# Trigonometry

Sub Topics

Trigonometry is a word derived from Greek meaning "The measurement of triangles". In simple words Trigonometry is the study of angles in triangles. Trigonometric ratios when viewed as functions provide powerful tools to solve many a real world problems. As ratios defined using right triangles it has classic applications in Astronomy and Navigation. Trigonometric functions as part of Calculus, are applied in many Physical situations involving rotations and vibrations.

Trigonometry page in tutor next, provides some of the important Trigonometric formulas, and give hints on identifying the situations which can be related to the formulas and provide few examples of Trigonometric problems.

## Trigonometry Formulas

Radian measure is introduced as a Linear measure of an angle. The formulas used for conversion of degree and radian measure of angles are as follows:

$\pi$ radians = 180º

To convert degrees into radian measure multiply by $\frac{\pi }{180}$.

To convert radian into degree measure multiply by $\frac{180}{\pi }$.

Right Triangle Trigonometry

 Sin $\theta$ = $\frac{Opposite Leg}{Hypotenuse}$ = $\frac{BC}{AB}$Cos $\theta$ = $\frac{Adjacent Leg}{Hypotenuse}$ = $\frac{AC}{AB}$tan $\theta$ = $\frac{Opposite Leg}{Adjacent Leg}$ = $\frac{BC}{AC}$Csc $\theta$ = $\frac{Hypotenuse}{Opposite Leg}$ = $\frac{AB}{BC}$Sec $\theta$ = $\frac{Hypotenuse}{Adjacent Leg}$ = $\frac{AB}{AC}$Cot $\theta$ = $\frac{Adjacent Leg}{Opposite Leg}$ = $\frac{AC}{BC}$

Unit Circle

The formulas to find the trigonometric function values using the coordinate of P (x,y) are as follows:

 Sin $\theta$ = yCos $\theta$ = xTan $\theta$ = $\frac{y}{x}$Csc $\theta$ = $\frac{1}{y}$Sec $\theta$ = $\frac{1}{x}$Cot $\theta$ = $\frac{x}{y}$

Trigonometric Identities:

 Quotient Identities Tan $\theta$ = $\frac{Sin\theta }{Cos\theta }$Cot $\theta$ = $\frac{Cos\theta }{Sin\theta }$ Reciprocal identities Csc $\theta$ = $\frac{1}{Sin \theta}$Sec $\theta$ = $\frac{1}{Cos \theta}$ Tan $\theta$ = $\frac{1}{Cot \theta}$ Pythagorean identities Sin2 $\theta$ + Cos2 $\theta$ = 11 + tan2 $\theta$ = Sec2  $\theta$1 + Cot2 $\theta$ = Csc2  $\theta$ Double Angle Formula Sin 2$\theta$ = 2 Sin$\theta$.Cos$\theta$Cos 2$\theta$ = Cos2 $\theta$ - Sin2 $\theta$Tan 2$\theta$ = $\frac{2tan\theta }{1-tan^{2}\theta }$

Sum and difference rule for Cosine

Cos($\alpha$ $\pm$ $\beta$) = Cos $\alpha$ Cos $\beta$ $\pm$ Sin $\alpha$ Sin $\beta$

Sum and difference rule for sine function

Sin($\alpha$ $\pm$ $\beta$) = Sin $\alpha$ Cos $\beta$ $\pm$ Cos $\alpha$ Sin $\beta$

Sine Rule

In a triangle ABC,

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$  where a, b and c are the sides BC, AC and AB of the triangle.

Cosine Rule

a2 = b2 + c2 -2bc.CosA
b2 = c2 + a2 - 2ca.CosB
c2 = a2 + b2 - 2ab.CosA

Area of triangle ABC

$\triangle$ ABC = $\frac{1}{2}$ bc Sin A

= $\frac{1}{2}$ ca Sin B

= $\frac{1}{2}$ ab Sin C

## How to do Trigonometry

I do know the trigonometric formulas. But how do I apply these in solving problems? This is the question many students ask when they practice Trigonometry. Few hints are given below to crack Trigonometric problems.

1. If the question is about finding the values of other Trigonometric functions when one or two function values are given, draw the right triangle and label it with the given values. Use the quadrant information to include signs in the label.

2.  Practice only helps you to prove Trigonometric identities.

3.  For real life applications related to Right triangle, relate the parts of the right triangle to the situation values given in the problem. Mark the angle of elevation or depression correct. Use the appropriate trig ratio that connects, the values given and the values to be found.

4.  The situations to apply sine and cosine rules are as follows:

Two sides and an angle opposite to one of the sides are given  -  Sine Rule
Two angles and a side are given - Sine Rule
Two sides and the angle contained by them are given - Cosine Rule
Three sides of the triangle are given - Cosine Rule

5. The bearings are generally given in terms of the angle any direction make with North - South Line.  When you read N 14º E , the line needs to start from North and move toward the East to make an angle = 14º.

## Trigonometry Word Problems

Let us solve a word problem an application involving right triangle.

A surveyor wanted to estimate the height of a mountain. He first measured the angle of elevation of the mountain top from a point A as 38º. He then moved 1000 ft closer to the mountain and now the angle of elevation of the mountain top was measured as 42º.  What is the estimated height of the mountain? Give the answer rounded to the nearest feet.

The sketch for the situation is given above. Two variables h and x are used to indicate the height of the mountain and the horizontal distance x.

Using the two right triangles ACD and BCD we can solve for 'h' eliminating the variable x. We use tan ratio as the lengths of the adjacent and opposite legs are involved.

In triangle ACD,

tan 38 = opposite leg / hypotenuse = $\frac{h}{x+1000}$       ------------(1)

In triangle BCD,

tan 42 = opposite leg / hypotenuse = $\frac{h}{x}$                -------------(2)

x = $\frac{h}{tan 42}$                                                         solved for x using equation (2)

Substituting the value of x in equation (1)

$tan$ 38 = $\frac{h}{\frac{h}{tan 42}+1000}$

$tan$ 38 = $\frac{h tan 42}{h+1000 tan 42}$

h tan 38 + 1000 tan 38 tan 42 = h tan 42                                      Cross multiplication

h(tan 42 - tan 38) = 1000 tan 38 tan 42

h = $\frac{1000 tan 38 tan 42}{tan 42-tan 38}$

≈ 5905.66 ft

The height of the mountain ≈ 5906 ft (rounded to the nearest feet).

Now let us solve a word problem applying Sine Rule

A Pilot during the course of flight, felt that the fuel may not be sufficient to continue the flight in the course of the destination. He went 15º off course to land in an Air Port which was at a distance of 80 miles from where he deviated. After filling the fuel, the plane headed toward the destination. The course of flight to the destination now formed an angle of 160º with the direction the plane came to land for fueling. Find the extra distance traveled for the purpose of fueling to the nearest mile.

The sketch for the situation looks like this.

AB is the original direction of the flight. AC is the changed direction for the purpose of fueling. CB is the second phase of flight after fueling.

The extra miles traveled = (AC + CB) - AB

Since measures of two angles and the length of one side are given, we can apply sine rule to find CB and AB.

m $\angle$B = 180 - (m $\angle$A + m $\angle$C) = 180 - (15 + 160)  = 5º

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$

$\frac{BC}{Sin15}=\frac{80}{Sin5}=\frac{AB}{Sin 160}$

Thus BC = $\frac{80}{Sin5}$ $\times Sin15$ ≈ 237.57 miles

AB = $\frac{80}{Sin5}$ $\times Sin160$ ≈ 313.94 miles

The extra miles traveled = (AC + CB) - AB = (80 + 237.57) - 313.94 ≈ 3.63 miles = 4 miles rounded to the nearest mile.

## Trigonometry Practice Problems

1. Verify the identity
1 + cot2 $\theta$ - cos2 $\theta$ - cos2 $\theta$.cot2 $\theta$ = 1                  (Hint: Factor and then use Pythagorean identities).

2. Solve: 2 Sec2 $\theta$ - tan4 $\theta$ = -1  for all real values of $\theta$

(Answer:  $\theta$ = $n\pi \pm$ $\frac{\pi }{3}$)    Hint: Use the Pythagorean identity to rewrite Sec2 $\theta$.)

3. A Satellite orbits at a height of 13,000 miles above the Earth's surface. Find the angle of depression to the horizon, assuming the radius of the Earth to be 3963 miles.

(Answer 17.75º.    Hint: The line to the Horizon from the Satellite is tangential to the Earth's surface)

4. A yacht leaves a dock and heads towards its destination at a bearing of S 2.5º E. If the Yacht averages a speed of 20 Knots over the 450 nautical mile trip

a)  Find the time required for the Yacht to make the trip.
b)  How far East and South, the Yacht hast to move to reach the destination.
c)  What is the bearing of the return trip?

( Answers : (a) 22.5 hours  (b) Towards East = 449.57 miles  Towards South = 19.63 miles  (c) N 87.5º W)