrac{\sqrt{2}}{2}\) \ (\ cos θ \) » style="text-align:center ; « >\(\dfrac{\sqrt{2}}{2}\)
    \ (θ \) θ » style="text-align:center ; « >\(\dfrac{π}{3}\) \ (\ sin θ \) » style="text-align:center ; « >\(\dfrac{\sqrt{3}}{2}\) \ (\ cos θ \) » style="text-align:center ; « >\(\dfrac{1}{2}\)
    \ (θ \) θ » style="text-align:center ; « >\(\dfrac{π}{2}\) \ (\ sin θ \) » style="text-align:center ; ">1 \ (\ cos θ \) » style="text-align:center ; « >0
    Exemple\(\PageIndex{2}\): Evaluating Trigonometric Functions

    Évaluez chacune des expressions suivantes.

    1. \(\sin \left(\dfrac{2π}{3} \right)\)
    2. \(\cos \left(−\dfrac{5π}{6} \right)\)
    3. \(\tan \left(\dfrac{15π}{4}\right)\)

    Solution :

    a) Sur le cercle unitaire, l'angle\(θ=\dfrac{2π}{3}\) correspond au point\(\left(−\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\). Par conséquent,

    \[ \sin \left(\dfrac{2π}{3}\right)=y=\left(\dfrac{\sqrt{3}}{2}\right). \nonumber \]

    Image d'un graphique. Un cercle est tracé sur le graphique, avec le centre du cercle à l'origine, là où se trouve un point. À partir de ce point, un segment de ligne s'étend horizontalement le long de l'axe x vers la droite jusqu'à un point situé sur le bord du cercle. Un autre segment de ligne s'étend en diagonale vers le haut et vers la gauche jusqu'à un autre point du bord du cercle. Ce point est étiqueté « (- (1/2), ((racine carrée de 3) /2))) ». Ces segments de ligne ont une longueur d'une unité. À partir du point « (- (1/2), ((racine carrée de 3) /2)) », il y a une ligne verticale qui s'étend vers le bas jusqu'à atteindre l'axe x. À l'intérieur du cercle se trouve une flèche incurvée qui part du segment de ligne horizontale et se déplace dans le sens antihoraire jusqu'à atteindre le segment de ligne diagonale. Cette flèche porte l'étiquette « theta = (2 pi) /3 ».

    b) Un angle\(θ=−\dfrac{5π}{6}\) corresponds to a revolution in the negative direction, as shown. Therefore,

    \[\cos \left(−\dfrac{5π}{6}\right)=x=−\dfrac{\sqrt{3}}{2}. \nonumber \]

    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the left to another point on the edge of the circle. This point is labeled “(-((square root of 3)/2)), -(1/2))”. These line segments have a length of 1 unit. From the point “(-((square root of 3)/2)), -(1/2))”, there is a vertical line that extends upwards until it hits the x axis. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels clockwise until it hits the diagonal line segment. This arrow has the label “theta = -(5 pi)/6”.

    c) An angle \(θ\)=\(\dfrac{15π}{4}\)=\(2π\)+\(\dfrac{7π}{4}\). Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of \(\dfrac{7π}{4}\) corresponds to the point \((\dfrac{\sqrt{2}}{2},-\dfrac{\sqrt{2}}{2})\), we can conclude that

    \[\tan \left(\dfrac{15π}{4}\right)=\dfrac{y}{x}=−1. \nonumber \]

    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment that extends diagonally downwards and to the right to another point on the edge of the circle. This point is labeled “(((square root of 2)/2), -((square root of 2)/2))”. These line segments have a length of 1 unit. From the point “(((square root of 2)/2), -((square root of 2)/2))”, there is a vertical line that extends upwards until it hits the x axis and thus the horizontal line segment. Inside the circle, there is a curved arrow that starts at the horizontal line segment and travels counterclockwise. The arrow makes one full rotation around the circle and then keeps traveling until it hits the diagonal line segment. This arrow has the label “theta = (15 pi)/4”.

    Exercise \(\PageIndex{2}\)

    Evaluate \(\cos(3π/4)\) and \(\sin(−π/6)\).

    Hint

    Look at angles on the unit circle.

    Answer

    \[\cos(3π/4) = −\sqrt{2}/2\nonumber \]

    \[ \sin(−π/6) =−1/2 \nonumber \]

    As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let \(θ\) be one of the acute angles. Let \(A\) be the length of the adjacent leg, \(O\) be the length of the opposite leg, and \(H\) be the length of the hypotenuse. By inscribing the triangle into a circle of radius \(H\), as shown in Figure \(\PageIndex{4}\), we see that \(A,H\), and \(O\) satisfy the following relationships with \(θ\):

    \(\sin θ=\dfrac{O}{H}\) \(\csc θ=\dfrac{H}{O}\)
    \(\cos θ=\dfrac{A}{H}\) \(\sec θ=\dfrac{H}{A}\)
    \(\tan θ=\dfrac{O}{A}\) \(\cot θ=\dfrac{A}{O}\)
    An image of a graph. The graph has a circle plotted on it, with the center of the circle at the origin, where there is a point. From this point, there is one line segment that extends horizontally along the x axis to the right to a point on the edge of the circle. There is another line segment with length labeled “H” that extends diagonally upwards and to the right to another point on the edge of the circle. From the point, there is vertical line with a length labeled “O” that extends downwards until it hits the x axis and thus the horizontal line segment at a point with a right triangle symbol. The distance from this point to the center of the circle is labeled “A”. Inside the circle, there is an arrow that points from the horizontal line segment to the diagonal line segment. This arrow has the label “theta”.
    Figure \(\PageIndex{4}\): By inscribing a right triangle in a circle, we can express the ratios of the side lengths in terms of the trigonometric functions evaluated at \(θ\).
    Example \(\PageIndex{3}\): Constructing a Wooden Ramp

    A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the top of the staircase is \(4\) ft from the ground and the angle between the ground and the ramp is to be \(10\)°, how long does the ramp need to be?

    Solution

    Let \(x\) denote the length of the ramp. In the following image, we see that \(x\) needs to satisfy the equation \(\sin(10°)=4/x\). Solving this equation for \(x\), we see that \(x=4/\sin(10°)\)\(23.035\) ft.

    An image of a ramp and a staircase. The ramp starts at a point and increases diagonally upwards and to the right at an angle of 10 degrees for x feet. At the end of the ramp, which is 4 feet off the ground, a staircase descends downwards and to the right.

    Exercise \(\PageIndex{3}\)

    A house painter wants to lean a \(20\)-ft ladder against a house. If the angle between the base of the ladder and the ground is to be \(60\)°, how far from the house should she place the base of the ladder?

    Hint

    Draw a right triangle with hypotenuse 20.

    Answer

    10 ft

    Trigonometric Identities

    A trigonometric identity is an equation involving trigonometric functions that is true for all angles \(θ\) for which the functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are listed next.

    Trigonometric Identities

    Reciprocal identities

    \[\tan θ=\dfrac{\sin θ}{\cos θ} \nonumber \]

    \[\cot θ=\dfrac{\cos θ}{\sin θ} \nonumber \]

    \[\csc θ=\dfrac{1}{\sin θ} \nonumber \]

    \[\sec θ=\dfrac{1}{\cos θ} \nonumber \]

    Pythagorean identities

    \[\begin{align} \sin^2θ+\cos^2θ &=1 \label{py1} \\[4pt] 1+\tan^2θ &=\sec^2θ \\[4pt] 1+\cot^2θ &=\csc^2θ \end{align} \]

    Addition and subtraction formulas

    \[\sin(α±β)=\sin α\cos β±\cos α \sin β \nonumber \]

    \[\cos(α±β)=\cos α\cos β∓\sin α \sin β \nonumber \]

    Double-angle formulas

    \[\sin(2θ)=2\sin θ\cos θ \label{double1} \]

    \[\begin{align} \cos(2θ) &=2\cos^2θ−1 \\[4pt] &=1−2\sin^2θ \\[4pt] &=\cos^2θ−\sin^2θ \end{align} \nonumber \]

    Example \(\PageIndex{4}\): Solving Trigonometric Equations

    For each of the following equations, use a trigonometric identity to find all solutions.

    1. \(1+\cos(2θ)=\cos θ\)
    2. \(\sin(2θ)=\tan θ\)

    Solution

    a) Using the double-angle formula for \(\cos(2θ)\), we see that \(θ\) is a solution of

    \[1+\cos(2θ)=\cos θ \nonumber \]

    if and only if

    \[1+2\cos^2θ−1=\cos θ, \nonumber \]

    which is true if and only if

    \[2\cos^2θ−\cos θ=0. \nonumber \]

    To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by \(\cos θ\). The problem with dividing by \(\cos θ\) is that it is possible that \(\cos θ\) is zero. In fact, if we did divide both sides of the equation by \(\cos θ\), we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that \(θ\) is a solution of this equation if and only if

    \[\cos θ(2\cos θ−1)=0. \nonumber \]

    Since \(\cos θ=0\) when

    \[θ=\dfrac{π}{2},\dfrac{π}{2}±π,\dfrac{π}{2}±2π,…, \nonumber \]

    and \(\cos θ=1/2\) when

    \[θ=\dfrac{π}{3},\dfrac{π}{3}±2π,…\mathrm{or}\ θ=−\dfrac{π}{3},−\dfrac{π}{3}±2π,…, \nonumber \]

    we conclude that the set of solutions to this equation is

    \[θ=\dfrac{π}{2}+nπ,\;θ=\dfrac{π}{3}+2nπ \nonumber \]

    and

    \[θ=−\dfrac{π}{3}+2nπ,\;n=0,±1,±2,….\nonumber \]

    b) Using the double-angle formula for \(\sin(2θ)\) and the reciprocal identity for \(\tan(θ)\), the equation can be written as

    \[2\sin θ\cos θ=\dfrac{\sin θ}{\cos θ}.\nonumber \]

    To solve this equation, we multiply both sides by \(\cos θ\) to eliminate the denominator, and say that if \(θ\) satisfies this equation, then \(θ\) satisfies the equation

    \[2\sin θ \cos^2θ−\sin θ=0. \nonumber \]

    However, we need to be a little careful here. Even if \(θ\) satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by \(\cos θ\). However, if \(\cos θ=0\), we cannot divide both sides of the equation by \(\cos θ\). Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor \(\sin θ\) out of both terms on the left-hand side instead of dividing both sides of the equation by \(\sin θ\). Factoring the left-hand side of the equation, we can rewrite this equation as

    \[\sin θ(2\cos^2θ−1)=0. \nonumber \]

    Therefore, the solutions are given by the angles \(θ\) such that \(\sin θ=0\) or \(\cos^2θ=1/2\). The solutions of the first equation are \(θ=0,±π,±2π,….\) The solutions of the second equation are \(θ=π/4,(π/4)±(π/2),(π/4)±π,….\) After checking for extraneous solutions, the set of solutions to the equation is

    \[θ=nπ \nonumber \]

    and

    \[θ=\dfrac{π}{4}+\dfrac{nπ}{2} \nonumber \]

    with \(n=0,±1,±2,….\)

    Exercise \(\PageIndex{4}\)

    Find all solutions to the equation \(\cos(2θ)=\sin θ.\)

    Hint

    Use the double-angle formula for cosine (Equation \ref{double1}).

    Answer

    \(θ=\dfrac{3π}{2}+2nπ,\dfrac{π}{6}+2nπ,\dfrac{5π}{6}+2nπ\)

    for \(n=0,±1,±2,…\).

    Example \(\PageIndex{5}\): Proving a Trigonometric Identity

    Prove the trigonometric identity \(1+\tan^2θ=\sec^2θ.\)

    Solution:

    We start with the Pythagorean identity (Equation \ref{py1})

    \[\sin^2θ+\cos^2θ=1. \nonumber \]

    Dividing both sides of this equation by \(\cos^2θ,\) we obtain

    \[\dfrac{\sin^2θ}{\cos^2θ}+1=\dfrac{1}{\cos^2θ}. \nonumber \]

    Since \(\sin θ/\cos θ=\tan θ\) and \(1/\cos θ=\sec θ\), we conclude that

    \[\tan^2θ+1=\sec^2θ. \nonumber \]

    Exercise \(\PageIndex{5}\)

    Prove the trigonometric identity \(1+\cot^2θ=\csc^2θ.\)

    Answer

    Divide both sides of the identity \(\sin^2θ+\cos^2θ=1\) by \(\sin^2θ\).

    Graphs and Periods of the Trigonometric Functions

    We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this pattern in the graphs of the functions. Let \(P=(x,y)\) be a point on the unit circle and let \(θ\) be the corresponding angle . Since the angle \(θ\) and \(θ+2π\) correspond to the same point \(P\), the values of the trigonometric functions at \(θ\) and at \(θ+2π\) are the same. Consequently, the trigonometric functions are periodic functions. The period of a function \(f\) is defined to be the smallest positive value \(p\) such that \(f(x+p)=f(x)\) for all values \(x\) in the domain of \(f\). The sine, cosine, secant, and cosecant functions have a period of \(2π\). Since the tangent and cotangent functions repeat on an interval of length \(π\), their period is \(π\) (Figure \(\PageIndex{5}\)).

    An image of six graphs. Each graph has an x axis that runs from -2 pi to 2 pi and a y axis that runs from -2 to 2. The first graph is of the function “f(x) = sin(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 0) and increases until the point (-((3 pi)/2), 1). After this point, the function decreases until the point (-(pi/2), -1). After this point, the function increases until the point ((pi/2), 1). After this point, the function decreases until the point (((3 pi)/2), -1). After this point, the function begins to increase again. The x intercepts shown on the graph are at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The second graph is of the function “f(x) = cos(x)”, which is a curved wave function. The graph of the function starts at the point (-2 pi, 1) and decreases until the point (-pi, -1). After this point, the function increases until the point (0, 1). After this point, the function decreases until the point (pi, -1). After this point, the function increases again. The x intercepts shown on the graph are at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0). The y intercept is at the point (0, 1). The graph of cos(x) is the same as the graph of sin(x), except it is shifted to the left by a distance of (pi/2). On the next four graphs there are dotted vertical lines which are not a part of the function, but act as boundaries for the function, boundaries the function will never touch. They are known as vertical asymptotes. There are infinite vertical asymptotes for all of these functions, but these graphs only show a few. The third graph is of the function “f(x) = csc(x)”. The vertical asymptotes for “f(x) = csc(x)” on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. Between the “x = -2 pi” and “x = -pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (-((3 pi)/2), 1). Between the “x = -pi” and “x = 0” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-(pi/2), -1). Between the “x = 0” and “x = pi” asymptotes, the function looks like an upward facing “U”, with a minimum at the point ((pi/2), 1). Between the “x = pi” and “x = 2 pi” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (((3 pi)/2), -1). The fourth graph is of the function “f(x) = sec(x)”. The vertical asymptotes for this function on this graph are at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. Between the “x = -((3 pi)/2)” and “x = -(pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (-pi, -1). Between the “x = -(pi/2)” and “x = (pi/2)” asymptotes, the function looks like an upward facing “U”, with a minimum at the point (0, 1). Between the “x = (pi/2)” and “x = (3pi/2)” asymptotes, the function looks like an downward facing “U”, with a maximum at the point (pi, -1). The graph of sec(x) is the same as the graph of csc(x), except it is shifted to the left by a distance of (pi/2). The fifth graph is of the function “f(x) = tan(x)”. The vertical asymptotes of this function on this graph occur at “x = -((3 pi)/2)”, “x = -(pi/2)”, “x = (pi/2)”, and “x = ((3 pi)/2)”. In between all of the vertical asymptotes, the function is always increasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-2 pi, 0), (-pi, 0), (0, 0), (pi, 0), and (2 pi, 0). The y intercept is at the origin. The sixth graph is of the function “f(x) = cot(x)”. The vertical asymptotes of this function on this graph occur at “x = -2 pi”, “x = -pi”, “x = 0”, “x = pi”, and “x = 2 pi”. In between all of the vertical asymptotes, the function is always decreasing but it never touches the asymptotes. The x intercepts on this graph occur at the points (-((3 pi)/2), 0), (-(pi/2), 0), ((pi/2), 0), and (((3 pi)/2), 0) and there is no y intercept.
    Figure \(\PageIndex{5}\): The six trigonometric functions are periodic.

    Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the following function:

    \[f(x)=A\sin(B(x−α))+C. \nonumber \]

    In Figure \(\PageIndex{6}\), the constant \(α\) causes a horizontal or phase shift. The factor \(B\) changes the period. This transformed sine function will have a period \(2π/|B|\). The factor \(A\) results in a vertical stretch by a factor of \(|A|\). We say \(|A|\) is the “amplitude of \(f\).” The constant \(C\) causes a vertical shift.

    An image of a graph. The graph is of the function “f(x) = Asin(B(x - alpha)) + C”. Along the y axis, there are 3 hash marks: starting from the bottom and moving up, the hash marks are at the values “C - A”, “C”, and “C + A”. The distance from the origin to “C” is labeled “vertical shift”. The distance from “C - A” to “A” and the distance from “A” to “C + A” is “A”, which is labeled “amplitude”. On the x axis is a hash mark at the value “alpha” and the distance between the origin and “alpha” is labeled “horizontal shift”. The distance between two successive minimum values of the function (in other words, the distance between two bottom parts of the wave that are next to each other) is “(2 pi)/(absolute value of B)” is labeled the period. The period is also the distance between two successive maximum values of the function.
    Figure \(\PageIndex{6}\): A graph of a general sine function.

    Notice in Figure \(\PageIndex{6}\) that the graph of \(y=\cos x\) is the graph of \(y=\sin x\) shifted to the left \(π/2\) units. Therefore, we can write

    \[\cos x=\sin(x+π/2). \nonumber \]

    Similarly, we can view the graph of \(y=\sin x\) as the graph of \(y=\cos x\) shifted right \(π/2\) units, and state that \(\sin x=\cos(x−π/2).\)

    A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function

    \[h(t)=3.7\sin \left(\dfrac{2π}{365}(x−80.5)\right)+12 \nonumber \]

    is a model for the number of hours of daylight \(h\) as a function of day of the year \(t\) (Figure \(\PageIndex{7}\)).

    An image of a graph. The x axis runs from 0 to 365 and is labeled “t, day of the year”. The y axis runs from 0 to 20 and is labeled “h, number of daylight hours”. The graph is of the function “h(t) = 3.7sin(((2 pi)/365)(t - 80.5)) + 12”, which is a curved wave function. The function starts at the approximate point (0, 8.4) and begins increasing until the approximate point (171.8, 15.7). After this point, the function decreases until the approximate point (354.3, 8.3). After this point, the function begins increasing again.
    Figure \(\PageIndex{7}\): The hours of daylight as a function of day of the year can be modeled by a shifted sine curve.
    Example \(\PageIndex{6}\): Sketching the Graph of a Transformed Sine Curve

    Sketch a graph of \(f(x)=3\sin(2(x−\frac{π}{4}))+1.\)

    Solution

    This graph is a phase shift of \(y=\sin (x)\) to the right by \(π/4\) units, followed by a horizontal compression by a factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of \(f\) is \(π\).

    An image of a graph. The x axis runs from -((3 pi)/2) to 2 pi and the y axis runs from -3 to 5. The graph is of the function “f(x) = 3sin(2(x-(pi/4))) + 1”, which is a curved wave function. The function starts decreasing from the point (-((3 pi)/2), 4) until it hits the point (-pi, -2). At this point, the function begins increasing until it hits the point (-(pi/2), 4). After this point, the function begins decreasing until it hits the point (0, -2). After this point, the function increases until it hits the point ((pi/2), 4). After this point, the function decreases until it hits the point (pi, -2). After this point, the function increases until it hits the point (((3 pi)/2), 4). After this point, the function decreases again.

    Exercise \(\PageIndex{6}\)

    Describe the relationship between the graph of \(f(x)=3\sin(4x)−5\) and the graph of \(y=\sin(x)\).

    Hint

    The graph of \(f\) can be sketched using the graph of \(y=\sin(x)\) and a sequence of three transformations.

    Answer

    To graph \(f(x)=3\sin(4x)−5\), the graph of \(y=\sin(x)\) needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function \(f\) will have a period of \(π/2\) and an amplitude of 3.

    Key Concepts

    • Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian measure 1. An angle with a degree measure of \(180\)° has a radian measure of \(\pi\) rad.
    • For acute angles \(θ\),the values of the trigonometric functions are defined as ratios of two sides of a right triangle in which one of the acute angles is \(θ\).
    • For a general angle \(θ\), let \((x,y)\) be a point on a circle of radius \(r\) corresponding to this angle \(θ\). The trigonometric functions can be written as ratios involving \(x\), \(y\), and \(r\).
    • The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period \(2π\). The tangent and cotangent functions have period \(π\).

    Key Equations

    • Generalized sine function

    \(f(x)=A \sin(B(x−α))+C\)

    Glossary

    periodic function
    a function is periodic if it has a repeating pattern as the values of \(x\) move from left to right
    radians
    for a circular arc of length \(s\) on a circle of radius 1, the radian measure of the associated angle \(θ\) is \(s\)
    trigonometric functions
    functions of an angle defined as ratios of the lengths of the sides of a right triangle
    trigonometric identity
    an equation involving trigonometric functions that is true for all angles \(θ\) for which the functions in the equation are defined