Trigonometry Formulas
Radian and Degree Measures
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 CircleThe formulas to find the trigonometric function values using the coordinate of P (x,y) are as follows:

Sin $\theta$ = y
Cos $\theta$ = x
Tan $\theta$ = $\frac{y}{x}$
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

Sin^{2} $\theta$ + Cos^{2} $\theta$ = 1
1 + tan^{2} $\theta$ = Sec^{2} $\theta$
1 + Cot^{2} $\theta$ = Csc^{2} $\theta$ 
Double Angle Formula
 Sin 2$\theta$ = 2 Sin$\theta$.Cos$\theta$
Cos 2$\theta$ = Cos^{2} $\theta$  Sin^{2} $\theta$
Tan 2$\theta$ = $\frac{2tan\theta }{1tan^{2}\theta }$ 
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
a
^{2} = b
^{2} + c
^{2} 2bc.CosA
b
^{2} = c
^{2} + a
^{2}  2ca.CosB
c
^{2} = a
^{2} + b
^{2}  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
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}$$tan$ 38 =
$\frac{h tan 42}{h+1000 tan 42}$h tan 38 + 1000 tan 38 tan 42 = h tan 42 Cross multiplication
h(tan 42  tan 38) = 1000 tan 38 tan 42
h =
$\frac{1000 tan 38 tan 42}{tan 42tan 38}$ ≈ 5905.66 ft
The height of the mountain ≈ 5906 ft (rounded to the nearest feet).
Now let us solve a word problem applying Sine Rule
A Pilot during the course of flight, felt that the fuel may not be sufficient to continue the flight in the course of the destination. He went 15º off course to land in an Air Port which was at a distance of 80 miles from where he deviated. After filling the fuel, the plane headed toward the destination. The course of flight to the destination now formed an angle of 160º with the direction the plane came to land for fueling. Find the extra distance traveled for the purpose of fueling to the nearest mile.
The sketch for the situation looks like this.
AB is the original direction of the flight. AC is the changed direction for the purpose of fueling. CB is the second phase of flight after fueling.
The extra miles traveled = (AC + CB)  AB
Since measures of two angles and the length of one side are given, we can apply sine rule to find CB and AB.
m $\angle$B = 180  (m $\angle$A + m $\angle$C) = 180  (15 + 160) = 5º
$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}$$\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 + cot
^{2} $\theta$  cos
^{2} $\theta$  cos
^{2} $\theta$.cot
^{2} $\theta$ = 1 (Hint: Factor and then use Pythagorean identities).
2. Solve: 2 Sec
^{2} $\theta$  tan
^{4} $\theta$ = 1 for all real values of $\theta$
(Answer: $\theta$ = $n\pi \pm$
$\frac{\pi }{3}$) Hint: Use the Pythagorean identity to rewrite Sec
^{2} $\theta$.)
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)