Vachmi
What is Trigonometry
Trigonometry is a study of relationship between the lengths of sides and angles of a triangle.It derives its name from Greek "trigonon" (triangle) and "metron" (measure).
In a triangle, there are 3 sides and 3 angles. In Trigonometry we learn about the ratios of the sides of a right triangle with respect to its acute angles.
Refer right angled △ ABC.
The hypotenuse refers to the side opposite to the right angle.
The opposite refers to the side opposite to an acute angle θ ('theta').
The adjacent refers to the side adjacent (next) to an acute angle θ ('theta').
The main functions or ratios in trigonometry are sine, cosine and tangent.
For an angle θ these ratios are defined as follows:
Sine function sin θ = $\frac{opposite}{hypotenuse}$
Cosine function cos θ = $\frac{adjacent}{hypotenuse}$
Tangent function tan θ = $\frac{opposite}{adjacent}$
There are 3 more Trigonometric functions which are reciprocals of the above 3 functions.
Cosecant function csc θ = $\frac{hypotenuse}{opposite}$
(Reciprocal of Sine function)
Secant function sec θ = $\frac{hypotenuse}{adjacent}$
(Reciprocal of Cosine function)
Cotangent function cot θ = $\frac{adjacent}{opposite}$
(Reciprocal of Tangent function)
We now know what trigonometric ratios mean. Let us now try to understand why they are so important in the world of mathematics.
Let us look at the triangles ABC, PQC and RSC.
As all the 3 triangles are similar, their corresponding sides are proportional
$\dfrac{QC}{SC}$ = $\dfrac{PC}{RC}$ = $\dfrac{PQ}{RS}$
From this we can derive that $\dfrac{RS}{RC}$ = $\dfrac{PQ}{PC}$ = sin θ
Same is true for all other trigonometric ratios such as cos, tan and others.
This shows that trigonometric ratios of △PQC and △RSC are same.
The same can be proved for △ABC and △PQC are they are also similar.
As all the 3 triangles are similar, their corresponding sides are proportional
$\dfrac{QC}{SC}$ = $\dfrac{PC}{RC}$ = $\dfrac{PQ}{RS}$
From this we can derive that $\dfrac{RS}{RC}$ = $\dfrac{PQ}{PC}$ = sin θ
Same is true for all other trigonometric ratios such as cos, tan and others.
This shows that trigonometric ratios of △PQC and △RSC are same.
The same can be proved for △ABC and △PQC are they are also similar.
This leads us to a very important conclusion that the values of the trigonometric ratios
of an angle do not vary with the lengths of the sides of the triangle,
if the angle remains the same.
This forms the basis of Trigonometry.
This forms the basis of Trigonometry.
Let us find out the values of all trigonometric functions for ∠ 53°
sin θ = $\frac{3.99}{5}$ $= 0.7986$
cos θ = $\frac{3.009}{5}$ $= 0.6018$
tan θ = $\frac{3.99}{3.009}$ $= 1.3270$
csc θ = $\frac{5}{3.99}$ $= 1.2522$
sec θ = $\frac{5}{3.009}$ $= 1.6617$
cot θ = $\frac{3.009}{3.99}$ $= 0.7536$
Before we move forward, take a look at relationship between degrees and radians.
Let us try to find the height of this building without actually measuring it.
Assume that you are standing at some distance from the building at point C. Let us say that distance between you and the building is 250m (BC). Assume that you can measure the angle of elevation from the point C to the top of the building i.e.∠C. Let this angle be 53°
Now it is very easy to calculate the height of the building (AB) by using tan function.
tan 53° = $\frac{AB}{BC} = \frac{AB}{250}$
We know that tan 53° = 1.327
∴ $1.327 = \frac{AB}{250} $ or $AB = 331.75 m$
Thus we have found out the height of the building with trigonometric function without actually measuring it.
Assume that you are standing at some distance from the building at point C. Let us say that distance between you and the building is 250m (BC). Assume that you can measure the angle of elevation from the point C to the top of the building i.e.∠C. Let this angle be 53°
Now it is very easy to calculate the height of the building (AB) by using tan function.
tan 53° = $\frac{AB}{BC} = \frac{AB}{250}$
We know that tan 53° = 1.327
∴ $1.327 = \frac{AB}{250} $ or $AB = 331.75 m$
Thus we have found out the height of the building with trigonometric function without actually measuring it.
Memorizing the values of trigonometric ratios
How do we memorize the values of sin, cos and tan for so many different angles?While it's not easy and not required to remember all the values, it helps if one can remember the ratios for some standard angles such as $0°$, $30°$, $45°$, $60°$ and $90°$
And thankfully, there is an easy way of doing it!!! 😇
See the sin, cos and tan values in the table below. Do you notice any pattern?
Yes ofcourse.
• For sin values, the numerator increases from 0 to 4 while the denominator remains the same.
• For cos values, the same values repeat but from right to left.
• As tan is just sin divided by cos, you need not remember the values for tan, but rather derive from sin and cos values.
• As cosec, sec and cot are reciprocals of sin, cos and tan respectively, it is not required to remember them separately.
| $0°$ | $30°$ | $45°$ | $60°$ | $90°$ | |
|---|---|---|---|---|---|
| sin θ | $\frac{\sqrt{0}}{2}$ | $\frac{\sqrt{1}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{4}}{2}$ |
| cos θ | $\frac{\sqrt{4}}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{1}}{2}$ | $\frac{\sqrt{0}}{2}$ |
| tan θ | $0$ | $\frac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | $\frac{\sqrt{4}}{\sqrt{0}}$$\text{(N/D)}$ |
| cosec θ | $\frac{2}{\sqrt{0}}$$\text{(N/D)}$ | $\frac{2}{\sqrt{1}}$ | $\frac{2}{\sqrt{2}}$ | $\frac{2}{\sqrt{3}}$ | $\frac{2}{\sqrt{4}}$ |
| sec θ | $\frac{2}{\sqrt{4}}$ | $\frac{2}{\sqrt{3}}$ | $\frac{2}{\sqrt{2}}$ | $\frac{2}{\sqrt{1}}$ | $\frac{2}{\sqrt{0}}$$\text{(N/D)}$ |
| cot θ | $\frac{\sqrt{4}}{\sqrt{0}}$$\text{(N/D)}$ | $\sqrt{3}$ | $1$ | $\frac{1}{\sqrt{3}}$ | $0$ |
N/D ⇒ Not defined as denominator is 0
I
$sin^2 θ + cos^2 θ = 1$
II
$sec^2 θ - tan^2 θ = 1$
III
$cosec^2 θ - cot^2 θ = 1$
IV
$sin2θ = 2 sinθ cosθ$
V
$cos2θ = cos^2θ - sin^2θ $
VI
$tan2θ = \dfrac{2tanθ}{1 - tan^2 θ}$
VII
$sin3θ = 3 sinθ - 4sin^3θ$
VIII
$cos3θ = 4cos^3θ - 3cosθ $
IX
$tan3θ = \dfrac{3tanθ - tan^3θ}{1 - 3tan^2 θ}$