Trigonometrical Ratios

There are six trigonometric ratios which relate the sides of a right triangle to its angles. Trigonometric ratios are generally used to calculate the unknown lengths and angles in a right triangle. The six trigonometric ratios are tabulated below: -

NOTE: - Since, Sinθ ≠ Sin×θ, Cosθ ≠ Cos×θ, etc.

Operation of Trigonometric Ratios

As we know that the trigonometric ratios are numbers like fraction, decimal or whole numbers. We can operate the operations of addition, subtraction, multiplication and division on them that we do in algebra.

Trigonometric Expressions

Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions.

Simplification and factorisation

We can simplify or factorise the trigonometric expressions as in algebra. For examples,

1. (a+b) (a−b) = a²− b²
   (SinA + CosA) (SinA − CosA) = Sin²A − Cos²A
2. (a+b)²= a²+ 2ab + b²
   (SinA + CosA)²= Sin²A + 2SinA.CosA + Cos²A
3. a³+b³= (a+b) (a²−ab+b²)
   Sin³A + Cos³A = (SinA + CosA)(Sin²A − 2SinA.CosA + Cos²A)
4. a³−b³= (a+b) (a²+ab+b²)
   Sin³A − Cos³A = (SinA − CosA) (Sin²A + SinA.CosA + Cos²A)

Relation between the Trigonometric Ratios of an Angle

Simply there are four fundamental relations of trigonometric ratios. They are,

1. Reciprocal Relation
2. Quotient Relation
3. Pythagoras Relation
4. Derived Relation
   
   

Reciprocal Relation

Reciprocal relations of trigonometric ratios are explained to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals. The reciprocals relations are given below: -

Sinθ = \(\frac{p}{h}\) and Cosecθ = \(\frac{h}{p}\)
Then, sinθ × cosecθ = \(\frac{p}{h}\) × \(\frac{h}{p}\) = 1

sinθ = \(\frac{1}{cosecθ }\), cosecθ = \(\frac{1}{sinθ }\)
Also, cosθ = \(\frac{b}{h}\) and secθ = \(\frac{h}{b}\)
Then, cosθ × secθ = \(\frac{b}{h}\) × \(\frac{h}{b}\) = 1

cosθ = \(\frac{1}{secθ}\), secθ \(\frac{1}{cosθ}\)
And, tanθ = \(\frac{p}{b}\), cotθ = \(\frac{b}{p}\)
Then tanθ × cotθ = 1

tanθ = \(\frac{1}{cotθ}\), cotθ = \(\frac{1}{tanθ}\)

Quotient Relation
We have, sinθ = \(\frac{p}{h}\), cosθ = \(\frac{b}{h}\)
Then, \(\frac{sinθ}{cosθ}\) = \(\frac{p}{h}\) × \(\frac{h}{b}\) = \(\frac{p}{b}\)
We have tanθ = \(\frac{p}{b}\)
∴ tanθ = \(\frac{sinθ}{cosθ}\)
Similarly,
cotθ = \(\frac{cosθ}{sinθ}\)

Pythagoras Relation
From the right angled triangle ABC,
CA²= AB² + BC²
or, h²= p² + b²
Dividing both sides
or, \(\frac{h^{2}}{h^{2}}\) = \(\frac{p^{2}+b^{2}}{h^{2}}\)
or, 1 = \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\)
or, \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\) = 1
or, (\(\frac{p^{2}}{h^{2}}\)) + (\(\frac{b^{2}}{h^{2}}\)) = 1
or, (sinθ)² + (cosθ)² = 1
∴ sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ
Also,
sinθ = \(\sqrt{1-cos^{2}}{θ}\)
or, cosθ = \(\sqrt{1−sin^{2}}{θ}\)

Again,
p²+ b² = h²
or,h²− p ²= b²
Dividing both sides by b²
or, \(\frac{h^{2}}{b{2}}\) - \(\frac{p^{2}}{b^{2}}\) = \(\frac{b^{2}}{b^{2}}\)
or, (\(\frac{h}{b}\))² + (\(\frac{p}{b}\))²= 1
or, (secθ)²− (tanθ)² = 1
or, sec²θ − tan²θ = 1
or, sec²θ = 1 + tan²θ
or, tan²θ = sec²θ − 1
Also,
secθ = \(\sqrt{1+tan^{2}}{θ}\)
or, tanθ = \(\sqrt{sec^{2} θ − 1}\)

Again,
p²+ b ²= h²
or, h²− b ²= p²
Dividing both sides by p²
\(\frac{h^{2}}{p{2}}\) − \(\frac{b^{2}}{p^{2}}\) = \(\frac{p^{2}}{p^{2}}\)
or, (cosecθ)²− (cotθ)² = 1
or, cosec²θ − cot²θ = 1
or, cosec²θ = 1 + cot²θ
or, cot²θ = cosec²θ − 1
Also, cotθ = \(\sqrt{cosec^{2} θ − 1}\)
or, cosecθ = \(\sqrt{1+cot^{2}}{θ}\)

Methods of proving Trigonometric Identities

To prove trigonometric identities, we may follow any one of the following: -

• Begin from the left-hand side (L.H.S.) and deduct it to the right-hand side (R.H.S.), if L.H.S. is more complex.
• Begin from R.H.S. and deduct it to L.H.S., if L.H.S. is complex.
• Reduce both the L.H.S. and the R.H.S. to the same expression if both the expression are complex.
• By transposition or cross multiplication, change the identity into the appropriate form. Then show that the new L.H.S. = the new R.H.S.
  
  

Conversion of a Trigonometric Ratios

With the use of two methods, we can express the trigonometric ratios in terms of other ratios of the same angle. The two methods are as followings: -

• Using trigonometric relations i.e Trigonometric formulae.
• Using Pythagoras theorem.

Trigonometric Ratios of some angles

As trigonometry have different angles, for those different angles trigonometric ratios also have different values. In the trigonometry the angles 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° are taken as standard angles.
We can find the trigonometric ratios of 0° and 90°. The following table shows the values of trigonometric ratios of the angles 0°, 30°, 45°, 60° and 90°.

Procedures to find the trigonometric ratios

The standard angles i.e 0°, 30°, 45°, 60°, 90° in prompt way is given below: -

Process 1: -
Put the number from 0 to 4 as follows: -

Process 2: -

Divide each number by 4

Process 3: -
Take square root of all the numbers

Process 4: -
The values which are obtained at first are the values of sine of the standard angles.

Process 5: -
Then reversing the order, the value of cosine of theb standard angles are obtained.

Process: -
After dividing the value of sine by the value of cosine, the values of tangent can be obtained: -

Complementary Angles

Two angles are called complementary when those angles are added up to 90° or the sum of two angles is 90°.
For example,
60° and 30° are complementary angles.
5° and 85° are complementary angles.
30° and 60°, 50° and 40°, (90° - θ) and θ are complementary angles.

Trigonometric Ratios of Complementary Angles

Here ABC is a right angle triangle right angled at B.
Let, ∠ABC = θ, then ∠CAB = 90° - θ
Now taking θ as an angle of reference,
AB = Perpendicular (p)
BC = Base (b)
CA = Hyptenuse (h)
Sinθ = \(\frac{p}{h}\) = \(\frac{AB}{CA}\) .............. .(i)
Cosθ = \(\frac{b}{h}\) = \(\frac{BC}{CA}\) ..............(ii)
Tanθ = \(\frac{p}{b}\) = \(\frac{AB}{BC}\) ..............(iii)
Cotθ = \(\frac{b}{p}\) = \(\frac{BC}{AB}\) ...............(iv)
Taking (90° - θ) as an angle of reference,
BC = Perpendicular (p)
AB = Base (b)
CA = Hypotenuse (h)
Now,
Sinθ(90° - θ) = \(\frac{p}{h}\) = \(\frac{BC}{CA}\) = Cosθ
Cosθ(90° - θ) = \(\frac{b}{h}\) = \(\frac{AB}{CA}\) = Sinθ
Tanθ(90° - θ) = \(\frac{p}{b}\) = \(\frac{BC}{AB}\) = Cotθ
On the other hand,
cosec(90° - θ) = secθ
sec(90° - θ) = cosecθ
cot(90° - θ) = tanθ