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Trigonometric Ratios: The sides and angles of a right-angled triangle are dealt with in Trigonometry. The crux of Trigonometry problems is that while some variables are provided, we have to find the remaining sides and angles of a triangle. This can be done by utilizing a suitable ratio of the side of a triangle with respect to its acute angle. The ratios of acute angles are called trigonometric ratios of angles. The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec).
When we delve into the etymology of the word ‘Trigonometry’, we get the Greek words ‘trigonon’ and ‘metron’. While the meaning of the word ‘trigonon’ is a triangle, ‘metron’ means to measure. Trigonometry deals with the measurement of the sides and the angles of a triangle and the Trigonometric Ratios problems allied with angles.
So, what is Trigonometric ratios definition? Ratios of the sides of a right triangle with respect to its acute angles are called trigonometric ratios of the angle.
There are six basic trigonometric ratios: sine, cosine, tangent, cotangent, secant, and cosecant
Trigonometry integrates memorisation, conceptual comprehension, and problem-solving skills and is a key component of Class 10 Mathematics. Given how frequently the natural formations of the planet resemble triangles, it aids pupils’ comprehension of the world. Want to learn this most crucial topic from the best content creator across the web? Click on the link below.
The above diagram has an acute angle \(\angle YAX = \theta \) with the initial side \(AX\) and terminal side \(AY.\) Draw \(PM\) perpendicular from \(P\) on \(AX\) to get the right-angled \(\Delta AMP\) in which \(\angle PAM = \theta .\)
In the right-angled \(\Delta AMP\) Base \( = AM\, = x,\) Perpendicular \( = PM = y\) and Hypotenuse \(AP\, = \,\,r.\)
We define the following six trigonometric ratios:
1. \({\mathop{\rm Sine}\nolimits} \,\theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Hypotenuse}}}} = \frac{y}{r},\) and is written as \(\sin \,\theta \)
2. \({\rm{Cosine}}\,\theta = \frac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}} = \frac{x}{r},\) and is written as \(\cos \,\theta \)
3. \({\rm{Tangent}}\,\theta = \frac{{{\rm{Perpendicular}}}}{{{\rm{Base}}}} = \frac{y}{x},\) and is written as \(\tan \,\theta \)
4. \({\rm{Cosecant}}\,\theta = \frac{{{\rm{Hypotenuse}}}}{{{\rm{Perpendicular}}}} = \frac{r}{y},\) and is written as \({\rm{cosec}}\,\theta \)
5. \({\rm{Secant}}\,\theta = \frac{{{\rm{Hypotenuse}}}}{{{\rm{Base}}}} = \frac{r}{x},\) and is written as \(\sec \,\theta \)
6. \({\rm{Cotangent}}\,\theta = \frac{{{\rm{Base}}}}{{{\rm{Perpendicular}}}} = \frac{x}{y},\) and is written as \(\cot \,\theta \)
Note: sin theta is an abbreviation for “sine of angle theta”; it is not the product of sin and theta. The same is the case for other trigonometric ratios.
Are there any Mnemonics to remembering these trigonometric ratios?
Yes, there is a mnemonic to remember these trigonometric ratios.
That is, Some People Have Curly Black Hair Through Proper Brushing.
Some People Have is for \({\mathop{\rm Sin}\nolimits} \,\,\theta = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Hypotenuse }}}}\)
Curly Black Hair is for \({\mathop{\rm Cos}\nolimits} \,\,\theta = \frac{{{\rm{ Base }}}}{{{\rm{ Hypotenuse }}}}\)
Through Proper Brushing is for \({\mathop{\rm Tan}\nolimits} \,\theta = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Base }}}}\)
History of \({\rm{sine}}\)
Usage of the word ‘sine’, the way we use it today, was in work Aryabhatiyam by Aryabhata, \(500\, AD.\) Aryabhata used the word ardha (half) -jya (chord) for the half-chord, which was used as jya or jiva in due course. When Aryabhatiyam was translated into Arabic, the word jiva was retained as it is. Jiva word was translated into the word sinus, which means curve when the Arabic version was translated into Latin. Soon the word sinus, also used as sine, became common in mathematical texts throughout Europe. English Professor of astronomy Edmund Gunter first abbreviated the notation \(‘\sin ‘.\)
History of \({\rm{cosine}}\) and \({\rm{tangent}}\)
The origin of the terms \({\rm{cosine}}\) and \({\rm{tangent}}\) came to the limelight much later. The \({\rm{cosine }}\) function comes to us from the need to compute the \({\rm{sine }}\) of the complementary angle. Aryabhata called it kotijya The name \({\rm{cosinus }}\) originated with Edmund Gunter. In \({\rm{1674}}\) the English Mathematician Sir Jonas Moore first used the abbreviated notation \(‘\cos ‘.\)
Students can refer to the image and table below to get a better understanding of trigonometric ratios and their functions.
In the above right angle triangle ABC, AB is the hypotenuse, BC is the side opposite to angle A, and AC is the adjacent side.
| Name of the trigonometric function | Abbreviation of the trigonometric function | Ratio of the trigonometric function |
| \({\rm{sine}}\) of \(\angle A\) | \(\sin A\) | \(\sin A = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Hypotenuse }}}}\) |
| \({\rm{cosine}}\) of \(\angle A\) | \(\cos A\) | \(\cos A = \frac{{{\rm{ Base }}}}{{{\rm{ Hypotenuse }}}}\) |
| \({\rm{tangent}}\) of \(\angle A\) | \(\tan A\) | \(\tan A = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Base }}}}\) |
| \({\rm{cosecant}}\) of \(\angle A\) | \({\rm{cosec}}\,A\) | \({\mathop{\rm cosec}\nolimits}\,A = \frac{{{\rm{Hypotenuse }}}}{{{\rm{ Perpendicular }}}}\) |
| \({\rm{secant}}\) of \(\angle A\) | \({\mathop{\rm sec}\nolimits}\,A\) | \(\sec A = \frac{{{\rm{Hypotenuse }}}}{{{\rm{ Base }}}}\) |
| \({\rm{cotangent}}\) of \(\angle A\) | \(\cot A\) | \(\cot A = \frac{{{\rm{ Base }}}}{{{\rm{ Perpendicular }}}}\) |
The three fundamental trigonometric ratios \(\sin \theta ,\cos \theta \) and \(\tan \theta \) of an angle \(\theta \) are very closely connected by relations. Thus, if any one of the ratios is known, the other two ratios can be easily calculated.
In the trigonometric ratios, for any acute angle \(\theta \), we have
1. \(\sin \,\theta = \frac{1}{{{\rm{cosec}}\,\theta }}\)
2. \({\mathop{\rm cosec}\nolimits}\,\theta = \frac{1}{{\sin \theta }}\)
3. \(\sin \,\theta \times {\rm{cosec}}\,\theta = 1\)
4. \(\cos \theta = \frac{1}{{\sec \theta }}\)
5. \(\sec \theta = \frac{1}{{\cos \theta }}\)
6. \(\cos \theta \,\, \times \,\,\sec\,\theta = 1\)
7. \(\tan \theta = \frac{1}{{\cot \theta }}\)
8. \(\cot \theta = \frac{1}{{\tan \theta }}\)
9. \(\tan \theta \times \cot \theta = 1\)
In geometry, we are already familiar with the construction of angles of \(30^\circ ,45^\circ ,60^\circ \) and \(90^\circ \) . The values of the trigonometric ratios for these angles and for \(0^\circ \) as well.
| \(\angle A\) | \({0^ \circ }\) | \({30^ \circ }\) | \({45^ \circ }\) | \({60^ \circ }\) | \({90^ \circ }\) |
| \(\sin \,A\) | \(0\) | \(\frac{1}{2}\) | \(\frac{1}{{\sqrt 2 }}\) | \(\frac{{\sqrt 3 }}{2}\) | \(1\) |
| \(\cos \,A\) | \(1\) | \(\frac{{\sqrt 3 }}{2}\) | \(\frac{1}{{\sqrt 2 }}\) | \(\frac{1}{2}\) | \(0\) |
| \(\tan \,A\) | \(0\) | \(\frac{1}{{\sqrt 3 }}\) | \(1\) | \(\sqrt 3 \) | Not Defined |
| \({\rm{cosec}}A\) | Not defined | \(2\) | \(\sqrt 2 \) | \(\frac{2}{{\sqrt 3 }}\) | \(1\) |
| \(\sec \,A\) | \(1\) | \(\frac{2}{{\sqrt 3 }}\) | \(\sqrt 2 \) | \(2\) | Not Defined |
| \(\cot \,A\) | Not defined | \(\sqrt 3 \) | \(1\) | \(\frac{1}{{\sqrt 3 }}\) | \(0\) |
Note: From the table, we can observe that as \(\angle A\) increases from \({0^ \circ }\) to \({90^ \circ }\), \(\sin A\) from \(0\) to \(1,\cos A\) decreases from \(1\) to \(0\) and \(\tan A\) increases from \(0\) to a very high value which is not defined.
1. \(\sin \,{0^ \circ }\, = \,0\)
2. \(\cos \,{0^ \circ }\, = \,1\)
3. \(\tan \,{0^ \circ }\, = 0\)
4. \(\sec \,{0^ \circ }\, = \,1\)
Note: \({\rm{cosec}}\,{0^{\rm{o}}}\) and \(\cot {0^ \circ }\) are not defined.
1. \(\sin \,{90^ \circ }\, = \,1\)
2. \(\cos \,{90^ \circ }\, = 0\)
3. \(\cos {\rm{ec}}\,{90^ \circ }\, = 1\)
4. \(\cot \,{90^ \circ } = 1\)
Note: \(\tan \,{90^ \circ }\) and \(\sec \,\,{90^ \circ }\) are not defined.
In the right triangle \(ABC,\angle CAB\) is an acute angle. Observe the position of the side \(BC\) with respect to angle \(A\). It faces \(\angle A.\) We call it the side perpendicular to angle \(A\). \(AC\) is the hypotenuse of the right-angled triangle, and the side \(AB\) is a part of \(\angle A.\) So, we call it the side base to \(\angle A.\)
1. \(\sin A = \frac{{{\rm{ perpendicular }}}}{{{\rm{ Hypotenuse }}}} = \frac{{BC}}{{AC}}\)
2. \(\cos A = \frac{{{\rm{ Base }}}}{{{\rm{ Hypotenuse }}}} = \frac{{AB}}{{AC}}\)
3. \(\tan A = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Base }}}} = \frac{{BC}}{{AB}}\)
4. \({\rm{cosec}}\,A = \frac{{{\rm{Hypotenuse}}}}{{{\rm{Perpendicular}}}} = \frac{{AC}}{{BC}}\)
5. \(\sec A = \frac{{{\rm{ Hypotenuse }}}}{{{\rm{ Base }}}} = \frac{{AC}}{{AB}}\)
6. \(\cot A = \frac{{{\rm{ Base }}}}{{{\rm{ perpendicular }}}} = \frac{{AB}}{{BC}}\)
Note: The value of the trigonometric ratio of an angle does not vary with the length of the side of the triangle if the angle remains the same.
Two angles are said to be allied when their sum of angles or difference of angles is either zero or a multiple of \({90^{\rm{o}}}.\) For example, the angles\({90^ \circ }\, \pm \,\,\theta ,\,{180^ \circ }\, \pm \,\theta ,{270^ \circ }\, + \,\theta ,{360^ \circ }\, – \,\theta \) etc., are angles allied to the angle \(\theta ,\)if \(\theta ,\) is measured in degrees.
Two angles are complementary if their sum equals \({90^ \circ }\)
Thus, \(\theta \) and \({90^ \circ } – \theta \) are complementary angles.
1. \(\sin \,\left( {{{90}^{\rm{o}}} – A} \right) = \cos \,A\)
2. \(\cos \,\left( {{{90}^{\rm{o}}} – A} \right) = \sin \,A\)
3. \(\tan \,\left( {{{90}^{\rm{o}}} – A} \right) = \cot \,A\)
4. \(\cot \,\left( {{{90}^{\rm{o}}} – A} \right) = \tan \,A\)
5. \(\sec \,\left( {{{90}^{\rm{o}}} – A} \right) = {\rm{cosec}}\,A\)
6. \({\rm{cosec}}\,\left( {{{90}^{\rm{o}}} – A} \right) = \sec \,A\)
Early astronomers used it to find out the distances of the stars and planets from the Earth. Most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometric concepts.
Q.1. Given \(15\cot A = 8{\rm{ , }}\) find \(\sin A\)
Ans: Given \(15\cot A = 8\)
\(\Rightarrow \cot A = \frac{8}{{15}} = \frac{{{\rm{ Base }}}}{{{\rm{ Perpendicular }}}}\)
By applying Pythagoras theorem to \(\Delta ABC\). we get
\(AC = \sqrt {A{B^2} + B{C^2}} \Rightarrow AC = \sqrt {{8^2} + {{15}^2}} \)
\(\Rightarrow AC = \sqrt {289} = 17{\rm{ units }}\)
Now, \({\mathop{\rm sin}\nolimits} \,A = \frac{{{\rm{Perpendicular}}}}{{{\rm{Hypotenuse}}}} = \frac{{15}}{{17}}\)
Hence, \(\sin \,A = \frac{{15}}{{17}}.\)
Q.2. Evaluate the following: \({\rm{sin}}\,{\rm{3}}{{\rm{0}}^{ \circ \,}}\,\cos \,{60^ \circ }\, + \,\sin \,{60^ \circ }\,\cos \,{30^ \circ }\)
Ans : Given \({\rm{sin}}\,{\rm{3}}{{\rm{0}}^{ \circ \,}}\,\cos \,{60^ \circ }\, + \,\sin \,{60^ \circ }\,\cos \,{30^ \circ }\)
\( = \frac{1}{2} \times \frac{1}{2} + \frac{{\sqrt 3 }}{2} \times \frac{{\sqrt 3 }}{2} = \frac{1}{4} + \frac{3}{4} = 1\)
Hence, the value of \(\sin \,{30^{\rm{o}}}\,\cos \,{60^{\rm{o}}} + \sin \,{60^{\rm{o}}}\,\cos \,{30^{\rm{o}}}\) is \(1.\)
Q.3. Evaluate: \(\frac{{\cot \,{{35}^{\rm{o}}}}}{{\tan \,{{55}^{\rm{o}}}}}\)
Ans. Given \(\frac{{\cot \,\,{{35}^ \circ }}}{{{\rm{tantan}}\,\,{{55}^ \circ }\,}}\)
We know that \(\cot \,\,\theta = \,\tan \,\left( {{{90}^ \circ } – \,\theta } \right)\)
So, \(\cot {35^ \circ }\, = \,\tan \,({95^ \circ }\, – \,{35^ \circ })\)
Hence, the value of \(\frac{{\cot \,{{35}^{\rm{o}}}}}{{\tan \,{{55}^{\rm{o}}}}}\) is \(1.\)
Q.4. Express \(\cos \,\,{85^ \circ }\, + \sin \,{75^ \circ }\) in terms of trigonometric ratios of angles between \({0^ \circ }\) and \({45^ \circ }\)
Ans: Given \(\cos \,\,{85^ \circ }\, + \sin \,{75^ \circ }\)
We know that \(\cos \,\theta = \sin \,\left( {{{90}^{\rm{o}}} – \theta } \right)\) and \(\sin \,\left( {90 – {{15}^{\rm{o}}}} \right)\)
Now, \(\cos \,\,{85^ \circ }\, + \,\sin \,{75^ \circ }\, = \,\cos \,\,(90\, – \,{5^ \circ })\, + \,\sin \,(90\,\, – \,{15^ \circ })\)\( = \,\sin \,{5^ \circ }\, + \,\cos \,{15^ \circ }\,\)
Hence, \(\cos \,\,{85^ \circ }\, + \,\sin \,{75^ \circ }\, = \sin \,\,{5^ \circ }\, + \,\cos \,\,{15^ \circ }\)
Q.5. In \(\Delta \,ABC\),right-angled at \(B,\) \(AB\, = \,5\,{\rm{cm}}\) and \(\angle ACB = {30^ \circ }\) Determine the lengths of the sides \(BC\)
Ans : Given \(\Delta \,ABC\) right-angled at \(B,AB\, = \,5\,{\rm{cm}}\) and \(ABC\, = {30^ \circ }\)
Now, \(\tan \,C = \frac{{AB}}{{BC}}\)
\(\Rightarrow \frac{5}{{BC}}\, = \,\tan \,{30^ \circ }\, = \frac{1}{{\sqrt 3 }}\)
\(\Rightarrow BC\, = \,5\sqrt 3 \,{\rm{cm}}\)
Hence, \(BC\, = \,5\sqrt 3 \,{\rm{cm}}\)
In this article, we learned about the definition of trigonometric ratios, trigonometric ratios formulas, trigonometric ratios table, trigonometric ratios of special angles, trigonometric ratios in right triangles, trigonometric ratios of allied angles, trigonometric ratios of complementary angles, and applications of trigonometry ratios.
Let’s look at some commonly asked questions about Trigonometric Ratios below:
Q.1. Can there be \({\rm{8}}\) trigonometric ratios?
Ans: There are six basic trigonometric ratios, \({\rm{sine, cosine, tangent, cosecant, secant }}\) and \({\rm{cotangent}}\) From these six basic ratios one can derive many other ratios.
Q.2. What are the \(3\) basic trigonometric ratios?
Ans: The three basic trigonometric ratios are \({\rm{sine,\,cosine \, \,and \, \,tangent }}\)
\(\sin \theta = \frac{{{\rm{Perpendicular }}}}{{{\rm{Hypotenuse }}}},\cos \theta = \frac{{{\rm{ Base }}}}{{{\rm{ Hypotenuse\,}}}}{\rm{ and\,}}\tan \theta = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Base }}}}\)
Q.3. How do you solve trigonometric ratios?
Ans: (a) Choose either by determining which side you know and which side you are looking for.
(b) Substitute your information into the trigonometric ratios.
(c) Solve the resulting equations to find the lengths of the sides.
Q.4. What are the formulae of trigonometry?
Ans: Six trigonometric ratios:
1.\(\sin A = \frac{{{\rm{ Perpendicular }}}}{{{\rm{ Hypotenuse }}}} = \frac{{BC}}{{AC}}\)
2. \(\cos A = \frac{{{\rm{ Base }}}}{{{\rm{Hypotenuse }}}} = \frac{{AB}}{{AC}}\)
3. \(\tan A = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Base }}}} = \frac{{BC}}{{AB}}\)
4. \({\mathop{\rm cosec}\nolimits}\,A = \frac{{{\rm{Hypotenuse}}}}{{{\rm{Perpendicular}}}} = \frac{{AC}}{{BC}}\)
5. \(\sec A = \frac{{{\rm{ Hypotenuse }}}}{{{\rm{ Base }}}} = \frac{{AC}}{{AB}}\)
6. \(\cot A = \frac{{{\rm{ Base }}}}{{{\rm{ Perpendicular }}}} = \frac{{AB}}{{BC}}\)
Relations between trigonometric ratios:
1.\({\rm{sin}}\,\theta = \frac{1}{{{\rm{cosec}}\,\theta }}\)
2.\({\rm{cosec}}\,\theta = \frac{1}{{\sin \theta }}\)
3.\({\rm{sin}}\,\theta \cdot {\rm{cosec}}\,\theta = 1\)
4.\({\rm{cos}}\,\theta \, = \frac{1}{{\sec \,\theta \,}}\)
5.\(\sec \,\theta \, = \,\frac{1}{{\cos \,\theta }}\)
6.\(\cos \,\theta \,.\,\sec \,\theta \, = \,1\)
7.\(\tan \,\theta = \frac{1}{{\cot \,\theta \,}}\)
8.\(\cot \,\theta = \frac{1}{{\tan \,\theta }}\)
9.\(\tan\,\theta .\,\cot\,\theta \, = 1\)
Trigonometric ratios of complementary angles:
1. \(\sin \,\left( {{{90}^{\rm{o}}} – A} \right) = \cos \,A\)
2. \(\cos \,\left( {{{90}^{\rm{o}}} – A} \right) = \sin \,A\)
3. \(\tan \,\left( {{{90}^{\rm{o}}} – A} \right) = \cot \,A\)
4. \(\cot \,\left( {{{90}^{\rm{o}}} – A} \right) = \tan \,A\)
5. \(\sec \,\left( {{{90}^{\rm{o}}} – A} \right) = {\rm{cosec}}\,A\)
6. \({\rm{cosec}}\,\left( {{{90}^{\rm{o}}} – A} \right) = \sec \,A\)
Q.5. What are the six trigonometric ratios?
Ans: Trigonometric table:
| Name of the trigonometric function | Abbreviation of the trigonometric function | Ratio of the trigonometric function |
| \({\rm{sine\,of }}\,\angle A\) | \(\sin A\) | \(\sin A = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Hypotenuse }}}}\) |
| \({\rm{cosine \, of }}\,\angle A\) | \(\cos A\) | \(\cos A = \frac{{{\rm{ Base }}}}{{{\rm{ Hypotenuse }}}}\) |
| \({\rm{tangent \,of }}\,\angle A\) | \(\tan A\) | \(\tan A = \frac{{{\rm{Perpendicular }}}}{{{\rm{ Base }}}}\) |
| \({\rm{cosecant\,of }}\,\angle A\) | \({\rm{cosec}}\,A\) | \({\mathop{\rm cosec}\nolimits}\,A = \frac{{{\rm{Hypotenuse }}}}{{{\rm{ Perpendicular }}}}\) |
| \({\rm{secant\,of }}\,\angle A\) | \({\mathop{\rm sec}\nolimits}\,A\) | \(\sec A = \frac{{{\rm{Hypotenuse }}}}{{{\rm{ Base }}}}\) |
| \({\rm{cotangent\,of }}\,\angle A\) | \(\cot A\) | \(\cot A = \frac{{{\rm{ Base }}}}{{{\rm{ Perpendicular }}}}\) |
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