Trigonometric Identities: Definition, Formula, Applications

Trigonometric Identities is an interesting and important concept that help students to solve trigonometry and geometry problems and understand various mathematical properties. In Maths, an “identity” is an equation that is always true. Similarly, a trigonometric identity is a trigonometric equation that is true for every possible value of the input variable on which it is defined. In other words, if an equation involving trigonometric ratios of angles is valid for all values of the angle(s) concerned, it is called a trigonometric identity.

What are Trigonometric Identities?

An algebraic equation is called an identity if it is true for all values of the variable(s) for which expressions involved are defined. In the same way, an equation involving trigonometric ratios of an angle 0(zero) is called a trigonometric identity if it is true for all values of 0(zero), for which the given trigonometric ratios are defined.

Trigonometric Ratios

Trigonometric ratios of the angle are certain ratios of the sides of a right triangle with respect to its acute angles. There are six trigonometric ratios, they are \({\text{sine,}}\,{\text{cosine,}}\,{\text{tangent,}}\,{\text{cotangent,}}\,{\text{sectant}}\) and \({\text{cosecant}}\).

History of sine

The trigonometric ratio ‘sine’ was first used in the work “Aryabhatiyam” by Aryabhata, in \(A.D.\) \(500\). He named half-chord as “ardha-jya”, which was later shortened to “jya” or “jiva”. The Arabic translation of “Aryabhatiyam” retained the word “jiva” as it is. When the Arabic version was translated to Latin, this word was translated as “sinus” (meaning “curve”).

Soon the word “sinus”, also written as “sine” became common in the mathematical books in Europe. An Astronomy Professor Edmund Gunter \((1581-1626)\)  was the first to use the abbreviated form \(^{”}{\text{si}}{{\text{n}}^{”}}\).

What are Trigonometric Identities Formula?

• Let us consider a right triangle \(ABC\), right angled at \(B\).

Then, Pythagoras theorem \(A{B^2} + B{C^2} = A{C^2}\)

Dividing both sides of the by \(A{C^2}\), we get

\(\Rightarrow \frac{{A{B^2}}}{{A{C^2}}} + \frac{{B{C^2}}}{{A{C^2}}} = \frac{{A{C^2}}}{{A{C^2}}}\)

\( \Rightarrow {(\cos A)^2} + {(\sin A)^2} = 1\)

\( \Rightarrow {\cos ^2}A + {\sin ^2}A = 1\)

This result is true for all values of \(A\).

Hence it is a trigonometric identity.

• Let us consider a right triangle \(ABC\), right angled at \(B\).

Then, Pythagoras theorem \(A{B^2} + B{C^2} = A{C^2}\)

Dividing both sides of \(A{B^2} + B{C^2} = A{C^2}\) by \(A{B^2}\), we get

\(\frac{{A{B^2}}}{{A{B^2}}} + \frac{{B{C^2}}}{{A{B^2}}} = \frac{{A{C^2}}}{{A{B^2}}}\)

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

\( \Rightarrow 1 + {(\tan A)^2} = {(\sec A)^2}\)

This relation is true for \(A = {0^ \circ }\).

For \(A = {90^ \circ },\,\tan A\), and \(\sec A\) are not defined.

Hence above equation is true for all value of \(A\) for which the trigonometric ratios \(\tan A\) and \(\sec A\) are defined.

• Let us consider a right triangle \(ABC\), right angled at \(B\).

Then, Pythagoras theorem \(A{B^2} + B{C^2} = A{C^2}\)

Dividing both sides of \(A{B^2} + B{C^2} = A{C^2}\) by \(B{C^2}\), we get

\(\frac{{A{B^2}}}{{B{C^2}}} + \frac{{B{C^2}}}{{B{C^2}}} = \frac{{A{C^2}}}{{B{C^2}}}\)

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

\( \Rightarrow {(\cot A)^2} + 1 = {(\operatorname{cosec} A)^2}\)

\( \Rightarrow {\cot ^2}A + 1 = {\operatorname{cosec} ^2}A\)

This is an identity as it is true all values of \(A\) for which \({\rm{cot}}\,A\) and \(cosecA\) are defined, \({\rm{cosec}}\,A\) and \({\rm{cot}}\,A\) are not defined for \(A = {0^\circ }\).

Basic Trigonometric Identities

There are three basic trigonometric identities:

\({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)

\(1 + {\tan ^2}\theta  = {\sec ^2}\theta \)

\(1 + {\cot ^2}\theta  = {\operatorname{cosec} ^2}\theta \)

The above identities and some more identities obtained from the above identities by performing simple algebraic addition, subtraction is listed below for ready reference.

1. \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)
2. \({\sin ^2}\theta  = 1 – {\cos ^2}\theta \)
3. \({\cos ^2}\theta  = 1 – {\sin ^2}\theta \)
4. \(1 + {\tan ^2}\theta  = {\sec ^2}\theta \)
5. \({\sec ^2}\theta  – {\tan ^2}\theta  = 1\)
6. \({\tan ^2}\theta  = {\sec ^2}\theta  – 1\)
7. \(1 + {\cot ^2}\theta  = {\operatorname{cosec} ^2}\theta \)
8. \({\cot ^2}\theta  = {\operatorname{cosec} ^2}\theta  – 1\)
9. \({\operatorname{cosec} ^2}\theta  – {\cot ^2}\theta  = 1\)
10. \(\sec \theta  + \tan \theta  = \frac{1}{{\sec \theta  – \tan \theta }}\)
11. \(\operatorname{cosec} \theta  + \cot \theta  = \frac{1}{{\operatorname{cosec} \theta  – \cot \theta }}\)

These identities are true for any angle \(θ\) for which the trigonometric ratios are meaningful.

All Trigonometric Identities

Reciprocal Identities

\(\sin \theta  = \frac{1}{{\operatorname{cosec} \theta }}\quad \operatorname{cosec} \theta  = \frac{1}{{\sin \theta }}\)

\(\cos \theta  = \frac{1}{{\sec \theta }}\quad \sec \theta  = \frac{1}{{\cos \theta }}\)

\(\tan \theta  = \frac{1}{{\cot \theta }}\quad \cot \theta  = \frac{1}{{\tan \theta }}\)

Cofunction Identities (Trigonometric Ratios of complementary angles)

\({\rm{sin}}\,\theta = {\rm{cos}}\,\left( {90^\circ – \theta } \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{cos}}\,\theta = {\rm{sin}}\left( {90^\circ – \theta } \right)\)

\({\rm{sec}}\,\theta = {\rm{cosec}}\,\left( {90^\circ – \theta } \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{cosec}}\,\theta = {\rm{sec}}\left( {90^\circ – \theta } \right)\)

\({\rm{tan}}\,\theta = {\rm{cot}}\,\left( {90^\circ – \theta } \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{cot}}\,\theta = {\rm{tan}}\left( {90^\circ – \theta } \right)\)

Quotient Identities

\(\tan \theta  = \frac{{\sin \theta }}{{\cos \theta }}\quad \cot \theta  = \frac{{\cos \theta }}{{\sin \theta }}\)

Pythagoras (square) Identities

\({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)

\(1 + {\tan ^2}\theta  = {\sec ^2}\theta \)

\(1 + {\cot ^2}\theta  = {\operatorname{cosec} ^2}\theta \)

Each of the relations given above; viz. reciprocal relations, quotient relations and square relations; is a trigonometrical identity.

Trigonometric Identities Table

This table expressing each trigonometric ratio in terms of other ratios is given below:

General Working Rule to Prove a Trigonometric Identity

1. To prove a trigonometric identity, we proceed either from the left-hand side or the right-hand side whichever is convenient and simplify it in such a manner so as to get the expression on the other side.
2. Sometimes, we simplify both the left-hand side and the right-hand side of the identity so that the same expression is obtained for both sides and thus equality between the left-hand side and the right-hand side is established.
3. In some cases, we may simplify the given identity with the condition if and only if between the results of each step and end the problem with some known result.
4. Sometimes we transfer some terms from the left side of identity to the right side or from the right side to the left side and prove the new identity obtained.
5. It is convenient to express \(\tan \theta ,\cot \theta ,\sec \theta \) and \(\operatorname{cosec} \theta \) in terms of \(\sin \theta \) and \(\cos \theta \).
6. If \(1 + \sin \theta ,1 – \sin \theta ,1 + \cos \theta ,1 – \cos \theta \) occur under square root sign in the denominator, then multiply numerator and denominator by \(1 + \sin \theta ,1 – \sin \theta ,1 + \cos \theta ,1 – \cos \theta \) respectively as the case may be.

Problems based on direct use of trigonometric and algebraic formula

Working rule:

1. Use the trigonometric formulae given in the table whichever are required.

2. Use the following algebraic formulae whichever are required:

1. \({(a + b)^2} = {a^2} + {b^2} + 2ab\)
2. \({(a – b)^2} = {a^2} + {b^2} – 2ab\)
3. \({a^2} – {b^2} = (a + b)(a – b)\)
4. \({a^2} + {b^2} = {(a + b)^2} – 2ab = {(a – b)^2} + 2ab\)
5. \({a^3} + {b^3} = (a + b)\left({{a^2} + {b^2} – ab}\right);\,{a^3} + {b^3} = {(a + b)^3} + 3ab(a – b)\)
6. \({a^3} – {b^3} = (a – b)\left({{a^2} + {b^2} + ab}\right);{a^3} – {b^3} = {(a – b)^3} + 3ab(a – b)\)

Applications of Trigonometric Identities

Trigonometric Identities have been used by astronomers in the early days to find the distances of stars and planets from the Earth. This is also used, even today, for most of the advanced methods in Engineering and Physical Sciences.

1. In music theory, trigonometric identities are applicable in the field of music for stringed instruments.
   For example, the music waves of a violin possess the same shape as a sine wave.
2. The trigonometric functions like sine and cosine are used to describe the sound waves and light waves.
3. Trigonometric functions are used in oceanography to calculate heights of waves.
4. Trigonometric functions are used in the creation of maps.
5. Trigonometric functions are used in satellite systems.

Solved Examples – Trigonometric Identities

Q.1. Prove that \(\left({1 – {{\sin }^2}\theta } \right){\sec ^2}\theta  = 1\).
Ans:
Given: \(\left({1 – {{\sin }^2}\theta }\right){\sec ^2}\theta  = 1\)
Consider \(LHS = \left( {1 – {{\sin }^2}\theta } \right){\sec ^2}\theta \)
\( = {\cos ^2}\theta  \cdot {\sec ^2}\theta \quad \because \left({{{\cos }^2}\theta  = 1 – {{\sin }^2}\theta } \right)\)
\( = \frac{1}{{{{\sec }^2}\theta }} \times {\sec ^2}\theta  = 1\quad \because \left({{{\cos }^2}\theta  = \frac{1}{{{{\sec }^2}\theta }}} \right)\)
\( = 1 = RHS\)
Hence, proved.

Q.2. Prove that \(\left({1 + {{\tan }^2}\theta } \right){\cos ^2}\theta  = 1\).
Ans:
Given: \(\left({1 + {{\tan }^2}\theta } \right){\cos ^2}\theta  = 1\)
Consider \(LHS = \left( {1 + {{\tan }^2}\theta } \right){\cos ^2}\theta \)
\(= {\sec ^2}\theta  \cdot {\cos ^2}\theta  = 1\quad \because \left({{{\sec }^2}\theta  = 1 + {{\tan }^2}\theta }\right)\)
\(= \frac{1}{{{{\cos }^2}\theta }} \times {\cos ^2}\theta  = 1\quad \because \left({{{\sec }^2}\theta  = \frac{1}{{{{\cos }^2}\theta }}}\right)\)
\(=1=RHS\)
Hence, proved.

Q.3. Prove that: \({\tan ^4}A + {\tan ^2}A = {\sec ^4}A – {\sec ^2}A\).
Ans:
Given: \({\tan ^4}A + {\tan ^2}A = {\sec ^4}A – {\sec ^2}A\)
Consider \(LHS = {\tan ^4}A + {\tan ^2}A\)
\( = {\tan ^2}A \cdot \left({{{\tan }^2}A + 1} \right)\)
\( = \left({{{\sec }^2}A – 1} \right) \cdot {\sec ^2}A\quad \left[{\because{{\sec }^2}A = 1 + {{\tan }^2}A} \right]\)
\( ={\sec ^4}A -{\sec ^2}A = RHS\)
Hence, proved.

Q.4. Prove that: \(\operatorname{cosec} \theta (1 – \cos \theta )(\operatorname{cosec} \theta  + \cot \theta ) = 1\).
Ans:
Given: \(\operatorname{cosec} \theta (1 – \cos \theta )(\operatorname{cosec} \theta  + \cot \theta ) = 1\)
Consider \(LHS = \operatorname{cosec} \theta (1 – \cos \theta )(\operatorname{cosec} \theta  + \cot \theta )\)
\( = (\operatorname{cosec} \theta  – \operatorname{cosec} \theta  \cdot \cos \theta )(\operatorname{cosec} \theta  + \cot \theta )\)
\(= \left({\operatorname{cosec} \theta  – \frac{1}{{\sin \theta }} \cdot \cos \theta } \right)(\operatorname{cosec} \theta  + \cot \theta )\quad \left({\because \operatorname{cosec} \theta  = \frac{1}{{\sin \theta }}} \right)\)
\( = \left({\operatorname{cosec} \theta  – \frac{{\cos \theta }}{{\sin \theta }}}\right)(\operatorname{cosec} \theta  + \cot \theta )\)
\( = (\operatorname{cosec} \theta  – \cot \theta )(\operatorname{cosec} \theta  + \cot \theta )\)
\( = \left({{{\operatorname{cosec} }^2}\theta  -{{\cot }^2}\theta } \right) = 1 = RHS\)
Hence, proved.

Q.5. Prove that:  \(\frac{{\cos A}}{{1 – \tan A}} + \frac{{\sin A}}{{1 – \cot A}} = \cos A + \sin A\).
Ans:
Given: \(\frac{{\cos A}}{{1 – \tan A}} + \frac{{\sin A}}{{1 – \cot A}} = \cos A + \sin A\)
Consider \(LHS = \frac{{\cos A}}{{1 – \tan A}} + \frac{{\sin A}}{{1 – \cot A}}\)
\( = \frac{{\cos A}}{{1 – \frac{{\sin A}}{{\cos A}}}} + \frac{{\sin A}}{{1 – \frac{{\cos A}}{{\sin A}}}}\,\,\,\,\,\,\left({\because \tan A = \frac{{\sin A}}{{\cos A}},\cot A = \frac{{\cos A}}{{\sin A}}} \right)\)
\( = \frac{{\cos A}}{{\frac{{\cos A – \sin A}}{{\cos A}}}} + \frac{{\sin A}}{{\frac{{\sin A – \cos A}}{{\sin A}}}}\)
\( = \frac{{{{\cos }^2}A}}{{\cos A – \sin A}} + \frac{{{{\sin }^2}A}}{{\sin A – \cos A}}\)
\( = \frac{{{{\cos }^2}A}}{{\cos A – \sin A}} – \frac{{{{\sin }^2}A}}{{\cos A – \sin A}}\)
\( = \frac{{{{\cos }^2}A – {{\sin }^2}A}}{{\cos A – \sin A}}\)
\( = \frac{{(\cos A + \sin A)(\cos A – \sin A)}}{{\cos A – \sin A}}\)
\( = \cos A + \sin A = RHS\)
Hence, proved.

PRACTICE QUESTIONS ON TRIGONOMETRIC IDENTITIES

Frequently Asked Questions (FAQs) on Trigonometric Identities

Q.1. Who is the father of trigonometry?
Ans: The word ‘trigonometry’ is derived from the Greek word ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact, trigonometry is the study of relationships between the sides and angles of a triangle. The earliest known work on trigonometry was recorded in Egypt and Babylon.
The first trigonometric table was apparently compiled by Hipparchus, who is consequently now known as the father of trigonometry.

Q.2.What is trigonometry formulae?
Ans: Trigonometry is that branch of mathematics in which the measurement of angles and the problems are allied with angles. There are six basic trigonometric ratios used in trigonometry.
Trigonometric ratios (\(T\)-ratios) of an acute angle of a right triangle.

1. \(\sin \theta  = \frac{{{\text{ Perpendicular }}}}{{{\text{ Hypotenuse }}}}\)
2. \(\cos \theta  = \frac{{{\text{ Base }}}}{{{\text{ Hypotenuse }}}}\)
3. \(\tan \theta  = \frac{{P{\text{ erpendicular }}}}{{{\text{ Base }}}}\)
4. \(\cot \theta  = \frac{{{\text{ Base }}}}{{{\text{ Perpendicular }}}}\)
5. \(\operatorname{cosec} \theta  = \frac{{{\text{ Hypotenuse }}}}{{P{\text{ erpendicular }}}}\)
6. \(\sec \theta  = \frac{{{\text{ Hypotenuse }}}}{{{\text{ Base }}}}\)

Q.3. What are the \(6\) trigonometric identities?
Ans: The Reciprocal Identities are as follows:

1. \(\sin \theta  = \frac{1}{{\operatorname{cosec} \theta }}\)            
2. \(\operatorname{cosec} \theta  = \frac{1}{{\sin \theta }}\)
3. \(\cos \theta  = \frac{1}{{\sec \theta }}\)             
4. \(\sec \theta  = \frac{1}{{\cos \theta }}\)
5. \(\tan \theta  = \frac{1}{{\cot \theta }}\)             
6. \(\cot \theta  = \frac{1}{{\tan \theta }}\)

Q.4. What is reciprocal of \({\text{sin}}\theta \) ?
Ans: Reciprocal of \({\text{sin}}\theta \) is \(\frac{1}{{{\text{cosec}}\theta }}\), converse of this is \(\operatorname{cosec} \theta  = \frac{1}{{\sin \theta }}\) is also true.

Q.5. What is reciprocal of \({\text{tan}}\theta \)?
Ans: Reciprocal of \({\text{tan}}\theta \) is \(\frac{1}{{\cot \theta }}\), converse of this is \(\cot \theta  = \frac{1}{{\tan \theta }}\) is also true.

Q.6. What are the \(3\) trigonometric identities?
Ans: There are three basic identities:

1. \({\sin ^2}\theta  + {\cos ^2}\theta  = 1\)
2. \(1 + {\tan ^2}\theta  = {\sec ^2}\theta \)
3. \(1 + {\cot ^2}\theta  = {\operatorname{cosec} ^2}\theta \)

Q.7. Does trigonometry apply to all triangles?
Ans: Most often trigonometric functions are used with right triangles. There are some situations when they can be used for any type of triangle. Trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude (height) into two right triangles.

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