1.3: Trigonometric Functions (2024)

Converting between Radians and Degrees

1. Express \(225°\) using radians.
2. Express \(5π/3\) rad using degrees.

Solution

Use the fact that \(180\)° is equivalent to \(\pi\) radians as a conversion factor (Table \(\PageIndex{1}\)):

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

1. \(225°=225°⋅\left(\dfrac{π}{180°}\right)=\left(\dfrac{5π}{4}\right)\) rad
2. \(\dfrac{5π}{3}\) rad = \(\dfrac{5π}{3}\)⋅\(\dfrac{180°}{π}\)=\(300\)°

Exercise \(\PageIndex{1}\)

1. Express \(210°\) using radians.
2. Express \(11π/6\) rad using degrees.

Hint

\(π\) radians is equal to 180°

Answer

1. \(7π/6\)
2. 330°

Definition: Trigonometric functions

Let \(P=(x,y)\) be a point on the unit circle centered at the origin \(O\). Let \(θ\) be an angle with an initial side along the positive \(x\)-axis and a terminal side given by the line segment \(OP\). The trigonometric functions are then defined as

If \(x=0, \sec θ\) and \(\tan θ\) are undefined. If \(y=0\), then \(\cot θ\) and \(\csc θ\) are undefined.

Example \(\PageIndex{2}\): Evaluating Trigonometric Functions

Evaluate each of the following expressions.

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

Solution:

a) On the unit circle, the angle \(θ=\dfrac{2π}{3}\) corresponds to the point \(\left(−\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)\). Therefore,

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

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

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

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

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

Exercise \(\PageIndex{2}\)

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

Hint

Look at angles on the unit circle.

Answer

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

\[ \sin(−π/6) =−1/2 \nonumber \]

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.

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

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 \(\sin^2θ\).

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 \(π\).

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.