8.4 Chapter Summary and Review

Key Concepts

Identities are useful for changing from one form to another when solving equations, for simplifying expressions, and for finding exact values for trigonometric functions.
it is not true in general that $\cos (α + β)$ is equal to $\cos (α) + \cos (β)$ for all angles $α$ and $β$ , or that $\sin (α + β)$ is equal to $\sin (α) + \sin (β)$ .
Note
Sum and Difference of Angles Identities
$\cos (α + β) = \cos (α) \cos (β) - \sin (α) \sin (β)$
$\sin (α + β) = \sin (α) \cos (β) + \cos (α) \sin (β)$
$\tan (α + β) = \frac{\tan (α) + \tan (β)}{1 - \tan (α) \tan (β)}$
$\cos (α - β) = \cos (α) \cos (β) + \sin (α) \sin (β)$
$\sin (α - β) = \sin (α) \cos (β) - \cos (α) \sin (β)$
$\tan (α - β) = \frac{\tan (α) - \tan (β)}{1 + \tan (α) \tan (β)}$
Note
Double Angle Identities
$\begin{aligned} \cos (2 θ) & = \cos^{2} (θ) - \sin^{2} (θ) & \sin (2 θ) = 2 \sin (θ) \cos (θ) \\ = 2 \cos^{2} (θ) - 1 & \tan (2 θ) = \frac{2 \tan (θ)}{1 - \tan^{2} (θ)} \\ = 1 - 2 \sin^{2} (θ) \end{aligned}$
Using one of the calculator keys $S I N^{- 1}, C O S^{- 1}$ , or $T A N^{- 1}$ performs the inverse operation for computing a sine, cosine or tangent.
Two functions are called inverse functions if each "undoes" the results of the other function.
If $y = f (x)$ is a function, we can often find a formula for the inverse function by interchanging $x$ and $y$ in the formula for the function, and then solving for $y$ .
The graphs of $y = f (x)$ and $y = f^{- 1} (x)$ are symmetric about the line $y = x$ .
The domain of $f^{- 1}$ is the same as the range of $f$ , and the range of $f^{- 1}$ is the same as the domain of $f$ .
Note
Horizontal Line Test
A function passes the Horizontal Line Test if every horizontal line intersects the graph at most once. In that case, there is only one $x$ -value for each $y$ -value, and the function is called one-to-one.
A function $f$ has an inverse function if and only if $f$ is one-to-one.
Note
Definitions of the inverse trig functions
$\sin^{- 1} (x) = θ if and only if \sin (θ) = x and \frac{- π}{2} \leq θ \leq \frac{π}{2}$
$\cos^{- 1} (x) = θ if and only if \cos (θ) = x and 0 \leq θ \leq π$
$\tan^{- 1} (x) = θ if and only if \tan (θ) = x and \frac{- π}{2} < θ < \frac{π}{2}$
Note
Inverse Trigonometric Functions
$\sin (\sin^{- 1} (x)) = x for - 1 \leq x \leq 1$
$\cos (\cos^{- 1} (x)) = x for - 1 \leq x \leq 1$
$\tan (\tan^{- 1} (x)) = x for all x$
However,
$\sin^{- 1} (\sin (x)) may not be equal to x$
$\cos^{- 1} (\cos (x)) may not be equal to x$
$\tan^{- 1} (\tan (x)) may not be equal to x$
The inverse sine function is also called the arcsine function and denoted by $\arcsin (x)$ . Similarly, the inverse cosine function is sometimes denoted by $\arccos (x)$ , and the inverse tangent function by $\arctan (x)$ .
When simplifying expressions involving inverse trigonometric functions, it can often clarify the computations if we assign a name such as $θ$ or $ϕ$ to the inverse trig value.
Note
Three More Functions
If $θ$ is an angle in standard position, and $P (x, y)$ is a point on the terminal side, then we define the following functions.
$The secant : \sec (θ) = \frac{r}{x}$
$The cosecant : \csc (θ) = \frac{r}{y}$
$The cotangent : \cot (θ) = \frac{x}{y}$
Note
Reciprocal Trigonometric Ratios
If $θ$ is one of the acute angles in a right triangle,
$\begin{aligned} \sec (θ) & = \frac{hypotenuse}{adjacent} \\ \csc (θ) & = \frac{hypotenuse}{opposite} \\ \cot (θ) & = \frac{adjacent}{opposite} \end{aligned}$
Note
Reciprocal Trigonometric Functions
$The secant function : \sec (θ) = \frac{1}{\cos (θ)}$
$The cosecant function : \csc (θ) = \frac{1}{\sin (θ)}$
$The cotangent function : \cot (θ) = \frac{1}{\tan (θ)}$
We can obtain graphs of the secant, cosecant, and cotangent functions as the reciprocals of the three basic functions.
We can solve equations of the form $\sec (θ) = k$ , $\csc (θ) = k$ , and $\cot (θ) = k$ by taking the reciprocal of both sides.
If we know one of the trigonometric ratios for an angle, we can use identities to find any of the others.
Note
Cotangent Identity
$\cot (θ) = \frac{1}{\tan (θ)} = \frac{\cos (θ)}{\sin (θ)}, \sin (θ) \neq 0$
Note
Two More Pythagorean Identities
$1 + \tan^{2} (θ) = \sec^{2} (θ) 1 + \cot^{2} (θ) = \csc^{2} (θ)$
We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.

Review Problems

For Problems 1–8, answer true or false.

$\sin (β + \frac{π}{4}) = \sin (β) + \frac{1}{\sqrt{2}}$

False

$\cos (\frac{π}{3} - t) = \frac{1}{2} - \cos (t)$

$\tan (z - w) = \frac{\sin (z - w)}{\cos (z - w)}$

True

$\sin (2 ϕ) = 1 - \cos (2 ϕ)$

$\sin (\frac{π}{2} - x) = 1 - \sin (x)$

False

$\sin (π - x) = \sin (x)$

$\cos^{2} (α) - \sin^{2} (α) = - 1$

False

$\tan^{- 1} (s) = \frac{1}{\tan (s)}$

If $\sin (x) = - 0.4$ and $cos ⁡ open parenthesis x close parenthesis greater than 0,$ find an exact value for $\cos (x + \frac{3 π}{4})$ .

$\frac{2 - \sqrt{21}}{5 \sqrt{2}}$

If $\cos (x) = - 0.75$ and $sin ⁡ open parenthesis x close parenthesis less than 0,$ find an exact value for $\cos (x - \frac{4 π}{3})$ .

If $\cos (θ) = \frac{- 3}{8}, π < θ < \frac{3 π}{2},$ and $\sin (ϕ) = \frac{1}{4}, \frac{π}{2} < ϕ < π,$ find exact values for

$\sin (θ + ϕ)$
$\tan (θ + ϕ)$

$\frac{5 \sqrt{33} - 3}{32}$
$\frac{5 \sqrt{33} - 3}{\sqrt{5} (3 \sqrt{3} + \sqrt{11})}$

If $\sin (ρ) = \frac{5}{6}, \frac{π}{2} < ρ < π,$ and $\cos (μ) = \frac{- 1}{3}, \frac{π}{2} < μ < π,$ find exact values for

$\cos (ρ - μ)$
$\tan (ρ - μ)$

If $\tan (x + y) = 2$ and $\tan (y) = \frac{1}{3},$ find $\tan (x)$ .

$1$

If $\tan (x - y) = \frac{1}{4}$ and $tan ⁡ open parenthesis x close parenthesis equals 4,$ find $\tan (y)$ .

For Problems 15-16, use the sum and difference formulas to expand each expression.

$\tan (t - \frac{5 π}{3})$

$\frac{\tan (t) + \sqrt{3}}{1 - \sqrt{3} \tan (t)}$

$\cos (s + \frac{7 π}{4})$

For Problems 17–18, use the figure to find the trigonometric ratios.

$\sin (θ)$
$\cos (θ)$
$\tan (θ)$
$sin ⁡ open parenthesis 2 θ close parenthesis$
$cos ⁡ open parenthesis 2 θ close parenthesis$
$tan ⁡ open parenthesis 2 θ close parenthesis$

$\frac{4}{5}$
$\frac{3}{5}$
$\frac{4}{3}$
$\frac{24}{25}$
$\frac{- 7}{25}$
$\frac{- 24}{7}$

$\sin (ϕ)$
$\cos (ϕ)$
$\tan (ϕ)$
$sin ⁡ open parenthesis 2 ϕ close parenthesis$
$cos ⁡ open parenthesis 2 ϕ close parenthesis$
$tan ⁡ open parenthesis 2 ϕ close parenthesis$

For Problems 19–24, use identities to simplify each expression.

$sin ⁡ open parenthesis 4 x close parenthesis cos ⁡ open parenthesis 5 x close parenthesis plus cos ⁡ open parenthesis 4 x close parenthesis sin ⁡ open parenthesis 5 x close parenthesis$

$sin ⁡ open parenthesis 9 x close parenthesis$

$cos ⁡ open parenthesis 3 β close parenthesis cos ⁡ open parenthesis 1.5 close parenthesis minus sin ⁡ open parenthesis 3 β close parenthesis sin ⁡ open parenthesis 1.5 close parenthesis$

$\frac{\tan (2 ϕ) - \tan (2)}{1 + \tan (2 ϕ) \tan (2)}$

$tan ⁡ open parenthesis 2 ϕ minus 2 close parenthesis$

$\frac{\tan (\frac{5 π}{9}) - \tan (\frac{2 π}{9})}{1 + \tan (\frac{5 π}{9}) \tan (\frac{2 π}{9})}$

$2 sin ⁡ open parenthesis 4 θ close parenthesis cos ⁡ open parenthesis 4 θ close parenthesis$

$sin ⁡ open parenthesis 8 θ close parenthesis$

$1 minus 2 sin squared ⁡ open parenthesis 3 ϕ close parenthesis$

For Problems 25–26,

Use identities to rewrite the equation in terms of a single angle.
Solve. Give exact solutions between $0$ and $2 π$ .

$\cos (2 θ) - \sin (θ) = 1$

$1 - 2 \sin^{2} (θ) - \sin (θ) = 1$
$0, π, \frac{7 π}{6}, \frac{11 π}{6}$

$\tan (2 z) + \tan (z) = 0$

For Problems 27–28, graph the function and decide if it has an inverse function.

$f (x) = 4 x - x^{3}$

$g (x) = 5 + \sqrt[3]{x - 2}$

For Problems 29–30, give exact values in radians.

$\tan^{- 1} (- \sqrt{3})$
$\arccos (- \frac{1}{2})$

$\frac{- π}{3}$
$\frac{2 π}{3}$

$arcsin ⁡ open parenthesis minus 1 close parenthesis$
$\cos^{- 1} (- 1)$

An IMAX movie screen is 52.8 feet high.

If your line of sight is level with the bottom of the screen, write an expression for the angle subtended by the screen when you sit $x$ feet away.
Evaluate your expression for $x equals 20$ feet and for $x equals 100$ feet.

$\tan^{- 1} (\frac{52.8}{x})$
$69.25 degrees, 27.83 degrees$

Rembrandt's painting The Night Watch measures 13 feet high by 16 feet wide.

Write an expression for the angle subtended by the width of the painting if you sit $d$ feet back from the center of the painting.
Evaluate your expression for $d equals 10$ feet and for $d equals 25$ feet.

For Problems 33–34, solve for $θ$ .

$v_{y} = v_{0} \sin (θ) - g t$

$θ = \sin^{- 1} (\frac{v_{y} + g t}{v_{0}})$

$Δ W = - q_{0} E \cos (π - θ) Δ l$

For Problems 35–36, find exact values without using a calculator.

$\cos [\tan^{- 1} (\frac{- \sqrt{5}}{2})]$

$\frac{2}{3}$

$\tan [\sin^{- 1} (\frac{2}{7})]$

For Problems 37–38, simplify the expression.

$\sin [\cos^{- 1} (2 t)]$

$\sqrt{1 - 4 t^{2}}$

$\tan [\cos^{- 1} (m)]$

Explain why one of the expressions $\sin^{- 1} (x)$ or $\sin^{- 1} (\frac{1}{x})$ must be undefined.

Because $| \sin (θ) | \leq 1, \sin^{- 1} (t)$ is undefined for $| t | greater than 1$ . If $x not equal to 0$ , then either $| x | greater than 1$ or $| \frac{1}{x} | > 1$ . If $x equals 0$ , then $\frac{1}{x}$ is undefined.

Does $\sin^{- 1} (- x) = - \sin^{- 1} (x)$ ? Does $\cos^{- 1} (- x) = - \cos^{- 1} (x)$ ?

For Problems 41–42, evaluate. Round answers to 3 decimal places if necessary.

$csc ⁡ open parenthesis 27 degrees close parenthesis$
$sec ⁡ open parenthesis 108 degrees close parenthesis$
$cot ⁡ open parenthesis 245 degrees close parenthesis$

$2.203$
$minus 3.236$
$0.466$

$csc ⁡ open parenthesis 5.3 close parenthesis$
$cot ⁡ open parenthesis 0.98 close parenthesis$
$sec ⁡ open parenthesis 2.17 close parenthesis$

For Problems 43–50, find all six trigonometric ratios for the angle $θ$ .

$\sin (θ) = \frac{13}{\sqrt{313}}$ , $\cos (θ) = \frac{12}{\sqrt{313}}$ , $\tan (θ) = \frac{13}{12}$ , $\sec (θ) = \frac{\sqrt{313}}{12}$ , $\csc (θ) = \frac{\sqrt{313}}{13}$ , $\cot (θ) = \frac{12}{13}$

$\sin (θ) = \frac{1}{3}$ , $\cos (θ) = \frac{- 2 \sqrt{2}}{3}$ , $\tan (θ) = \frac{- 1}{2 \sqrt{2}}$ , $\sec (θ) = \frac{- 3}{2 \sqrt{2}}$ , $csc ⁡ open parenthesis θ close parenthesis equals 3$ , $\cot (θ) = - 2 \sqrt{2}$

$\sin (θ) = \frac{- 9}{\sqrt{106}}$ , $\cos (θ) = \frac{- 5}{\sqrt{106}}$ , $\tan (θ) = \frac{9}{5}$ , $\sec (θ) = \frac{- \sqrt{106}}{5}$ , $\csc (θ) = \frac{- \sqrt{106}}{9}$ , $\cot (θ) = \frac{5}{9}$

$6 \cos (α) = - 5$ , $180 degrees less than α less than 270 degrees$

$\sin (α) = \frac{- \sqrt{11}}{6}$ , $\cos (α) = \frac{- 5}{6}$ , $\tan (α) = \frac{\sqrt{11}}{5}$ , $\sec (α) = \frac{- 6}{5},$ $\csc (α) = \frac{- 6}{\sqrt{11}}$ , $\cot (α) = \frac{5}{\sqrt{11}}$

$4 sin ⁡ open parenthesis θ close parenthesis equals 3, θ$ is obtuse

For Problems 51–56, write algebraic expressions for the six trigonometric ratios of the angle.

$\sin (θ) = \frac{s}{4},$ $\cos (θ) = \frac{\sqrt{16 - s^{2}}}{4},$ $\tan (θ) = \frac{s}{\sqrt{16 - s^{2}}},$ $\sec (θ) = \frac{4}{\sqrt{16 - s^{2}}},$ $\csc (θ) = \frac{4}{s},$ $\cot (θ) = \frac{\sqrt{16 - s^{2}}}{s}$

$\sin (θ) = \frac{w}{\sqrt{w^{2} + 144}},$ $\cos (θ) = \frac{- 12}{\sqrt{w^{2} + 144}},$ $\tan (θ) = \frac{- w}{12},$ $\sec (θ) = \frac{- \sqrt{w^{2} + 144}}{12},$ $\csc (θ) = \frac{\sqrt{w^{2} + 144}}{w},$ $\cot (θ) = \frac{- 12}{w}$

$2 \sin (α) - k = 0, \frac{π}{2} < α < π$

$\sin (α) = \frac{k}{2},$ $\cos (α) = \frac{- \sqrt{4 - k^{2}}}{2},$ $\tan (α) = \frac{- k}{\sqrt{4 - k^{2}}},$ $\sec (α) = \frac{- 2}{\sqrt{4 - k^{2}}},$ $\csc (α) = \frac{2}{k},$ $\cot (α) = \frac{- \sqrt{4 - k^{2}}}{k}$

$h \cos (β) - 3 = 0, \frac{3 π}{2} < β < 2 π$

For Problems 57–58, find all six trigonometric ratios of the arc $θ$ . Round to two places.

$sin ⁡ open parenthesis θ close parenthesis equals 0.3$ , $\cos (θ) = - 0.4$ , $\tan (θ) = - 0.75$ , $\sec (θ) = - 2.5$ , $csc ⁡ open parenthesis θ close parenthesis approximately equals 3.33$ , $\cot θ (\approx) - 1.33$

For Problems 59–62, evaluate exactly.

$4 \cot (\frac{3 π}{4}) - \sec^{2} (\frac{π}{3})$

$minus 8$

$\frac{1}{2} \csc (\frac{2 π}{3}) + \tan^{2} (\frac{5 π}{6})$

$\csc (\frac{7 π}{6}) \cos (\frac{5 π}{4})$

$the square root of 2$

$\sec (\frac{7 π}{4}) \cot (\frac{4 π}{3})$

For Problems 63–64, find all solutions between $0$ and $2 π$ . Round your solutions to tenths.

$3 \csc (θ) + 2 = 12$

$θ \approx 2.8, θ \approx 0.30$

$5 \cot (θ) + 15 = - 3$

For Problems 65–70, sketch a graph of each function. Then choose the function or functions described by each statement.

$y = \sec (x) y = \csc (x) y = \cot (x)$

$y = \cos^{- 1} (x) y = \sin^{- 1} (x) y = \tan^{- 1} (x)$

The graph has vertical asymptotes at multiples of $π$ .

$y = \csc (x)$ or $y = \cot (x)$

The graph has a horizontal asymptote at $\frac{π}{2}$ .

The function values are the reciprocals of $y = \cos (x)$ .

$y = \sec (x)$

The function is defined only for $x$ -values between $minus 1$ and $1$ , inclusive.

None of the function values lie between $minus 1$ and $1$ .

$y = \sec (x)$ or $y = \csc (x)$

The graph includes the origin.

For Problems 71–74,

Graph the function on the interval $minus 2 π less than or equal to x less than or equal to 2 π$ , and use the graph to write the function in a simpler form.
Verify your conjecture algebraically.

$f (x) = \tan (x) [\cos (x) - \cot (x)]$

$f (x) = \sin (x) - 1$

$g (x) = \csc (x) - \cot (x) \cos (x)$

$G (x) = \sin x (\sec (x) - \csc (x))$

$G (x) = \tan (x) - 1$

$F (x) = \frac{1}{2} (\frac{\cos (x)}{1 + \sin (x)} + \frac{1 + \sin (x)}{\cos (x)})$

For Problems 75–78, simplify the expression.

$1 - \frac{\sin (x)}{\csc (x)}$

$\cos^{2} (x)$

$\frac{\sin (x)}{\csc (x)} + \frac{\cos (x)}{\sec (x)}$

$\frac{2 + \tan^{2} (B)}{\sec^{2} (B)} - 1$

$\cos^{2} (B)$

$\frac{\csc (t)}{\tan (t) + \cot (t)}$

For Problems 79–82, use the suggested substitution to simplify the expression.

$\frac{\sqrt{16 + x^{2}}}{x}, x = 4 \tan (θ)$

$\csc (θ)$

$x \sqrt{4 - x^{2}}, x = 2 \sin (θ)$

$\frac{x^{2} - 3}{x}, x = \sqrt{3} \sec (θ)$

$the square root of 3 tan ⁡ open parenthesis θ close parenthesis sin ⁡ open parenthesis θ close parenthesis$

$\frac{x}{\sqrt{x^{2} + 2}}, x = \sqrt{2} \tan (θ)$

This problem outlines a geometric proof of difference of angles formula for tangent.

In the figure below left, $α equals ∠ A B C$ and $β equals ∠ D B C$ . Write expressions in terms of $α$ and $β$ for the sides $A C, D C,$ and $A D$ .
In the figure above right, explain why $△ A B C$ is similar to $△ F B E$ .
Explain why $∠ F D C equals α$ .
Write an expression in terms of $α$ and $β$ for side $C F$ .
Explain why $△ F B E$ is similar to $△ A D E$ .
Justify each equality in the statement
$\tan (α - β) = \frac{D E}{B E} = \frac{A D}{B F} = \frac{\tan (α) - \tan (β)}{1 + \tan (α) \tan (β)}$

$A C = \tan (α), D C = \tan (β), A D = \tan (α) - \tan (β)$
They are right triangles that share $∠ B$ .
$∠ A equals ∠ F, ∠ B$ is the complement of $∠ A,$ and $∠ F D C$ is the complement of $∠ F$ .
$\frac{C F}{C D} = \tan (α),$ so $C F = \tan (α) \tan (β)$
They are right triangles with $∠ A equals ∠ F$ .
$∠ E B D = α - β,$ so $\tan (α - β) = \frac{opp}{adj} = \frac{D E}{B E}; \frac{D E}{B E}$ and $\frac{A D}{B F}$ are ratios of corresponding sides of similar triangles; $A D = \tan (α) - \tan (β)$ by part (a), $B F = B C + C F = 1 + \tan (α) \tan (β)$ by part (d).

Let $L sub 1$ and $L sub 2$ be two lines with slopes $m sub 1$ and $m sub 2$ , respectively, and let $θ$ be the acute angle formed between the two lines. Use an identity to show that

$\tan (θ) = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}$

For Problems 85–86, use the fact that if $θ$ is one angle of a triangle and $s$ is the length of the opposite side, then the diameter of the circumscribing circle is

$d = s \csc (θ)$

Round your answers to the nearest hundredth.

In the figure above, find the diameter of the circumscribing circle, the angle $α$ , and the sides $a$ and $b$ .

$d = 25 \csc (112^{\circ}), α = 45^{\circ}, a \approx 19.07, b \approx 10.54$

A triangle has one side of length 17cm and the angle opposite is $26 degrees$ . Find the diameter of the circle that circumscribes the triangle.

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.