Trigonometric, Identities & Formulas

Trigonometric, Identities & Formulas
Relation between angle and sides of the triangle, values of trigonometric functions at standard angles, transformation and trigonometric identities

Maths Assignment Chapter 8 Class 10

Maths Assignment Chapter 3 Class 11

Trigonometry:

It is the branch of mathematics dealing with the relations of the sides and angles of right triangles.

\[sin\theta =\frac{Perpendicular}{Hypotenuse}=\frac{P}{H}\]\[Cos\theta =\frac{Base}{Hypotenuse}=\frac{B}{H}\]\[Tan\theta =\frac{Perpendicular}{Base}=\frac{P}{B}\]\[Cot\theta =\frac{Base}{Perpendicular}=\frac{B}{P}\]\[Sec\theta =\frac{Hypotenuse}{Base}=\frac{H}{B}\]\[Cosec\theta =\frac{Hypotenuse}{Perpendicular}=\frac{H}{P}\]From these six formulas we also find that \[sin\theta =\frac{1}{cosec\theta }\Rightarrow cosec\theta =\frac{1}{sin\theta }\]\[cos\theta =\frac{1}{sec\theta }\Rightarrow sec\theta =\frac{1}{cos\theta }\]\[tan\theta =\frac{1}{cot\theta }\Rightarrow cot\theta =\frac{1}{tan\theta }\]\[Tan\theta =\frac{Sin\theta }{Cos\theta }\; \; and\; \; \; Cot\theta =\frac{Cos\theta }{Sin\theta }\] Shortcut method to learn these formulas
First of all learn this line: "Pandit Badri Parsad Har Har Bole Sona Chandi Tole" and then write these words in a tabular form as shown below

Sona

Chandi

Tole

Pandit

Badri

Parsad

Har

Har

Bole

Convert the first letter of the each word as shown below, and in the last row write the reciprocal of sinθ, cosθ, and tanθ

sin θ

cos θ

tan θ

P

B

P

H

H

B

cosec θ

sec θ

cot θ

VALUE OF THE TRIGONOMETRIC FUNCTIONS WITH STANDARD ANGLE

	0^o	30^o	45^o	60^o	90^o
Sin	0	1/2	\[\frac{1}{\sqrt{2}}\]	\[\frac{\sqrt{3}}{2}\]	1
Cos	1	\[\frac{\sqrt{3}}{2}\]	\[\frac{1}{\sqrt{2}}\]	1/2	0
Tan	0	\[\frac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	\[\infty\]
Cot	\[\infty\]	\[\sqrt{3}\]	1	\[\frac{1}{\sqrt{3}}\]	0
sec	1	\[\frac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	\[\infty\]
cosec	\[\infty\]	2	\[\sqrt{2}\]	\[\frac{2}{\sqrt{3}}\]	1

Method of finding these values

Sin0^o	Sin30^o	Sin45^o	Sin60^o	Sin90^o
0	1	2	3	4
\[\frac{0}{4}\]	\[\frac{1}{4}\]	\[\frac{2}{4}\]	\[\frac{3}{4}\]	\[\frac{4}{4}\]
0	\[\frac{1}{4}\]	\[\frac{1}{2}\]	\[\frac{3}{4}\]	1
\[\sqrt{0}\]	\[\sqrt{\frac{1}{4}}\]	\[\sqrt{\frac{1}{2}}\]	\[\sqrt{\frac{3}{4}}\]	\[\sqrt{1}\]
0	\[\frac{1}{2}\]	\[\frac{1}{\sqrt{2}}\]	\[\frac{\sqrt{3}}{2}\]	1

Steps For finding the values of sinθ

1) Write counting from 0 to 4

2) Divide all the numbers by 4 and simplify these numbers

3) Taking square root of all these numbers

4) The values we get are the values on the sin function at different standard angles

For values of other trigonometric ratios
write all these values for sinθ in the reverse order(from right to left)

For values of tanθ use the formula tanθ = sinθ/cosθ

For values the values of cotθ use cotθ = 1/tanθ

For the values of secθ use secθ = 1/cosθ

For the values of cosecθ use cosecθ = 1/sinθ

TRANSFORMATION OF TRIGONOMETRIC FUNCTIONS

sin(90 - θ) = cosθ, cos(90 - θ) = sinθ,

tan(90 - θ) = cotθ, cot(90 - θ) = tanθ,

sec(90 - θ) = cosecθ, cosec(90 - θ) = secθ

TRIGONOMETRIC IDENTITIES

Derivation of Sin²θ + cos²θ = 1

Let us take an right angled triangle ABC as shown below

In triangle ABC, by Pythagoras theorem

AB² = AC² + BC² ................ (1)

Dividing on both side by AB²

$\frac{AB^{2}}{AB^{2}}=\frac{AC^{2}}{AB^{2}}+\frac{BC^{2}}{AB^{2}}$

$1=\left ( \frac{AC}{AB} \right )^{2}+\left ( \frac{BC}{AB} \right )^{2}$

$1=\left ( \frac{P}{H} \right )^{2}+\left ( \frac{B}{H} \right )^{2}$

$1=sin^{2}\theta +cos^{2}\theta$

Derivation of cosec²θ - cot²θ = 1

Dividing equation(1) by AC²

$\frac{AB^{2}}{AC^{2}}=\frac{AC^{2}}{AC^{2}}+\frac{BC^{2}}{BC^{2}}$

$\left ( \frac{AB}{AC} \right )^{2}=1+\left ( \frac{BC}{AC} \right )^{2}$

$\left ( \frac{H}{P} \right )^{2}=1+\left ( \frac{B}{P} \right )^{2}$

$cosec^{2}\theta =1+cot^{2}\theta \Rightarrow cosec^{2}\theta-cot^{2}\theta=1$

Derivation of Sec²θ - tan²θ = 1

Dividing on both side by BC²

$\frac{AB^{2}}{BC^{2}}=\frac{AC^{2}}{BC^{2}}+\frac{BC^{2}}{BC^{2}}$

$\left(\frac{AB}{BC} \right )^{2}=\left ( \frac{AC}{BC} \right )^{2}+1$

$\left(\frac{H}{B} \right )^{2}=\left ( \frac{P}{B} \right )^{2}+1$

$sec^{2}\theta =tan^{2}\theta +1\Rightarrow sec^{2}\theta -tan^{2}\theta =1$

So we have the following results

$Sin^{2}\theta +Cos^{2}\theta = 1,\; \; \; Sin^{2}\theta=1-Cos^{2}\theta,\; \; \; Cos^{2}\theta =1-Sin^{2}\theta$

$Sec^{2}\theta -Tan^{2}\theta = 1,\; \; \; Sec^{2}\theta=1+Tan^{2}\theta,\; \; \; Tan^{2}\theta =Sec^{2}\theta-1$

$Cosec^{2}\theta -Cot^{2}\theta = 1,\; \; Cosec^{2}\theta=1+Cot^{2}\theta,\; \; \; Cot^{2}\theta = Cosec^{2}\theta-1$

Geometrical Representation of Trigonometric Ratios with the 60^o .

Let ΔABC is an equilateral triangle. Therefore its each angle is = 60^o . Draw AD⊥ BC so that AD bisect the base as well as the vertex angle A.

Let each side of ΔABC is a. That is AB = BC = CA = a and BD = a/2 and ㄥA= 30^o

In ΔABD by Pythagoras theorem

$AD^{2}=\sqrt{AB^{2}-BD^{2}}$

$AD^{2}=\sqrt{a^{2}-\left (\frac{a}{2} \right )^{2}}$

$AD^{2}=\sqrt{\frac{4a^{2}-a^{2}}{4}}$

$AD=\frac{\sqrt{3}a}{2}$

Now In ΔABD

$Sin 60^{o}=\frac{P}{H}=\frac{\frac{\sqrt{3}}{2}a}{a}=\frac{\sqrt{3}}{2}$

$Cos 60^{o}=\frac{B}{H}=\frac{a/2}{a}=\frac{1}{2}$

$tan 60^{o}=\frac{P}{B}=\frac{\frac{\sqrt{3}}{2}a}{\frac{1}{2}a}=\sqrt{3}$

$cot60^{o}=\frac{B}{P}=\frac{\frac{1}{2}a}{\frac{\sqrt{3}}{2}a}=\frac{1}{\sqrt{3}}$

$Sec 60^{o}=\frac{H}{B}=\frac{a}{a/2}=2$

$Cosec60^{o}=\frac{H}{P}=\frac{a}{\frac{\sqrt{3}}{2}a}=\frac{2}{\sqrt{3}}$

Geometrical Representation of Trigonometric Ratios with the 30^o .

Now In ΔABD

$Sin 30^{o}=\frac{P}{H}=\frac{a/2}{a}=\frac{1}{2}$

$Cos 30^{o}=\frac{B}{H}=\frac{\left ( \sqrt{3}/2 \right )a }{a}=\frac{\sqrt{3}}{2}$

$tan 30^{o}=\frac{P}{B}=\frac{a/2 }{\left ( \sqrt{3}/2 \right )a}=\frac{1}{\sqrt{3}}$

$cot 30^{o}=\frac{B}{P}=\frac{\frac{\sqrt{3}}{2}a}{\frac{1}{2}a}=\sqrt{3}$

$sec30^{o}=\frac{H}{B}=\frac{a}{\frac{\sqrt{3}}{2}a}=\frac{2}{\sqrt{3}}$

$Cosec \: 30^{o}=\frac{H}{p}=\frac{a}{a/2}=2$

Geometrical Representation of Trigonometric Ratios with the 45^o.

In order to find the geometrical representation of trigonometric functions at 45^o , we should make an isosceles right angled triangle . And then find the values with the same method as shown above.

Here let each equal side of the triangle = a

Then by Pythagoras theorem

$AB=\sqrt{BC^{2}+AC^{2}}$

$AB =\sqrt{a^{2}+a^{2}} =\sqrt{2a^{2}}=\sqrt{2}a$

$Sin45^{o} =\frac{P}{H}=\frac{a}{\sqrt{2}a} =\frac{1}{\sqrt{2}}$

$Cos 45^{o} =\frac{B}{H}=\frac{a}{\sqrt{2}a}=\frac{1}{\sqrt{2}}$

$Tan45^{o} =\frac{P}{B}=\frac{a}{a}=1$

$Cot 45^{o} =\frac{B}{P}=\frac{a}{a}=1$

$Cosec45^{o} =\frac{H}{P}=\frac{\sqrt{2}a}{a}=\sqrt{2}$

$Sec45^{o} =\frac{H}{B}=\frac{\sqrt{2}a}{a}=\sqrt{2}$

Angle of elevation:

Angle of Depression:

THANKS FOR YOUR VISIT

PLEASE COMMENT BELOW

TRIGONOMETRY-CBSE Mathematics

Search This Blog

Featured Posts

Mathematics Class 09 Lab Manual | 17 Lab Activities