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Math Assignment Class VIII | Square & Square Root

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  Math Assignment  Class VIII | Square & Square Root Download or Print free  assignment with answer key  for   Class  8 Squares and  Square Roots.   Important and extra questions that cover all topics of square and square root and is useful and helpful for the students. Math Assignment  Class VIII | Square & Square Root LEVEL -1

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

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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

0o

30o

45o

60o

90o

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

Sin0o

Sin30o

Sin45o

Sin60o

Sin90o

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

equation

equation

equation

equation

Derivation of cosec2θ - cot2θ = 1

Dividing equation(1) by AC2  

equation

equation

equation

equation 

Derivation of Sec2θ - tan2θ = 1

Dividing on both side by BC2

equation

equation

equation

equation

So we have the following results

equation

equation

equation

Geometrical Representation of Trigonometric Ratios with the  60o .



Let ΔABC is an equilateral triangle. Therefore its each angle is = 60o . 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= 30o 

In   ΔABD by Pythagoras theorem

equation

equation

equation

equation



Now In   ΔABD

equation

equation

equation

equation


equation


equation

Geometrical Representation of Trigonometric Ratios with the  30o .


Now In   ΔABD

equation

equation

equation

equation

equation

equation

Geometrical Representation of Trigonometric Ratios with the 45o.

In order to find the geometrical representation of trigonometric functions at 45o , 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

equation

equation

equation

equation

equation

equation

equation

equation

Angle of elevation:


Angle of Depression:

THANKS FOR YOUR VISIT
PLEASE COMMENT BELOW
TRIGONOMETRY-CBSE Mathematics


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