A triangle in which one angle is a right angle (90 degree) is called a right triangle. The side opposite to the right angle is called hypotenuse and the remaining two sides are called the legs of the triangle. In the following figure, angle \theta is an acute angle that is, 0^\circ < \theta < 90^\circ . The laws of trigonometry are derived based on the acute angles of a right angle triangle, opposite arm, adjacent arm and hypotenuse.

Properties:

\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{OB}{OA} \\

\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{OA} \\

\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{OB}{AB} \\

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

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

\cot \theta = \frac{1}{\tan \theta} = \frac{AB}{OB} \\

The following properties can be proved from the above mentioned properties

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

\csc^2\theta = 1 + \cot^2\theta \\

\sec^2\theta = 1 + \tan^2\theta \\

Proof:

\sin^2\theta + \cos^2\theta \\

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

= \frac{OB^2}{OA^2} + \frac{AB^2}{OA^2} \\

= \frac{OB^2 + AB^2}{OA^2} \\

= \frac{OA^2}{OA^2} \: \: [From Pythagorean theorem]

= 1

Similarly we can prove other two properties

Quadrant Formula of Trigonometric Ratios: [Fig]

• In the first quadrant all (sin, cos, tan and so cosec, sec and cot) are positive
• In the second quadrant only sin and so cosec are positive, all others are negative
• In the third quadrant only tan and so cot are positive, all others are negative
• In the fourth quadrant only cos and so sec are positive, all others are negative

• If the angles are determined by addition or subtraction from integer multiples of 90^o ,

Then sin converts to cos

         cos converts to sin

         tan converts to cot

cot converts to tan

sec converts to csc

csc converts to sec

Example:

Determine the value of sin 120^o

\sin 120^\circ = \sin(90^\circ + 30^\circ) [ falls in second quadrant so positive for sin] \\

= \cos 30^\circ [angle derived from addition of integer multiple of 90, so cos] \\

= 0.866

Example:

Determine the value of cos 120^o

\cos 120^\circ = \cos(180^\circ - 60^\circ) \: \: [ falls in second quadrant so negative for cos] \\

= - \cos 60^\circ \\

= – 0.5

Complementary Angle Theorem:

Two acute angles are called complementary if their sum is a right angle. Since the sum of the angles of a triangle is 180 degree, it follows that for a right triangle, sum of the acute angles are complementary. From the above triangle OAB,

\sin O = \sin(90-A) = \cos A

Practical Example:

A tree breaks at 20 feet above the ground and the broken part makes 30^o angle with the standing part. What is the original height of the tree?         

\cos \theta = \frac{\text{adjacent arm}}{\text{hypotenuse}}

or, \cos 30^\circ = \frac{x}{h-x}

or, \frac{\sqrt{3}}{2} = \frac{x}{h-x}

or, 2x = \sqrt{3}h - \sqrt{3}x

or, 2x + 1.732x = 1.732x

or, h = \frac{3.732\times20}{1.732} = 43

Practical Example:

Suppose that you make an angle of 30^o when you are watching the top point of a building. After walking 20 feet toward the building, you make 45^o angle with the same point of the building. Determine the height of the building.          

From triangle OCB

tan 45^\circ = \frac{h}{x}

or, 1 = \frac{h}{x}

or, h = x

From triangle OAB,

tan 30^\circ = \frac{h}{x+20}

or, \frac{1}{\sqrt{3}} = \frac{h}{x+20}

or, \frac{1}{\sqrt{3}} = \frac{h}{h+20}

or, \sqrt{3}h = h + 20

or, h = 27.32 \text{feet}