1.1Convert 150 Degrees to Radians
Convert $150^\circ$ to radians.
Solution
Use the degree-to-radian conversion rule:
$$
150^\circ \cdot \frac{\pi}{180^\circ} = \frac{150\pi}{180} = \frac{5\pi}{6}
$$
Trigonometry Practice
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Triangle area formulas
If two sides and the included angle are known:
$$
\text{Area} = \frac{1}{2}ab\sin C
$$
This can be combined with the law of sines or cosines.
Difficulty
Direct
1.2Convert 7 Pi Over 6 to Degrees
Convert $\frac{7\pi}{6}$ to degrees.
Solution
Multiply by $\frac{180^\circ}{\pi}$:
$$
\frac{7\pi}{6} \cdot \frac{180^\circ}{\pi} = 7 \cdot 30^\circ = 210^\circ
$$
1.3Read a Unit Circle Coordinate
On the unit circle, what point corresponds to the angle $\frac{\pi}{3}$?
Solution
On the unit circle, a point at angle $\theta$ has coordinates
$$
(\cos \theta, \sin \theta)
$$
For $\theta = \frac{\pi}{3}$,
$$
\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
$$
So the point is
$$
\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).
$$
1.4Evaluate Sine Using a Reference Angle
Evaluate $\sin\left(\frac{5\pi}{4}\right)$.
Solution
The angle $\frac{5\pi}{4}$ is in Quadrant III, and its reference angle is $\frac{\pi}{4}$.
Since sine is negative in Quadrant III,
$$
\sin\left(\frac{5\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}
$$
1.5Find a Trig Ratio from a Right Triangle
A right triangle has opposite side $7$ and hypotenuse $25$ relative to angle $\theta$.
Find $\sin \theta$.
Solution
Use the definition of sine:
$$
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}
$$
1.6Evaluate a Reciprocal Trig Function
If $\cos \theta = \frac{1}{2}$, what is $\sec \theta$?
Solution
Secant is the reciprocal of cosine:
$$
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{1/2} = 2
$$
1.7Evaluate a Special Angle Tangent
Evaluate $\tan\left(\frac{\pi}{6}\right)$.
Solution
From the $30^\circ$-$60^\circ$-$90^\circ$ triangle,
$$
\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$
1.8Use Periodicity to Simplify an Expression
Simplify $\sin(x + 2\pi)$.
Solution
Sine has period $2\pi$, so adding $2\pi$ does not change its value:
$$
\sin(x + 2\pi) = \sin x
$$
1.9Identify Where Tangent Is Undefined
For which values of $x$ is $\tan x$ undefined?
Solution
Tangent is undefined when cosine is zero.
That happens at
$$
x = \frac{\pi}{2} + k\pi
$$
for any integer $k$.
1.10Use the Pythagorean Identity to Find Cosine
If $\sin x = \frac{3}{5}$ and $x$ is in Quadrant I, find $\cos x$.
Solution
Use the Pythagorean identity:
$$
\sin^2 x + \cos^2 x = 1
$$
Substitute $\sin x = \frac{3}{5}$:
$$
\left(\frac{3}{5}\right)^2 + \cos^2 x = 1
$$
$$
\frac{9}{25} + \cos^2 x = 1
$$
$$
\cos^2 x = \frac{16}{25}
$$
Since $x$ is in Quadrant I, cosine is positive:
$$
\cos x = \frac{4}{5}
$$
Difficulty
Integrated
2.1Evaluate Cosine in Quadrant III
Evaluate $\cos\left(\frac{7\pi}{6}\right)$.
Solution
The angle $\frac{7\pi}{6}$ is in Quadrant III, with reference angle $\frac{\pi}{6}$.
Cosine is negative in Quadrant III, so
$$
\cos\left(\frac{7\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}
$$
2.2Find a Trig Ratio After Using the Triangle
A right triangle has legs $7$ and $24$.
If $\theta$ is opposite the $7$-unit leg, find $\sec \theta$.
Solution
First find the hypotenuse:
$$
7^2 + 24^2 = 49 + 576 = 625
$$
So the hypotenuse is
$$
\sqrt{625} = 25
$$
Now use cosine:
$$
\cos \theta = \frac{24}{25}
$$
Therefore,
$$
\sec \theta = \frac{25}{24}
$$
2.3Identify Amplitude, Period, and Midline
For
$$
y = -3\sin(2x - \frac{\pi}{3}) + 4,
$$
find the amplitude, period, and midline.
Solution
For $y = A\sin(Bx - C) + D$:
$$
\text{Amplitude} = |A| = 3
$$
$$
\text{Period} = \frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi
$$
$$
\text{Midline} = y = D = 4
$$
2.4Find a Phase Shift from Standard Form
For
$$
y = 2\cos(3x + \pi) - 1,
$$
what is the phase shift?
Solution
Rewrite the angle expression:
$$
3x + \pi = 3\left(x + \frac{\pi}{3}\right)
$$
So the graph is shifted left by $\frac{\pi}{3}$.
The phase shift is
$$
-\frac{\pi}{3}
$$
or, equivalently, $\frac{\pi}{3}$ to the left.
2.5Evaluate an Exact Value with a Sum Identity
Evaluate
$$
\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right).
$$
Solution
Use the sum identity:
$$
\sin(a+b) = \sin a \cos b + \cos a \sin b
$$
So,
$$
\sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right)
= \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right)
+ \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)
$$
Substitute the exact values:
$$
= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}
+ \frac{\sqrt{2}}{2}\cdot\frac{1}{2}
$$
$$
= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}
= \frac{\sqrt{6}+\sqrt{2}}{4}
$$
2.6Interpret an Inverse Trig Expression
Evaluate $\arcsin\left(\frac{\sqrt{3}}{2}\right)$.
Solution
The principal range of $\arcsin$ is
$$
\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$
The angle in that interval whose sine is $\frac{\sqrt{3}}{2}$ is
$$
\frac{\pi}{3}
$$
2.7Solve a Sine Equation on an Interval
Solve for $x$ on $[0, 2\pi)$:
$$
2\sin x - 1 = 0
$$
Solution
First isolate sine:
$$
2\sin x = 1
$$
$$
\sin x = \frac{1}{2}
$$
On $[0, 2\pi)$, sine equals $\frac{1}{2}$ at
$$
x = \frac{\pi}{6} \quad \text{and} \quad x = \frac{5\pi}{6}
$$
2.8Solve a Tangent Equation
Solve for all real $x$:
$$
\tan x = -1
$$
Solution
One solution is
$$
x = -\frac{\pi}{4}
$$
Because tangent has period $\pi$, the full solution set is
$$
x = -\frac{\pi}{4} + k\pi
$$
for any integer $k$.
Difficulty
Applied
3.1Find a Height from an Angle of Elevation
A drone is $20$ m horizontally from an observer.
The angle of elevation is $30^\circ$.
How high is the drone?
Solution
Use tangent:
$$
\tan 30^\circ = \frac{\text{height}}{20}
$$
So
$$
\text{height} = 20\tan 30^\circ
$$
Since $\tan 30^\circ = \frac{\sqrt{3}}{3}$,
$$
\text{height} = 20\cdot\frac{\sqrt{3}}{3} = \frac{20\sqrt{3}}{3}\text{ m}
$$
3.2Use the Law of Sines
In triangle $ABC$, $A = 30^\circ$, $B = 45^\circ$, and $a = 10$.
Find $b$.
Solution
Use the law of sines:
$$
\frac{a}{\sin A} = \frac{b}{\sin B}
$$
Substitute the known values:
$$
\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}
$$
Solve for $b$:
$$
b = 10 \cdot \frac{\sin 45^\circ}{\sin 30^\circ}
$$
$$
b = 10 \cdot \frac{\sqrt{2}/2}{1/2} = 10\sqrt{2}
$$
3.3Use the Law of Cosines
In triangle $ABC$, the sides are $a = 7$, $b = 9$, and the included angle $C = 60^\circ$.
Find $c$.
Solution
Use the law of cosines:
$$
c^2 = a^2 + b^2 - 2ab\cos C
$$
Substitute the values:
$$
c^2 = 7^2 + 9^2 - 2(7)(9)\cos 60^\circ
$$
$$
c^2 = 49 + 81 - 126\cdot\frac{1}{2}
$$
$$
c^2 = 130 - 63 = 67
$$
So
$$
c = \sqrt{67}
$$
3.4Interpret a Sinusoidal Model
The height of a tide is modeled by
$$
h(t) = 3\sin\left(\frac{\pi t}{6}\right) + 8,
$$
where $t$ is measured in hours.
What are the maximum height and the period?
Solution
The amplitude is $3$, so the maximum height is
$$
8 + 3 = 11
$$
The period is
$$
\frac{2\pi}{\pi/6} = 12
$$
So the maximum height is $11$ and the period is $12$ hours.
3.5Find the Area of an Oblique Triangle
Two sides of a triangle are $8$ and $12$ with included angle $45^\circ$.
Find the area.
Solution
Use the area formula for two sides and the included angle:
$$
\text{Area} = \frac{1}{2}ab\sin C
$$
Substitute the values:
$$
\text{Area} = \frac{1}{2}(8)(12)\sin 45^\circ
$$
$$
\text{Area} = 48\cdot\frac{\sqrt{2}}{2} = 24\sqrt{2}
$$
Difficulty
Challenge
4.1Use a Double-Angle Identity with a Given Sine Value
If $\sin x = \frac{3}{5}$ and $x$ is in Quadrant I, find $\cos(2x)$.
Solution
First find $\cos x$ using the Pythagorean identity:
$$
\cos x = \frac{4}{5}
$$
Now use the double-angle identity:
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$
Substitute:
$$
\cos(2x) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2
$$
$$
\cos(2x) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}
$$
4.2Rewrite a Product as a Sum
Rewrite $\sin(3x)\cos(x)$ using a product-to-sum identity.
Solution
Use the identity
$$
\sin a \cos b = \frac{1}{2}[\sin(a+b) + \sin(a-b)]
$$
with $a = 3x$ and $b = x$:
$$
\sin(3x)\cos(x) = \frac{1}{2}[\sin(4x) + \sin(2x)]
$$
4.3Determine the Number of Triangles in an SSA Case
In a triangle, $A = 30^\circ$, $a = 10$, and $b = 14$.
How many triangles are possible?
Solution
Use the SSA height test.
The height is
$$
h = b\sin A = 14\sin 30^\circ = 14\cdot\frac{1}{2} = 7
$$
Since
$$
h < a < b
$$
that is,
$$
7 < 10 < 14,
$$
there are two possible triangles.
4.4Evaluate a Half-Angle Exactly
Find the exact value of $\cos\left(\frac{\pi}{12}\right)$.
Solution
Use the half-angle formula:
$$
\cos^2\left(\frac{x}{2}\right) = \frac{1+\cos x}{2}
$$
Let $x = \frac{\pi}{6}$. Then $\frac{x}{2} = \frac{\pi}{12}$, so
$$
\cos^2\left(\frac{\pi}{12}\right) = \frac{1+\cos\left(\frac{\pi}{6}\right)}{2}
$$
$$
= \frac{1+\frac{\sqrt{3}}{2}}{2} = \frac{2+\sqrt{3}}{4}
$$
Since $\frac{\pi}{12}$ is in Quadrant I, cosine is positive:
$$
\cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{2+\sqrt{3}}}{2}
$$