Trigonometry Practice Questions — ACT — Page 4

Question 61Question →

For an angle $\theta$ satisfying $\pi < \theta < \frac{3\pi}{2}$ , if $\cos \theta = -\frac{12}{13}$ , what is the value of $\frac{\sec \theta - \cos \theta}{\tan \theta}$ ?

-\frac{12}{13}

-\frac{5}{13}

\frac{5}{13}

\frac{12}{13}

-\frac{25}{169}

Show answer & explanation

Answer: $-\frac{5}{13}$

Answer

The expression simplifies to

\sin \theta

, which equals

-\frac{5}{13}

.

Using identity substitutions

\sec \theta = \frac{1}{\cos \theta}

,

1 - \cos^2 \theta = \sin^2 \theta

, and

\tan \theta = \frac{\sin \theta}{\cos \theta}

, the expression simplifies directly to

\sin \theta

. Since

\cos \theta = -\frac{12}{13}

in Quadrant III where sine is negative,

\sin \theta = -\sqrt{1 - \left(-\frac{12}{13}\right)^2} = -\frac{5}{13}

.

Step-by-Step Solution

1

Simplify the given trigonometric expression using fundamental identities.

\frac{\sec \theta - \cos \theta}{\tan \theta} = \frac{\frac{1}{\cos \theta} - \cos \theta}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1 - \cos^2 \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{\sin^2 \theta}{\sin \theta} = \sin \theta

Applying the reciprocal identity

\sec \theta = \frac{1}{\cos \theta}

, the quotient identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

, and the Pythagorean identity

1 - \cos^2 \theta = \sin^2 \theta

simplifies the expression directly to

\sin \theta

.

2

Calculate the magnitude of

\sin \theta

using the Pythagorean identity.

|\sin \theta| = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

The Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

allows finding the absolute value of

\sin \theta

.

3

Determine the correct sign for

\sin \theta

based on the given quadrant.

\sin \theta = -\frac{5}{13}

Since

\pi < \theta < \frac{3\pi}{2}

, the angle lies in Quadrant III, where the sine function is negative.

Key Concept

Simplifying expressions using fundamental Pythagorean, reciprocal, and quotient identities while applying quadrant sign rules.

Question 62Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the figure into two triangles, $\triangle ABC$ and $\triangle ACD$ . It is given that $AB = 6$ , $BC = 10$ , $\angle ABC = 120^\circ$ , $\angle CAD = 45^\circ$ , and $\angle ADC = 60^\circ$ . What is the length of side $CD$ ?

\frac{14\sqrt{6}}{3}

14

7\sqrt{3}

\frac{14\sqrt{2}}{3}

\sqrt{76}

Show answer & explanation

Answer: $\frac{14\sqrt{6}}{3}$

Answer

The length of side

CD

is

\frac{14\sqrt{6}}{3}

.

First, apply the Law of Cosines to

\triangle ABC

to find the length of diagonal

AC

:

AC^2 = 6^2 + 10^2 - 2(6)(10)\cos(120^\circ) = 36 + 100 - 120(-0.5) = 196

, which yields

AC = 14

. Next, use the Law of Sines in

\triangle ACD

:

\frac{CD}{\sin(45^\circ)} = \frac{14}{\sin(60^\circ)}

. Solving for

CD

gives

CD = 14 \cdot \frac{\sqrt{2}/2}{\sqrt{3}/2} = \frac{14\sqrt{2}}{\sqrt{3}} = \frac{14\sqrt{6}}{3}

.

Step-by-Step Solution

1

Apply the Law of Cosines in

\triangle ABC

to calculate the length of diagonal

AC

.

AC^2 = 6^2 + 10^2 - 2(6)(10)\cos(120^\circ) = 36 + 100 - 120\left(-\frac{1}{2}\right) = 136 + 60 = 196

, so

AC = 14

.

Two side lengths and the included angle of

\triangle ABC

are known.

2

Apply the Law of Sines in

\triangle ACD

to set up a proportion for side

CD

.

\frac{CD}{\sin(\angle CAD)} = \frac{AC}{\sin(\angle ADC)} \implies \frac{CD}{\sin(45^\circ)} = \frac{14}{\sin(60^\circ)}

.

The Law of Sines relates opposite sides and angles in

\triangle ACD

.

3

Solve for

CD

and rationalize the denominator.

CD = 14 \cdot \frac{\sin(45^\circ)}{\sin(60^\circ)} = 14 \cdot \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = \frac{14\sqrt{2}}{\sqrt{3}} = \frac{14\sqrt{6}}{3}

.

Evaluating the exact trigonometric values yields the final simplified length.

Key Concept

Law of Sines and Law of Cosines in Composite Triangles

Estimated Time:2m 0s

Question 63Question →

In right triangle $XYZ$ , the right angle is at vertex $Y$ , and line segment $YW$ is an altitude perpendicular to hypotenuse $XZ$ at point $W$ . If the length of side $XY$ is $15$ units and $\cos(X) = \frac{4}{5}$ , what is the length of line segment $ZW$ ?

9

12

\frac{27}{4}

\frac{45}{4}

16

Show answer & explanation

Answer: $\frac{27}{4}$

Answer

The length of line segment

ZW

is

\frac{27}{4}

units.

In right triangle

XYZ

,

\cos(X) = \frac{XY}{XZ}

. Given

XY = 15

and

\cos(X) = \frac{4}{5}

, we solve for hypotenuse

XZ = \frac{75}{4}

. In right triangle

XYW

,

\cos(X) = \frac{XW}{XY} = \frac{XW}{15}

, yielding

XW = 12

. Subtracting

XW

from total hypotenuse

XZ

yields

ZW = \frac{75}{4} - 12 = \frac{27}{4}

.

Step-by-Step Solution

1

Find the length of hypotenuse

XZ

using

\cos(X)

in

\triangle XYZ

.

XZ = \frac{75}{4}

In

\triangle XYZ

,

\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{XZ}

. Substituting

\cos(X) = \frac{4}{5}

and

XY = 15

gives

\frac{4}{5} = \frac{15}{XZ}

, so

XZ = \frac{15 \times 5}{4} = \frac{75}{4}

.

2

Find the length of segment

XW

using

\cos(X)

in right triangle

XYW

.

XW = 12

In right triangle

XYW

(with right angle at

W

),

\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XW}{XY}

. Substituting

\cos(X) = \frac{4}{5}

and

XY = 15

gives

\frac{4}{5} = \frac{XW}{15}

, so

XW = 12

.

3

Calculate segment

ZW

by subtracting

XW

from total hypotenuse

XZ

.

ZW = \frac{27}{4}

Since point

W

lies on segment

XZ

,

ZW = XZ - XW = \frac{75}{4} - 12 = \frac{75}{4} - \frac{48}{4} = \frac{27}{4}

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) in Nested Right Triangles

Question 64Question →

In right triangle $ABC$ , the right angle is located at vertex $C$ . Point $M$ is the midpoint of leg $BC$ . The length of leg $AC$ is $12$ units, and $\tan(\angle MAC) = \frac{1}{3}$ . What is the value of $\sin(\angle BAC)$ ?

\frac{2\sqrt{13}}{13}

\frac{3\sqrt{13}}{13}

\frac{2}{3}

\frac{\sqrt{10}}{10}

\frac{2\sqrt{5}}{5}

Show answer & explanation

Answer: $\frac{2\sqrt{13}}{13}$

Answer

The value of

\sin(\angle BAC)

is

\frac{2\sqrt{13}}{13}

.

In right triangle

ACM

,

\tan(\angle MAC) = \frac{MC}{AC} = \frac{1}{3}

. Since

AC = 12

, we find

MC = 4

. Because

M

is the midpoint of side

BC

,

BC = 2 \times 4 = 8

. In right triangle

ABC

, the hypotenuse is

AB = \sqrt{12^2 + 8^2} = \sqrt{208} = 4\sqrt{13}

. The sine of angle

BAC

is defined as

\frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2\sqrt{13}}{13}

.

Step-by-Step Solution

1

Find the length of segment

MC

using right triangle

ACM

.

MC = 4

In right triangle

ACM

with right angle at

C

,

\tan(\angle MAC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{MC}{AC}

. Given

\tan(\angle MAC) = \frac{1}{3}

and

AC = 12

,

\frac{MC}{12} = \frac{1}{3} \implies MC = 4

.

2

Determine the length of side

BC

.

BC = 8

Since

M

is the midpoint of leg

BC

,

BC = 2 \times MC = 2 \times 4 = 8

.

3

Calculate hypotenuse

AB

of right triangle

ABC

.

AB = 4\sqrt{13}

By the Pythagorean theorem in

\triangle ABC

:

AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}

.

4

Calculate

\sin(\angle BAC)

.

\sin(\angle BAC) = \frac{2\sqrt{13}}{13}

In right triangle

ABC

,

\sin(\angle BAC) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}

.

Key Concept

Applying SOHCAHTOA definitions and the Pythagorean theorem in multi-step right triangle geometry.

Estimated Time:2m 0s

Question 65Question →

If $\sin \theta + \cos \theta = \sqrt{1.5}$ , what is the value of $\tan \theta + \cot \theta$ ?

Show answer & explanation

Answer: 4

Answer

4

Squaring both sides of

\sin \theta + \cos \theta = \sqrt{1.5}

yields

\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5

. Using the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

, we get

1 + 2\sin \theta \cos \theta = 1.5

, which simplifies to

\sin \theta \cos \theta = 0.25

. Rewriting

\tan \theta + \cot \theta

using quotient identities gives

\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

. Substituting

\sin \theta \cos \theta = 0.25

yields

\frac{1}{0.25} = 4

.

Step-by-Step Solution

1

Square both sides of the given equation

\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5

Squaring both sides allows the expansion of the binomial

(\sin \theta + \cos \theta)^2

to reveal the product term

\sin \theta \cos \theta

.

2

Apply the Pythagorean identity to solve for

\sin \theta \cos \theta

\sin \theta \cos \theta = 0.25

Since

\sin^2 \theta + \cos^2 \theta = 1

, substituting

1

into

1 + 2\sin \theta \cos \theta = 1.5

gives

2\sin \theta \cos \theta = 0.5

, so

\sin \theta \cos \theta = 0.25

.

3

Rewrite

\tan \theta + \cot \theta

using quotient identities

\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

Using

\tan \theta = \frac{\sin \theta}{\cos \theta}

and

\cot \theta = \frac{\cos \theta}{\sin \theta}

, finding a common denominator yields

\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}

. Applying

\sin^2 \theta + \cos^2 \theta = 1

simplifies the numerator to

1

.

4

Substitute the numerical value of

\sin \theta \cos \theta

to find the answer

\tan \theta + \cot \theta = \frac{1}{0.25} = 4

Dividing

1

by

0.25

evaluates to the exact integer

4

.

Key Concept

Fundamental Trigonometric Identities (Pythagorean and Quotient Identities)

Estimated Time:1m 30s

Question 66Question →

What is the exact value of $\cos\left(2\arctan(-3)\right)$ expressed as a decimal?

Show answer & explanation

Answer: -0.8

Answer

The exact decimal value of

\cos\left(2\arctan(-3)\right)

is

-0.8

.

Letting

\theta = \arctan(-3)

gives

\tan(\theta) = -3

. Utilizing the double-angle formula

\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}

, we substitute

\tan(\theta) = -3

to obtain

\frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8

.

Step-by-Step Solution

1

Define an angle variable for the inverse trigonometric expression.

Let

\theta = \arctan(-3)

, which means

\tan(\theta) = -3

in Quadrant IV where

-\frac{\pi}{2} < \theta < 0

.

Applying the standard definition and principal range of the arctangent function.

2

Apply the double-angle identity for cosine expressed in terms of tangent.

\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}

This identity directly connects

\cos(2\theta)

to the known value of

\tan(\theta)

without needing radicals.

3

Substitute

\tan(\theta) = -3

into the identity and evaluate.

\cos(2\theta) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8

Simplifying the numerical expression produces the exact decimal value.

Key Concept

Inverse Trigonometric Functions and Double-Angle Identities

Estimated Time:1m 30s

Question 67Question →

If $\alpha = \arccos\left(-\frac{5}{13}\right)$ , what is the exact value of $\tan\left(\alpha - \frac{\pi}{4}\right)$ ?

\frac{17}{7}

\frac{7}{17}

-\frac{17}{7}

-\frac{7}{17}

-\frac{17}{5}

Show answer & explanation

Answer: $\frac{17}{7}$

Answer

The exact value of

\tan\left(\alpha - \frac{\pi}{4}\right)

is

\frac{17}{7}

.

Evaluating

\alpha = \arccos\left(-\frac{5}{13}\right)

places

\alpha

in Quadrant II (

\frac{\pi}{2} < \alpha < \pi

). In this quadrant, cosine is negative (

-\frac{5}{13}

) and sine is positive (

\frac{12}{13}

), giving

\tan\alpha = -\frac{12}{5}

. Substituting

\tan\alpha = -\frac{12}{5}

and

\tan\left(\frac{\pi}{4}\right) = 1

into the identity

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

yields

\frac{-\frac{12}{5} - 1}{1 - \frac{12}{5}} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}

.

Step-by-Step Solution

1

Determine the quadrant and reference values for

\alpha = \arccos\left(-\frac{5}{13}\right)

Since the principal range of

\arccos(x)

is

[0, \pi]

and

-\frac{5}{13} < 0

,

\alpha

lies in Quadrant II where

\cos\alpha = -\frac{5}{13}

and

\sin\alpha > 0

.

Inverse cosine returns angles in Quadrant II for negative arguments.

2

Calculate

\sin\alpha

and

\tan\alpha

\sin\alpha = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}

, so

\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{-5/13} = -\frac{12}{5}

.

Using the Pythagorean identity and definition of tangent.

3

Apply the tangent angle subtraction identity

\tan\left(\alpha - \frac{\pi}{4}\right) = \frac{\tan\alpha - \tan\left(\frac{\pi}{4}\right)}{1 + \tan\alpha \tan\left(\frac{\pi}{4}\right)} = \frac{-\frac{12}{5} - 1}{1 + \left(-\frac{12}{5}\right)(1)} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}

.

Evaluating

\tan\left(\frac{\pi}{4}\right) = 1

and simplifying the complex fraction.

Key Concept

Inverse Trigonometric Functions and Compound Angle Identities

Estimated Time:2m 0s

Question 68Question →

In right triangle $PQR$ , the right angle is at vertex $Q$ . Point $S$ lies on leg $PQ$ such that the length of $PS$ is $7$ units and the length of $SQ$ is $5$ units. If $\tan(\angle PRQ) = \frac{4}{3}$ , what is the value of $\cos(\angle SRQ)$ ?

\frac{5\sqrt{106}}{106}

\frac{9\sqrt{106}}{106}

\frac{9\sqrt{130}}{130}

\frac{16\sqrt{281}}{281}

\frac{5}{9}

Show answer & explanation

Answer: $\frac{9\sqrt{106}}{106}$

Answer

The value of

\cos(\angle SRQ)

is

\frac{9\sqrt{106}}{106}

.

The total length of leg

PQ

is

7 + 5 = 12

. Using the tangent definition in right triangle

PQR

,

\tan(\angle PRQ) = \frac{PQ}{QR} = \frac{12}{QR} = \frac{4}{3}

, which gives

QR = 9

. Next, considering right triangle

SQR

with legs

SQ = 5

and

QR = 9

, the hypotenuse

SR = \sqrt{5^2 + 9^2} = \sqrt{106}

. Finally,

\cos(\angle SRQ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{QR}{SR} = \frac{9}{\sqrt{106}} = \frac{9\sqrt{106}}{106}

.

Step-by-Step Solution

1

Find the length of leg

PQ

PQ = PS + SQ = 7 + 5 = 12

units

Point

S

lies on segment

PQ

, so the total length is the sum of its parts.

2

Calculate the length of leg

QR

using

\tan(\angle PRQ)

\tan(\angle PRQ) = \frac{PQ}{QR} \implies \frac{4}{3} = \frac{12}{QR} \implies QR = 9

units

In right triangle

PQR

, tangent is opposite side over adjacent side relative to

\angle PRQ

.

3

Find hypotenuse

SR

of right triangle

SQR

SR = \sqrt{SQ^2 + QR^2} = \sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}

units

Apply the Pythagorean theorem to right triangle

SQR

with right angle at

Q

.

4

Determine

\cos(\angle SRQ)

and rationalize the denominator

\cos(\angle SRQ) = \frac{QR}{SR} = \frac{9}{\sqrt{106}} = \frac{9\sqrt{106}}{106}

Cosine is defined as the ratio of adjacent side over hypotenuse in right triangle

SQR

.

Key Concept

Applying SOHCAHTOA and the Pythagorean Theorem in composite right triangle figures

Estimated Time:2m 0s

Question 69Question →

Which of the following expressions is equivalent to $\frac{\tan \theta + \cot \theta}{\csc \theta}$ for all values of $\theta$ where the expression is defined?

\sec \theta

\cos \theta

\sin \theta

\csc \theta

\tan^2 \theta

Show answer & explanation

Answer: $\sec \theta$

Answer

\sec \theta

Substituting

\tan \theta = \frac{\sin \theta}{\cos \theta}

and

\cot \theta = \frac{\cos \theta}{\sin \theta}

gives

\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

in the numerator. Dividing by

\csc \theta = \frac{1}{\sin \theta}

cancels the

\sin \theta

term, leaving

\frac{1}{\cos \theta}

, which is equal to

\sec \theta

.

Step-by-Step Solution

1

Express tangent and cotangent using sine and cosine

\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}

Quotient identities allow rewriting all functions in terms of sine and cosine.

2

Combine the fractions in the numerator over a common denominator

\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}

Finding a common denominator of

\sin \theta \cos \theta

allows application of the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

.

3

Divide by the denominator

\csc \theta

\frac{\frac{1}{\sin \theta \cos \theta}}{\frac{1}{\sin \theta}} = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin \theta}{1} = \frac{1}{\cos \theta} = \sec \theta

The cosecant function is the reciprocal of sine, so dividing by

\csc \theta

cancels out the

\sin \theta

factor in the denominator.

Key Concept

Fundamental Trigonometric Identities

Question 70Question →

If $\tan \theta = 2$ , what is the value of $\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}$ ?

Show answer & explanation

Answer: 1.25

Answer

The value of the expression is 1.25.

Using the trigonometric quotient identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

, dividing both the numerator and denominator of

\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}

by

\cos \theta

transforms the fraction into

\frac{3\tan \theta + 4}{5\tan \theta - 2}

. Substituting

\tan \theta = 2

gives

\frac{3(2) + 4}{5(2) - 2} = \frac{10}{8} = 1.25

. Alternatively, constructing a right triangle with opposite side 2 and adjacent side 1 gives a hypotenuse of

\sqrt{5}

, yielding

\sin \theta = \frac{2}{\sqrt{5}}

and

\cos \theta = \frac{1}{\sqrt{5}}

, which produces the exact same ratio of

\frac{10/\sqrt{5}}{8/\sqrt{5}} = 1.25

.

Step-by-Step Solution

1

Divide every term in both the numerator and the denominator by

\cos \theta

.

The expression becomes

\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 4\left(\frac{\cos \theta}{\cos \theta}\right)}{5\left(\frac{\sin \theta}{\cos \theta}\right) - 2\left(\frac{\cos \theta}{\cos \theta}\right)}

.

Dividing by

\cos \theta

allows us to convert sine-and-cosine terms into tangent terms using the quotient identity.

2

Apply the quotient identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

.

The expression simplifies to

\frac{3\tan \theta + 4}{5\tan \theta - 2}

.

This reduces the trigonometric expression to an algebraic expression containing only

\tan \theta

.

3

Substitute

\tan \theta = 2

into the simplified expression and compute the result.

\frac{3(2) + 4}{5(2) - 2} = \frac{6 + 4}{10 - 2} = \frac{10}{8} = 1.25.

Performing basic arithmetic yields the exact numeric answer.

Key Concept

Quotient Identity of Tangent

Question 71Question →

In the standard $(x,y)$ coordinate plane, what is the distance along the $x$ -axis between any two consecutive maximum points on the graph of the function $f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4$ ?

\frac{\pi}{3}

\frac{\pi}{2}

\frac{2\pi}{3}

2\pi

6\pi

Show answer & explanation

Answer: $\frac{2\pi}{3}$

Answer

The distance between any two consecutive maximum points is

\frac{2\pi}{3}

.

The distance along the

x

-axis between consecutive peaks of a cosine wave represents one full period. For the function

f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4

, the coefficient of

x

is 3. The period formula for cosine is

\frac{2\pi}{|B|}

, which evaluates to

\frac{2\pi}{3}

.

Step-by-Step Solution

1

Relate consecutive maximum points to the function's period

The distance along the

x

-axis between consecutive maximum points of a periodic trigonometric function equals one period length,

T

.

Cosine graphs repeat their peak values once per complete wave cycle.

2

Identify the coefficient

B

of the variable

x

In

f(x) = 5 \cos\left(3x - \frac{\pi}{2}\right) + 4

, the coefficient of

x

is

B = 3

.

The standard transformation model is

y = A \cos(Bx - C) + D

.

3

Calculate the period using

T = \frac{2\pi}{|B|}

T = \frac{2\pi}{3}

.

Dividing the base cosine period of

2\pi

by

|B| = 3

gives the period of the transformed function.

Key Concept

The period of a transformed cosine function

y = A \cos(Bx - C) + D

is

\frac{2\pi}{|B|}

, which measures the horizontal distance between consecutive peak values.

Question 72Question →

What is the period of the trigonometric function $f(x) = 3 \cos\left(\frac{\pi}{5}x + \frac{\pi}{2}\right) - 4$ ?

10

5

2\pi

\frac{\pi}{5}

\frac{2}{5}

Show answer & explanation

Answer: $10$

Answer

10

The period of a function of the form

f(x) = A \cos(Bx + C) + D

is given by

T = \frac{2\pi}{|B|}

. Substituting

B = \frac{\pi}{5}

yields

T = \frac{2\pi}{\frac{\pi}{5}} = 2\pi \cdot \frac{5}{\pi} = 10

.

Step-by-Step Solution

1

Identify the coefficient of

x

(

B

) in the given trigonometric function.

B = \frac{\pi}{5}

The standard form for a transformed cosine function is

f(x) = A \cos(Bx + C) + D

, where

B

controls the horizontal stretch or compression.

2

Apply the period formula for cosine,

T = \frac{2\pi}{|B|}

.

T = \frac{2\pi}{\frac{\pi}{5}}

The fundamental period of

\cos(x)

is

2\pi

, which is scaled inversely by

|B|

.

3

Simplify the fraction by multiplying by the reciprocal.

T = 2\pi \cdot \frac{5}{\pi} = 10

Dividing by a fraction is equivalent to multiplying by its reciprocal, and the factor of

\pi

cancels out.

Key Concept

Period of Transformed Cosine Functions

Question 73Question →

Match each transformed trigonometric function on the left with its set of defining graphical properties on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

f(x) = 4\sin(3x) + 2

matches with 'Amplitude of 4, period of 2π/3, and a midline at y = 2';

g(x) = -2\cos(2x - π)

matches with 'Amplitude of 2, period of π, and a phase shift of π/2 units to the right';

h(x) = 3\tan(x/2) - 1

matches with 'Period of 2π, vertical shift of 1 unit down, with vertical asymptotes at x = π + 2kπ'; and

k(x) = \cos(x + π/3) - 4

matches with 'Midline at y = -4, amplitude of 1, and a phase shift of π/3 units to the left'.

Each trigonometric function is matched according to its standard parameter transformation rules:

y = A\sin(B(x - C)) + D

or

y = A\cos(B(x - C)) + D

, where

|A|

is amplitude, period is

\frac{2\pi}{|B|}

for sine/cosine and

\frac{\pi}{|B|}

for tangent,

C

is horizontal phase shift, and

y = D

is the midline.

Step-by-Step Solution

1

Analyze

f(x) = 4\sin(3x) + 2

using the standard form

y = A\sin(B(x - C)) + D

Amplitude

= |A| = 4

, period

= \frac{2\pi}{B} = \frac{2\pi}{3}

, midline

= D \implies y = 2

.

Direct extraction of parameters for sine graphs.

2

Factor out

B = 2

from

g(x) = -2\cos(2x - \pi)

g(x) = -2\cos\left(2\left(x - \frac{\pi}{2}\right)\right)

, so amplitude

= |-2| = 2

, period

= \frac{2\pi}{2} = \pi

, phase shift

= \frac{\pi}{2}

to the right.

Factoring

B

is necessary to correctly identify the horizontal phase shift.

3

Analyze tangent function parameters for

h(x) = 3\tan\left(\frac{1}{2}x\right) - 1

Period

= \frac{\pi}{B} = \frac{\pi}{1/2} = 2\pi

, shifted down

1

unit (

y = -1

). Asymptotes occur when

\frac{1}{2}x = \frac{\pi}{2} + k\pi \implies x = \pi + 2k\pi

.

Tangent period uses

\frac{\pi}{B}

instead of

\frac{2\pi}{B}

, and asymptotes occur where tangent arguments equal odd multiples of

\frac{\pi}{2}

.

4

Analyze transformation parameters for

k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4

Amplitude

= 1

, midline

= y = -4

, phase shift

= \frac{\pi}{3}

to the left.

Addition inside the function argument

(x + C)

corresponds to a horizontal shift to the left.

Key Concept

Identifying amplitude, period, midline, phase shift, and asymptotes from transformed trigonometric equations

Question 74Question →

In the standard $(x,y)$ coordinate plane, a cosine function of the form $y = a \cos(b(x - c)) + d$ , where $a > 0$ and $b > 0$ , has a local maximum at $(2, 9)$ and the very next local minimum at $(6, 1)$ . What is the value of $b$ ?

\frac{\pi}{8}

\frac{\pi}{4}

\frac{\pi}{2}

\pi

2\pi

Show answer & explanation

Answer: $\frac{\pi}{4}$

Answer

\frac{\pi}{4}

The distance along the

x

-axis from a maximum to the consecutive minimum is half of one full period of the cosine wave. Here, that distance is

6 - 2 = 4

, which means the full period is

2 \times 4 = 8

. Using the relationship

\text{Period} = \frac{2\pi}{b}

, we solve for

b

to get

b = \frac{2\pi}{8} = \frac{\pi}{4}

.

Step-by-Step Solution

1

Determine the horizontal distance between the consecutive maximum and minimum points.

The horizontal distance is

6 - 2 = 4

units.

The

x

-coordinates of the maximum and minimum points are

2

and

6

, respectively.

2

Calculate the period of the cosine function.

\text{Period} = 2 \times 4 = 8

units.

The horizontal distance between a peak and the immediately following trough of a cosine wave represents exactly half of one full period.

3

Solve for the coefficient

b

using the period formula

b = \frac{2\pi}{\text{Period}}

.

b = \frac{2\pi}{8} = \frac{\pi}{4}

.

For a trigonometric function of the form

y = a \cos(b(x - c)) + d

, the relationship between the period and

b

is

\text{Period} = \frac{2\pi}{b}

.

Key Concept

Determining Period and Frequency Coefficient of a Cosine Function

Question 75Question →

In the standard $(x,y)$ coordinate plane, a sinusoidal function $f(x) = A \sin(Bx - C) + D$ reaches a maximum value of $9$ at $x = 1$ and its immediate next minimum value of $-1$ at $x = 4$ . What is the period of $f(x)$ ?

Show answer & explanation

Answer: 6

Answer

The period of the sinusoidal function is 6.

For any sinusoidal graph, the horizontal distance between a maximum point and the immediate next minimum point corresponds to one-half of the period. Given that the maximum occurs at

x = 1

and the next minimum occurs at

x = 4

, the half-period is

4 - 1 = 3

. Multiplying this half-period by

2

gives the full period of

6

.

Step-by-Step Solution

1

Identify the horizontal distance between the consecutive maximum and minimum points.

The horizontal distance is

4 - 1 = 3

.

The maximum occurs at

x = 1

and the consecutive minimum occurs at

x = 4

.

2

Relate the horizontal distance between consecutive extrema to the period of the function.

Half of the period is equal to

3

.

In any sinusoidal function, the horizontal distance between a peak (maximum) and the adjacent trough (minimum) represents exactly one-half of a complete cycle.

3

Calculate the full period of the function.

Period =

2 \times 3 = 6

.

Multiplying the half-period by 2 yields the full period of the function.

Key Concept

The horizontal distance between consecutive maximum and minimum points of a sinusoidal graph is half of the function's period.

Estimated Time:1m 15s

Question 76Question →

Which of the following values represents the period, in radians, of the trigonometric function $f(x) = 3 \tan\left(\frac{2}{3}x - \frac{\pi}{6}\right) + 5$ ?

\frac{3\pi}{2}

3\pi

\frac{2\pi}{3}

\frac{\pi}{6}

\frac{3\pi}{4}

Show answer & explanation

Answer: $\frac{3\pi}{2}$

Answer

The period of the given tangent function is

\frac{3\pi}{2}

radians.

The parent function

y = \tan(x)

repeats every

\pi

radians. For a function in the form

f(x) = A \tan(Bx - C) + D

, the horizontal scale factor

B

alters the period according to

\text{Period} = \frac{\pi}{|B|}

. With

B = \frac{2}{3}

, dividing

\pi

by

\frac{2}{3}

gives

\frac{3\pi}{2}

.

Step-by-Step Solution

1

Identify the standard form of the transformed tangent function and its parameters.

For

f(x) = A \tan(Bx - C) + D

, the coefficient of

x

is

B = \frac{2}{3}

.

The horizontal stretch/compression factor

B

determines the period of the function.

2

Apply the period formula for the tangent function.

\text{Period} = \frac{\pi}{|B|} = \frac{\pi}{\frac{2}{3}} = \frac{3\pi}{2}

Unlike sine and cosine functions which have a fundamental period of

2\pi

, the parent tangent function

y = \tan(x)

has a period of

\pi

radians.

Key Concept

Period of Transformed Tangent Functions

Estimated Time:1m 0s

Question 77Question →

Match each trigonometric function on the left with its correct combination of amplitude and period on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

f(x) = 4 \sin(3x)

matches Amplitude = 4, Period =

\frac{2\pi}{3}

;

g(x) = 2 \cos\left(\frac{1}{2}x\right)

matches Amplitude = 2, Period =

4\pi

;

h(x) = -3 \sin(2x)

matches Amplitude = 3, Period =

\pi

.

Each trigonometric function is correctly evaluated using the general properties: Amplitude equals

|A|

and Period equals

\frac{2\pi}{|B|}

.

Step-by-Step Solution

1

Identify the standard trigonometric form parameters.

For equations of the form

y = A \sin(Bx)

or

y = A \cos(Bx)

, Amplitude

= |A|

and Period

= \frac{2\pi}{|B|}

.

Applying the definitions of amplitude and period for sine and cosine functions.

2

Calculate amplitude and period for

f(x) = 4 \sin(3x)

.

Amplitude

= |4| = 4

, Period

= \frac{2\pi}{3}

.

Here

A = 4

and

B = 3

.

3

Calculate amplitude and period for

g(x) = 2 \cos\left(\frac{1}{2}x\right)

.

Amplitude

= |2| = 2

, Period

= \frac{2\pi}{1/2} = 4\pi

.

Here

A = 2

and

B = \frac{1}{2}

.

4

Calculate amplitude and period for

h(x) = -3 \sin(2x)

.

Amplitude

= |-3| = 3

, Period

= \frac{2\pi}{2} = \pi

.

Here

A = -3

(so

|A| = 3

) and

B = 2

.

Key Concept

Amplitude and Period of Sine and Cosine Graphs

Question 78Question →

In the standard $(x,y)$ coordinate plane, the graph of a trigonometric function is given by $f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2$ . What is the horizontal distance between any two consecutive points where the graph intersects its midline $y = 2$ ?

\frac{\pi}{6}

\frac{\pi}{3}

\frac{\pi}{2}

\frac{2\pi}{3}

\frac{4\pi}{3}

Show answer & explanation

Answer: $\frac{\pi}{3}$

Answer

The horizontal distance between consecutive midline intersections is

\frac{\pi}{3}

.

The midline of

f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2

is the horizontal line

y = 2

. Intersections with this line occur when

\sin\left(3x - \frac{\pi}{2}\right) = 0

. Since the sine function equals zero at integer multiples of

\pi

, the difference in the argument between consecutive zero points is

\pi

. Setting

3\Delta x = \pi

gives

\Delta x = \frac{\pi}{3}

, which is half the period of the function.

Step-by-Step Solution

1

Identify the frequency parameter

B

from the function

f(x) = 4 \sin\left(3x - \frac{\pi}{2}\right) + 2

.

The frequency coefficient inside the sine expression is

B = 3

.

The standard form of a transformed sine function is

f(x) = A \sin(Bx - C) + D

.

2

Calculate the full period

T

of the function.

T = \frac{2\pi}{|B|} = \frac{2\pi}{3}

.

The standard period

2\pi

of a sine function is scaled horizontally by a factor of

\frac{1}{B}

.

3

Determine the horizontal distance between consecutive intersections with the midline

y = 2

.

\text{Distance} = \frac{T}{2} = \frac{\frac{2\pi}{3}}{2} = \frac{\pi}{3}.

A sinusoidal wave completes one full period over length

T

and intersects its midline twice during each full cycle, making consecutive midline crossings separated by half of the period.

Key Concept

Distance between consecutive midline intersections of a sinusoidal function

Estimated Time:1m 0s

Question 79Question →

A telecommunications company connects two remote cell towers, Tower X and Tower Y, to a central relay station R. The cable path from Relay Station R to Tower X is $7$ kilometers long, and the cable path from Relay Station R to Tower Y is $8$ kilometers long. The angle formed between the two cables at Relay Station R, $\angle XRY$ , measures $120^\circ$ . A straight wireless backup link is established directly between Tower X and Tower Y. What is the distance, in kilometers, of this direct wireless link?

Show answer & explanation

Answer: 13

Answer

The distance of the direct wireless link between Tower X and Tower Y is 13 kilometers.

Using the Law of Cosines with two side lengths of 7 km and 8 km and an included angle of 120° gives

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ) = 49 + 64 - 112(-0.5) = 169

. Taking the square root gives 13 km.

Step-by-Step Solution

1

Identify known components of the triangle formed by the relay station and towers

Side

RX = 7\text{ km}

, side

RY = 8\text{ km}

, and included angle

\angle R = 120^\circ

Two sides and the included angle (SAS) are known, indicating that the Law of Cosines must be used to find the third side.

2

Apply the Law of Cosines formula for side

XY

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ)

The Law of Cosines relates three sides of a triangle to the cosine of one of its angles:

c^2 = a^2 + b^2 - 2ab\cos(C)

.

3

Calculate the numeric value of

XY^2

XY^2 = 49 + 64 + 56 = 169

Since

\cos(120^\circ) = -0.5

, the term

-2(56)(-0.5)

evaluates to

+56

.

4

Take the principal square root of

169

XY = 13\text{ km}

Distance must be a positive length.

Key Concept

Law of Cosines (Side-Angle-Side configuration)

Question 80Question →

In $\triangle ABC$ , the length of side $a$ (opposite angle $A$ ) is $9$ inches, the length of side $b$ (opposite angle $B$ ) is $12$ inches, and the measure of angle $B$ is $45^\circ$ . What is the exact value of $\sin A$ ?

\frac{3\sqrt{2}}{8}

\frac{2\sqrt{2}}{3}

\frac{3\sqrt{2}}{4}

\frac{3}{8}

\frac{4\sqrt{2}}{3}

Show answer & explanation

Answer: $\frac{3\sqrt{2}}{8}$

Answer

The exact value of

\sin A

is

\frac{3\sqrt{2}}{8}

.

According to the Law of Sines,

\frac{\sin A}{a} = \frac{\sin B}{b}

. Substituting

a = 9

,

b = 12

, and

B = 45^\circ

yields

\frac{\sin A}{9} = \frac{\sin 45^\circ}{12}

. Multiplying both sides by

9

and substituting

\sin 45^\circ = \frac{\sqrt{2}}{2}

gives

\sin A = \frac{9 \cdot \frac{\sqrt{2}}{2}}{12} = \frac{9\sqrt{2}}{24} = \frac{3\sqrt{2}}{8}

.

Step-by-Step Solution

1

State the Law of Sines for the given triangle components.

\frac{\sin A}{a} = \frac{\sin B}{b}

The Law of Sines relates the sines of angles to their opposite side lengths in any triangle.

2

Substitute the known values (

a = 9

,

b = 12

, and

B = 45^\circ

) into the formula.

\frac{\sin A}{9} = \frac{\sin 45^\circ}{12}

Plugging in the given numbers isolates the unknown quantity

\sin A

.

3

Solve for

\sin A

and substitute the exact value of

\sin 45^\circ = \frac{\sqrt{2}}{2}

.

\sin A = \frac{9 \cdot \sin 45^\circ}{12} = \frac{9 \cdot \frac{\sqrt{2}}{2}}{12} = \frac{9\sqrt{2}}{24} = \frac{3\sqrt{2}}{8}

Simplifying the fraction gives the exact trigonometric ratio.

Key Concept

Law of Sines

Trigonometry

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept