Precalculus by Richard Wright

Let someone else praise you, and not your own mouth; an outsider, and not your own lips. Proverbs 27:2 NIV

Summary: In this section, you will:

- Use right triangles to evaluate trigonometric functions.
- Use special right triangles to evaluate trigonometric functions of common angles.

SDA NAD Content Standards (2018): PC.5.1, PC.5.3

Using right triangles and measurements of angles of elevation and depression, heights of towers and buildings can be calculated. But first, an introduction to trigonometric identities.

Trigonometric identities are equations that are true for every value of the variable in the domain. Identities provide ways rewrite and simplify complex trigonometric expressions and equations.

*Reciprocal Identities*

$\begin{array}{lll}sinu=\frac{1}{cscu}& cosu=\frac{1}{secu}& tanu=\frac{1}{cotu}\\ cscu=\frac{1}{sinu}& secu=\frac{1}{cosu}& cotu=\frac{1}{tanu}\end{array}$

*Quotient Identities*

$\begin{array}{ll}tanu=\frac{sinu}{cosu}& cotu=\frac{cosu}{sinu}\end{array}$

*Pythagorean Identities*

$\begin{array}{ll}{sin}^{2}u+{cos}^{2}u=1& 1+{tan}^{2}u={sec}^{2}u\\ & {cot}^{2}u+1={csc}^{2}u\end{array}$

**Note**: sin^{2} *u* = (sin *u*)^{2}.

$\begin{array}{ll}sin\left(90\xb0-u\right)=cosu& cos\left(90\xb0-u\right)=sinu\\ tan\left(90\xb0-u\right)=cotu& cot\left(90\xb0-u\right)=tanu\\ sec\left(90\xb0-u\right)=cscu& csc\left(90\xb0-u\right)=secu\end{array}$

The reciprocal identities were already introduced in previous lessons. Refer back to lessons 4-02 and 4-03. The other identities are explained below.

On the unit circle, sin *θ* = *y* and cos *θ* = *x*. Divide these equations.

$$\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}=\frac{y}{x}=\mathrm{tan}\theta $$

Thus,

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

And, the reciprocal is also true.

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

Think of drawing a right triangle on the unit circle so that one leg is on the *x*-axis, the other leg is vertical, and the hypotenuse is a radius of the circle.

The horizontal leg is the *x*-coordinate of the point, but on the unit circle cos *θ* = *x*. The vertical leg is the *y*-coordinate of the point, but on the unit circle sin *θ* = *y*. The hypotenuse is a radius of the unit circle, so its length is 1. Apply the Pythagorean Theorem.

*x*^{2} + *y*^{2} = 1^{2}

cos^{2} *θ* + sin^{2} *θ* = 1

Divide this by either sin^{2} *θ* or cos^{2} *θ* to get the other two Pythagorean identities.

Imagine a right triangle where one acute angle measure is *u*. The two acute angles in a right triangle are complementary. If the other acute angle is *v*, see figure 3, then

*u* + *v* = 90°

Solve for *v*.

*v* = 90° − *u*

If the sides of the triangle are *a*, *b*, and *c* as in figure 3, then $\mathrm{sin}u=\frac{a}{c}$ and $\mathrm{cos}v=\frac{a}{c}$. Thus sin *u* = cos *v*. Since *v* = 90° − *u*,

sin *u* = cos (90° − *u*)

The other cofunction identities have similar logic.

Let *θ* be an acute angle and cos *θ* = 0.3. Find the values of a) sin *θ* and b) tan *θ* using trigonometric identities.

We know cosine and want to find sine, so pick an identity that has both:

sin

^{2}*θ*+ cos^{2}*θ*= 1

sin^{2}*θ*+ (0.3)^{2}= 1

sin^{2}*θ*= 0.91

sin*θ*≈ 0.9539We know cosine and want to find tangent. From part a we also know sine, so we can use a quotient identity.

$\begin{array}{c}tan\theta =\frac{sin\theta}{cos\theta}\\ tan\theta =\frac{0.9539}{0.3}\\ tan\theta \approx 3.1783\end{array}$

Let α be an acute angle and csc α = 2. Find a) sin α and b) cot α using trigonometric identities.

We know cosecant and want to find sine, so use an identity with both of those.

$\begin{array}{c}sin\alpha =\frac{1}{csc\alpha}\\ sin\alpha =\frac{1}{2}\end{array}$

We know cosecant and want to find cotangent, so use an identity with both of them.

$\begin{array}{c}{cot}^{2}\alpha +1={csc}^{2}\alpha \\ {cot}^{2}\alpha +1={2}^{2}\\ cot\alpha =\sqrt{3}\end{array}$

Let *β* be an acute angle and tan *β* = ½. Find a) cot *β* and b) sec *β*.

2; $\frac{\sqrt{5}}{2}$

Some real-world problems can be solved by drawing right triangles and finding unknown lengths. Other problems use angles of elevation and depression.

The angle of elevation is the angle between the horizontal up to an object.

The angle of depression is the angle between the horizontal down to an object.

A 8-foot step ladder actually is not 8-feet high. The size of a step ladder is actually the length of the rails that the steps are attached to. When the ladder is in use the rails are slanted so the height is less. If the rails of an 8-foot step ladder make an angle of 50° with the ground, how high is the top of the ladder from the ground?

Draw a right triangle using the rails of the ladder as the hypotenuse. The hypotenuse is 8 feet long and the angle with the ground is 50°.

The hypotenuse and one angle are known and the opposite leg is the unknown. The formula of sine from lesson 4-03 has those three parts. Use sine to solve the problem.

$\begin{array}{c}sinL=\frac{opp}{hyp}\\ sin50\xb0=\frac{h}{8}\\ 8sin50\xb0=h\\ h\approx 6.12\end{array}$

The height of the ladder is actually about 6.12 feet.

Find the length of side *a* in figure 6.

8.57

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of elevation of 57° between a line of sight to the top of the tree and the ground. Find the height of the tree.

The adjacent leg and angle are known. The height is the unknown. Use a trigonometric formula that has those three quantities.

$\begin{array}{c}tan\theta =\frac{opp}{adj}\\ tan57\xb0=\frac{h}{30}\\ 30tan57\xb0=h\\ h\approx 46.20\end{array}$

The height of the tree is about 46 feet.

A hiker is standing on mesa 80 feet above the desert floor. The angle of depression to a creek is 50°. Farther away, the angle of depression to a horse is 30°. How far apart are the horse and the creek?

Draw a picture to visualize the situation.

This will be a two part problem. First, since $\stackrel{\u203e}{\mathrm{AG}}$ and $\stackrel{\u203e}{\mathrm{DF}}$ are parallel, then ∠α and ∠ADF are congruent. So are ∠β and ∠AEF.

Start by finding the length of DF using ΔADF.

$\begin{array}{c}tanu=\frac{opp}{adj}\\ tan30\xb0=\frac{80}{DF}\\ DF=\frac{80}{tan30\xb0}\\ DF\approx 138.56\end{array}$

Now find the length of EF using ΔAEF.

$\begin{array}{c}tanu=\frac{opp}{adj}\\ tan50\xb0=\frac{80}{EF}\\ EF=\frac{80}{tan50\xb0}\\ EF\approx 67.13\end{array}$

Now subtract to find *x*.

*x* = DF − EF

*x* = 138.56 − 67.13 = 71.43

The horse is approximately 71 feet from the creek.

A student is observing a radio tower. Two sets of guy wires are attached at the same spot on the ground 50 feet away from the tower. The other ends of the guy wires are attached to the tower at different points. If the angles of elevation of the guy wires are 50° and 70°, how far apart are they attached on the tower?

77.8 ft

*Reciprocal Identities*

$\begin{array}{lll}sinu=\frac{1}{cscu}& cosu=\frac{1}{secu}& tanu=\frac{1}{cotu}\\ cscu=\frac{1}{sinu}& secu=\frac{1}{cosu}& cotu=\frac{1}{tanu}\end{array}$

*Quotient Identities*

$\begin{array}{ll}tanu=\frac{sinu}{cosu}& cotu=\frac{cosu}{sinu}\end{array}$

*Pythagorean Identities*

$\begin{array}{ll}{sin}^{2}u+{cos}^{2}u=1& 1+{tan}^{2}u={sec}^{2}u\\ & {cot}^{2}u+1={csc}^{2}u\end{array}$

**Note**: sin^{2} *u* = (sin *u*)^{2}.

$\begin{array}{ll}sin\left(90\xb0-u\right)=cosu& cos\left(90\xb0-u\right)=sinu\\ tan\left(90\xb0-u\right)=cotu& cot\left(90\xb0-u\right)=tanu\\ sec\left(90\xb0-u\right)=cscu& csc\left(90\xb0-u\right)=secu\end{array}$

The angle of elevation is the angle between the horizontal up to an object.

The angle of depression is the angle between the horizontal down to an object.

Helpful videos about this lesson.

- Explain the cofunction identity.
- If sin
*θ*= 0.9, find a) cos*θ*and b) csc*θ*. - If sin
*θ*= 0.25, find a) sin(90° −*θ*) and b) tan*θ*. - If sec
*θ*= 1.45, find a) cos*θ*and b) tan*θ*. - If cos
*θ*= 0.6, find a) sin*θ*and b) cot*θ*. - If csc
*θ*= 10, find a) sin*θ*and b) csc (90° −*θ*). - A 20-ft ladder leans against a building so that the angle between the ladder and the ground is 75°. How high up the building does the ladder reach?
- A 30-ft ladder leans against a building so that the angle between the ladder and the ground is 70°. How far from the building is the base of the ladder?
- The angle of elevation to the top of the Willis Tower is 33.2° when you are a half-mile from the base of the tower. How high is the tower?
- *If the Empire State Building is 1250 ft high and the angle of the elevation to the top is 52°, how far from the building are you?
- A group of civil engineers wants to build a bridge over a canyon, but they do not know how wide the canyon is. They raise different tall objects up beside the canyon until one of them casts a shadow to the other side of the canyon. The height of the object is 80 ft and they estimate the angle of elevation of the sun is 35°. Roughly, how wide is the canyon? (Ben P)
- A tall pine tree grows vertically. If Sam is 50 feet from the tree and measures the angle of elevation as 80°, how tall is the tree?
- A large advertising banner hangs on the side of a building. Duane works in a neighboring building 75 feet away and measures to angle of elevation to the top of the banner as 50° and the angle of depression to the bottom as 20°. How long is the banner?
- Marie is standing on a platform waiting to ride a roller coaster. She measures the angle of depression to the bottom of the long hill as 13° and the angle of elevation to the top of the hill as 52°. If she is 110 feet away, how high is the hill?
- A steeple is on top of a church. Marco stands 52 ft from the church and measures the angle of elevation to the base of the steeple as 44°. He measures the angle of elevation to the top of the steeple as 56°. How tall is the steeple?
- Philip is standing on Inspiration Point in Arcadia Scenic Turnout 800 feet above Lake Michigan. He can see two ship, one behind the other. If the angle of depression to the closer ship is 18° and the farther ship is 15°, how far apart are the ships?
- (4-03) Use a special right triangle to evaluate a) tan 30° and b) $sec\frac{\pi}{4}$.
- (4-03) Evaluate the six trigonometric functions for the given angle.

- (4-02) Evaluate the six trigonometric functions for $\frac{4\pi}{3}$ using the unit circle.
- (4-01) A car with a 30-inch diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min. How many revolutions per minute do the wheels make?
- (3-02) What is the intensity of a loud stereo blaring music at 95 dB?

Let *θ* be an acute angle. Use the given function value with trigonometric identities to evaluate the given function.

Problem Solving

Mixed Review

- For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.
- 0.4359; 10⁄9
- 0.9682; 0.2582
- 0.6897; 1.05
- 0.8; 0.75
- 0.1; 1.0050
- 19.3 ft
- 10.3 ft
- 1728 ft
- 977 ft
- 114 ft
- 284 ft
- 116.7 ft
- 166.2 ft
- 26.9 ft
- 523 ft
- $\frac{\sqrt{3}}{3}$; \(\sqrt{2}\)
- $sin\theta =\frac{3\sqrt{34}}{34},cos\theta =\frac{5\sqrt{34}}{34},tan\theta =\frac{3}{5},csc\theta =\frac{\sqrt{34}}{3},sec\theta =\frac{\sqrt{34}}{5},cot\theta =\frac{5}{3}$
- $sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2},cos\frac{4\pi}{3}=-\frac{1}{2},tan\frac{4\pi}{3}=\sqrt{3},csc\frac{4\pi}{3}=-\frac{2\sqrt{3}}{3},sec\frac{4\pi}{3}=-2,cot\frac{4\pi}{3}=\frac{\sqrt{3}}{3}$
- 3520 rad/min; 560.2 rev/min
- 0.00316 W/m
^{2}