Sin Cos Formulas: Trigonometric identities are essential for students to comprehend because it is a crucial part of the syllabus as well. The sides of a right-angled triangle serve as the foundation for sin and cos formulae. Along with the tan function, the fundamental trigonometric functions in trigonometry are sin and cos.

In contrast to the cosine of an angle, which corresponds to the ratio of the nearby side to the hypotenuse, the sine of an angle is the ratio of the opposite side to the hypotenuse. These are a few of the fundamental trigonometric formulas that help us comprehend the various properties of triangles. This article has provided an in-depth explanation about the formulas and trigonometric identities. Students can have a good understanding of the formulas by referring to this article. Continue reading this article to learn about the sin cos formulas and the basics to solve any trigonometry question.

Trigonometry: Definitions

Just visualise if you are on top of a tower, and at a certain distance from the foot of the tower, you find your bicycle parked. Can you tell the length between the foot of a tower and the bike parked? A right-angled triangle is imagined to be made in this condition. In such a condition, the distances or heights can be found using mathematical techniques, which come under Trigonometry’s branch.

Earlier astronomers used trigonometry to find out the distance of the stars and planets from the other planets. Most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

Trigonometric Ratios

As discussed, trigonometry is an idea to find the persisting sides and angles of a triangle when some sides and angles are given. This problem is answered by using some ratios of the sides of a triangle discussing its acute angles. These ratios of acute angles are called trigonometric ratios of angles. Let us now define numerous trigonometric ratios and develop a clear understanding of the sine and cosine rules.

Let us take a right-angled triangle \(\Delta PQR\) as shown.

Here, \(\angle RPQ\) or \(\angle P\) is an acute angle. The side \(QR\) is facing \(\angle P\). So, we say it the side opposite to \(\angle P\). \(PR\) is the hypotenuse of the \(\Delta PQR\), and the side \(PQ\) is a part of \(\angle P\). So, we say it the side adjacent to \(\angle P\). The trigonometric ratios of the angle \(P\) in the right triangle \(PQR\) are defined as follows:

• Sine of \(\angle P = \frac{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{QR}}{{PR}}\)
• Cosine of \(\angle P = \frac{{{\rm{ the\, side\, adjacent\, to\, the\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{PQ}}{{PR}}\)
• Tangent of \(\angle P = \frac{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}}{{{\rm{ the\, side\, adjacent\, to\, the\, angle\, }}P}} = \frac{{QR}}{{PQ}}\)
• Cosecant of \( = \frac{{{\rm{ hypotenuse }}}}{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}} = \frac{{PR}}{{QR}}\)
• Secant of \(\angle P = \frac{{{\rm{ hypotemuse }}}}{{{\rm{ the\, side\, adjacent\, to\, the\, angle\, }}P}} = \frac{{PR}}{{PQ}}\)
• Cotangent of \(\angle P = \frac{{{\rm{ the\, side\, adjacent\, to\, the\, angle\, }}P}}{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}} = \frac{{PQ}}{{QR}}\)

The ratios mentioned above are shortened as \(\sin \,P,\,\cos \,P,\,\tan \,P,\,{\rm{cosec}}\,P,\,\sec \,P\) and \(\cot P\) respectively. The ratios \(\cos ecP,\sec P\) and \(\cot P\) are, in particular, the reciprocals of the ratios \(\sin P,\cos P\), and \(\tan P\).

Therefore, the trigonometric ratios of an acute angle in a right-angled triangle express the correlation between the angle and length of its sides.

Trigonometric Formulas & Ratios: Table

Below is the sin, cos formula table. In this part, we will know the values of the trigonometric ratios of the angles \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }\) which are obtained using Pythagoras theorem in a right-angled triangle.Below is the table of all the values of trigonometric ratios of \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }\).

\(\angle P\)	\({0^ \circ }\)	\({30^ \circ }\)	\({45^ \circ }\)	\({60^ \circ }\)	\({90^ \circ }\)
\(\sin P\)	\(0\)	\(\frac{1}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{{\sqrt 3 }}{2}\)	\(1\)
\(\cos P\)	\(1\)	\(\frac{{\sqrt 3 }}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{1}{2}\)	\(0\)
\(\tan P\)	\(0\)	\(\frac{1}{{\sqrt 3 }}\)	\(1\)	\(\sqrt 3\)	Not defined
\({\rm{cosec}}\,P\)	Not defined	\(2\)	\(\sqrt 2 \)	\(\frac{2}{{\sqrt 3 }}\)	\(1\)
\(\sec P\)	\(1\)	\(\frac{2}{{\sqrt 3 }}\)	\(\sqrt 2 \)	\(2\)	Not defined
\(\cot P\)	Not defined	\(\sqrt 3 \)	\(1\)	\(\frac{1}{{\sqrt 3 }}\)	\(0\)

Sin, Cos: History

The first use of the motive of ‘sine’ in the way we use it today was in work Aryabhatiyam by Aryabhata, in A.D. \(500\). Aryabhata used the name Ardha-jya for the half-chord, which was shortened to jya or jiva over time. When the Aryabhatiyam was rendered into Arabic, the word jiva was retained as it is. The name jiva was rendered into the sinus, which means curve, when the Arabic model was rendered to Latin.

Soon the name sinus, also used as sine, became synonymus in mathematical texts throughout Europe. An English academician of astronomy Edmund Gunter \((1581–1626)\), first used the shortened symbol ‘sin’.

The birth of the terms ‘cosine’ and ‘tangent’ was much later. The cosine function comes to light from the need to calculate the sine of the complementary angle. Aryabhatta named it kotijya. The word cosinus derived from Edmund Gunter. In \(1674\), the English Mathematician Sir Jonas Moore first used the abbreviated notation ‘cos’.

Sin, Cos: Formula

Let us take a right-angled triangle \(\Delta PQR\) as shown.

Here, \(\angle RPQ\) or \(\angle P\) is an acute angle. The side \(QR\) is facing \(\angle P\). So, we say it the side opposite to \(\angle P\). \(PR\) is the hypotenuse of the \(\Delta PQR\), and the side \(PQ\) is a part of \(\angle P\) So, we say it the side adjacent to \(\angle P.\)

The trigonometric ratios of the angle \(P\) in the right triangle \(PQR\) are defined as follows:

• Sine of \(\angle P = \frac{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{QR}}{{PR}}\)
• Cosine of \(\angle P = \frac{{{\rm{ the\, side\, adjacent\, to\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{PQ}}{{PR}}\)

Sin and Cos: Relationship

The equation is identity when it is valid for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle(s) involved. This section will prove one trigonometric identity involving sin and cos and use it further to prove other useful trigonometric identities.

In \(\Delta PQR\), right angled at \(Q\), we have
\(P{Q^2} + Q{R^2} = P{R^2}…….(i)\)

Dividing each term of (i) by \(P{R^2}\), we get
\(\frac{{P{Q^2}}}{{P{R^2}}} + \frac{{Q{R^2}}}{{P{R^2}}} = \frac{{P{R^2}}}{{P{R^2}}}\)
\( \Rightarrow {\left( {\frac{{PQ}}{{PR}}} \right)^2} + {\left( {\frac{{QR}}{{PR}}} \right)^2} = 1\)
\( \Rightarrow {(\cos P)^2} + {(\sin P)^2} = 1\)
\( \Rightarrow {\cos ^2}P + {\sin ^2}P = 1\)

This is true for all \(P\) such that \({0^ \circ } \le P \le {90^ \circ }\) So, this is a trigonometric identity.

Sin, Cos and other Trigonometric Ratios: Relationship

In the above section, we have found the ratios of all trigonometric functions using the below right-angles triangle. Using the same, let us find out the relationship between sin cos and the other trigonometric ratios.

We know that, from the definition of \(\sin \) and \(\cos \), we have 

Sine of \(\angle P = \frac{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{QR}}{{PR}} \ldots  \ldots (i)\)

Cosine of \(\angle P = \frac{{{\rm{ the\, side\, adjacent\, to\, angle\, }}P}}{{{\rm{ hypotenuse }}}} = \frac{{PQ}}{{PR}} \ldots  \ldots (ii)\)

Also, from the definition of \(\tan ,\,{\rm{cosec}},\,\sec \) and \(\cot \) we have,

Tangent of \(\angle P = \frac{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}}{{{\rm{ the\, side\, adjacent\, to\, angle\, }}P}} = \frac{{QR}}{{PQ}} \ldots  \ldots  \ldots (iii)\)

Cosecant of \(\angle P = \frac{{{\rm{ hypotenuse }}}}{{{\rm{ the side opposite to angle }}P}} = \frac{{PR}}{{QR}} \ldots  \ldots  \ldots (iv)\)

Secant of \(\angle P = \frac{{{\rm{ hypotenuse }}}}{{{\rm{ the\, side\, adjacent\, to\, angle\, }}P}} = \frac{{RR}}{{PQ}} \ldots  \ldots  \ldots (v)\)

Cotangent of \(\angle P = \frac{{{\rm{ the\, side\, adjacent\, to\, angle\, }}P}}{{{\rm{ the\, side\, opposite\, to\, angle\, }}P}} = \frac{{PQ}}{{QR}} \ldots  \ldots .(vi)\)

We can write equation (iii) as 

\(\tan P = \frac{{QR}}{{PQ}} = \frac{{\frac{{OR}}{{PR}}}}{{\frac{{PQ}}{{PR}}}} = \frac{{\sin P}}{{\cos P}}\)

Therefore, the relationship between \(\sin ,\cos \) and \(\tan \) can be written as
\(\tan P = \frac{{\sin P}}{{\cos P}}\)

Now, from (iv), we have 

\({\mathop{\rm cosec}\nolimits} P = \frac{{PR}}{{QR}} = \frac{{\frac{1}{{QR}}}}{{PR}} = \frac{1}{{\sin P}}\)

Therefore, the relationship between \(\sin \) and \({\rm{cosec}}\) can be written as \({\mathop{\rm cosec}\nolimits} P = \frac{1}{{\sin P}}\)

Also, from equation (v), we have 

\(\sec P = \frac{{PR}}{{PQ}} = \frac{1}{{\frac{{PQ}}{{PR}}}} = \frac{1}{{\cos P}}\)

Therefore, the relationship between \(\cos \) and \(\sec \) can be written as \(\sec P = \frac{1}{{\cos P}}\)

Now, from equation (vi), we have

\(\cot P = \frac{{PQ}}{{QR}} = \frac{1}{{\frac{{OR}}{{PQ}}}} = \frac{1}{{\tan P}} = \frac{1}{{\frac{{\cos p}}{{\cos P}}}} = \frac{{\cos P}}{{\sin P}}\)

Therefore, the relationship between \(\sin ,\cos \) and \(\cot \) can be written as \(\cot P = \frac{{\cos P}}{{\sin P}}\)

Sin Cos: List of Formulas

The list of important formulas are mentioned below:

• \(\sin ( – x) = – \sin x\)
• \(\cos ( – x) = \cos x\)
• \(\cos (x + y) = \cos x\cos y – \sin x\sin y\)
• \(\cos (x – y) = \cos x\cos y + \sin x\sin y\)
• \(\cos \left( {{{90}^ \circ } – x} \right) = \sin x\)
• \(\sin \left( {{{90}^ \circ } – x} \right) = \cos x\)
• \(\sin (x + y) = \sin x\cos y + \cos x\sin y\)
• \(\sin (x – y) = \sin x\cos y – \cos x\sin y\)
• \(\cos \left( {{{90}^ \circ } + x} \right) = – \sin x\)
• \(\sin \left( {{{90}^ \circ } + x} \right) = \cos x\)
• \(\cos \left( {{{180}^ \circ } – x} \right) = – \cos x\)
• \(\sin \left( {{{180}^ \circ } – x} \right) = \sin x\)
• \(\cos \left( {{{180}^ \circ } + x} \right) = – \cos x\)
• \(\sin \left( {{{180}^ \circ } + x} \right) = \sin x\)
• \(\cos x + \cos y = 2\cos \frac{{x + y}}{2}\cos \frac{{x – y}}{2}\)
• \(\cos x – \cos y = – 2\sin \frac{{x + y}}{2}\sin \frac{{x – y}}{2}\)
• \(\sin x + \sin y = 2\sin \frac{{x + y}}{2}\cos \frac{{x – y}}{2}\)
• \(\sin x – \sin y = 2\cos \frac{{x + y}}{2}\sin \frac{{x – y}}{2}\)
• \(2\cos x\cos y = \cos (x + y) + \cos (x – y)\)
• \( – 2\sin x\sin y = \cos (x + y) – \cos (x – y)\)
• \(2\sin x\cos y = \sin (x + y) + \sin (x – y)\)
• \(2\cos x\sin y = \sin (x + y) – \sin (x – y)\)

Sin 2x Formula (Double Angle Formula)

We know that, \(\sin (x + y) = \sin x\cos y + \cos x\sin y\)

Replacing \(y\) by \(x\), we get \(\sin 2x = 2\sin x\cos x\)

Therefore, we get \(\sin 2x = 2\sin x\cos x\)

We can write, \(\sin 2x = \frac{{2\sin x\cos x}}{1} = \frac{{2\sin x\cos x}}{{{{\cos }^2}x + {{\sin }^2}x}}\)

Dividing each term by \({\cos ^2}x\) we get

\(\sin 2x = \frac{{2\tan x}}{{1 + {{\tan }^2}x}}\)

Cos 2x Formula (Double Angle Formula)

We know that, \(\cos (x + y) = \cos x\cos y – \sin x\sin y\)

Replacing \(y\) by \(x\), we get \(\cos 2x = {\cos ^2}x – {\sin ^2}x\)

\( = {\cos ^2}x – \left( {1 – {{\cos }^2}x} \right) = 2{\cos ^2}x – 1\)

So, \(\cos 2x = 2{\cos ^2}x – 1\)

Again, \(\cos 2x = {\cos ^2}x – {\sin ^2}x \Rightarrow \cos 2x = 1 – {\sin ^2}x – {\sin ^2}x = 1 – 2{\sin ^2}x\)

Also, \(\cos 2x = {\cos ^2}x – {\sin ^2}x = \frac{{{{\cos }^2}x – {{\sin }^2}x}}{{{{\cos }^2}x + {{\sin }^2}x}}\) we get

Dividing numerator and denominator by \({\cos ^2}x\) we get

\(\cos 2x = \frac{{1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}}\)

Sin 3x and Cos 3x Formulas (Triple Angle Formula)

We can write \(\sin 3x\) as \(\sin (2x + x)\)

We know that \(\sin (x + y) = \sin x\cos y + \cos x\sin y\)

So, \(\sin (2x + x) = \sin 2x\cos x + \cos 2x\sin x\)
\( = 2\sin x\cos x\cos x + \left( {1 – 2{{\sin }^2}x} \right)\sin x\)
\( = 2\sin x\left( {1 – {{\sin }^2}x} \right) + \sin x – 2{\sin ^3}x\)
\( = 2\sin x – 2{\sin ^3}x + \sin x – 2{\sin ^3}x\)
\( = 3\sin x – 4{\sin ^3}x\)

Therefore, \(\sin 3x = 3\sin x – 4{\sin ^3}x\)

Now, we can write \(\cos 3x\) as \(\cos (2x + x)\)

We know that \(\cos (x + y) = \cos x\cos y – \sin x\sin y\)

So, \(\cos (2x + x) = \cos 2x\cos x – \sin 2x\sin x\)
\( = \left( {2{{\cos }^2}x – 1} \right)\cos x – 2\sin x\cos x\sin x\)
\( = \left( {2{{\cos }^2}x – 1} \right)\cos x – 2\cos x\left( {1 – {{\cos }^2}x} \right)\)
\( = 2{\cos ^3}x – \cos x – 2\cos x + 2{\cos ^3}x\)
\( = 4{\cos ^3}x – 3\cos x\)

Therefore, \(\cos 3x = 4{\cos ^3}x – 3\cos x\)

Sin and Cos: Integration

The integration formula for \(\sin \) and \(\cos \) is given below:

• \(\int {\sin \,x} \,dx = \, – \cos \,x + c\)
• \(\int {\cos } xdx = \sin x + c\)

Here \(c\) is constant

Solved Examples: Sin Cos Formulas

Q.1. Prove that \(\frac{{\cos 7x + \cos 5x}}{{\sin 7x – \sin 5x}} = \cot x\)
Ans: We know that \(\cos x + \cos y = 2\cos \frac{{x + y}}{2}\cos \frac{{x – y}}{2}\) and \(\sin x – \sin y = 2\cos \frac{{x + y}}{2}\sin \frac{{x – y}}{2}\)
So, \(\frac{{\cos 7x + \cos 5x}}{{\sin 7x – \sin 5x}} = \frac{{2\cos \frac{{7x + 5x}}{2}\cos \frac{{7x – 5x}}{2}}}{{2\cos \frac{{7x + 5x}}{2}\sin \frac{{7x – 5x}}{2}}} = \frac{{\cos x}}{{\sin x}} = \cot x\)
Therefore, \(\frac{{\cos 7x + \cos 5x}}{{\sin 7x – \sin 5x}} = {\mathop{\rm cotx}\nolimits} \)

Q.2. Prove that \(\frac{{\sin 5x – 2\sin 3x + \sin x}}{{\cos 5x – \cos x}} = \tan x\)
Ans: We know that \(\sin x + \sin y = 2\sin \frac{{x + y}}{2}\cos \frac{{x – y}}{2}\) and \(\cos x – \cos y = – 2\sin \frac{{x + y}}{2}\sin \frac{{x – y}}{2}\)
So, \(\frac{{\sin 5x – 2\sin 3x + \sin x}}{{\cos 5x – \cos x}} = \frac{{\sin 5x + {\mathop{\rm sin}\nolimits} x – 2\sin 3x}}{{\cos 5x – \cos x}}\)
\( = \frac{{2\sin 3x\cos 2x – 2\sin 3x}}{{ – 2\sin 3x\sin 2x}} = – \frac{{\sin 3x(\cos 2x – 1)}}{{\sin 3x\sin 2x}}\)
\( = \frac{{1 – \cos 2x}}{{\sin 2x}} = \frac{{2{{\sin }^2}x}}{{2\sin x\cos x}} = \tan x\)
Therefore, \(\frac{{\sin 5x – 2\sin 3x + \sin x}}{{\cos 5x – \cos x}} = \tan x\)

Q.3. Prove that \(\frac{{\cot \,A – \cos \,A}}{{\cot \,A + \cos \,A}} = \frac{{{\rm{cosec}}\,A – 1}}{{{\rm{cosec}}\,A + 1}}\)
Ans: We know that \(\cot A = \frac{{\cos A}}{{\sin A}}\)
So, \(\frac{{\cot A – \cos A}}{{\cot A + \cos A}} = \frac{{\frac{{\cos A}}{{\sin A}} – \cos A}}{{\frac{{\cos A}}{{\sin A}} + \cos A}}\)
\( = \frac{{\cos A\left( {\frac{1}{{\sin A}} – 1} \right)}}{{\cos A\left( {\frac{1}{{\sin A}} + 1} \right)}}\)
\( = \frac{{\left( {\frac{1}{{\sin A}} – 1} \right)}}{{\left( {\frac{1}{{\sin A}} + 1} \right)}}\)
\( = \frac{{{\mathop{\rm cosec}\nolimits} A – 1}}{{{\mathop{\rm cosec}\nolimits} A + 1}}\)
Therefore, \(\frac{{\cot A – \cos A}}{{\cot A + \cos A}} = \frac{{{\mathop{\rm cosec}\nolimits} A – 1}}{{{\mathop{\rm cosec}\nolimits} A + 1}}\)

Q.4. Express the ratios \({\rm{cosA,tanA}}\) and \({\rm{secA}}\) in terms of \({\rm{sinA}}\)
Ans: We know that \({\cos ^2}A + {\sin ^2}A = 1\)
\(\Rightarrow {\cos ^2}A = 1 – {\sin ^2}A\)
\( \Rightarrow \cos A = \sqrt {1 – {{\sin }^2}A} \)
Now, \(\tan A = \frac{{\sin A}}{{\cos A}} = \frac{{\sin A}}{{\sqrt {1 – {{\sin }^2}A} }}\)
and \(\sec A = \frac{1}{{\cos A}} = \frac{1}{{\sqrt {1 – {{\sin }^2}A} }}\)

Q.5. Prove that \(\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\)
Ans: \(\sin 2x + 2\sin 4x + \sin 6x = \sin 2x + \sin 6x + 2\sin 4x\)
We Know that \(\sin x + \sin y = 2\sin \frac{{x + y}}{2}\cos \frac{{x – y}}{2}\)
So, \(\sin 2x + \sin 6x + 2\sin 4x = 2\sin \left( {\frac{{8x}}{2}} \right)\cos \left( {\frac{{4x}}{2}} \right) + 2\sin 4x\)
\( = 2\sin 4x\cos 2x + 2\sin 4x\)
\( = 2\sin 4x(\cos 2x + 1)\)
Also, \(\cos 2x = 2{\cos ^2}x – 1\)
So, \(2\sin 4x(\cos 2x + 1) = 2\sin 4x\left( {2{{\cos }^2}x – 1 + 1} \right)\)
\( = 2\sin 4x\left( {2{{\cos }^2}x} \right)\)
\( = 4\sin 4x{\cos ^2}x\)
Therefore, \(\sin 2x + 2\sin 4x + \sin 6x = 4{\cos ^2}x\sin 4x\)

Summary

In this article, we studied the definition of sine and cosine, the history of sine and cosine and formulas of sin and cos. Also, we have learnt the relationship between sin and cos with the other trigonometric ratios and the sin, cos double angle and triple angle formulas.

FAQs on Sin Cos Formulas

Frequently asked questions related to sin cos formulas are mentioned as follows:

Q.1: What is the commonly used identity between \(\sin \) and \(\cos \)?
Ans: The identity between \(\sin \) and \(\cos \) is \({\cos ^2}A + {\sin ^2}A = 1\)

Q.2: What is the \(\cos \) double angle formula?
Ans: The double angle formula for \(\cos \) is \(\cos 2x = \frac{{1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}} = {\cos ^2}x – {\sin ^2}x = 2{\cos ^2}x – 1 = 1 – 2{\sin ^2}x\)

Q.3: What is the relationship between the ratio \(\tan ,\sin \) and \(\cos \)?
Ans: The relationship between \(\sin ,\cos \) and \(\tan \) is \(\tan A = \frac{{\sin A}}{{\cos A}}\)

Q.4: Write the formula for \(2\sin a\cos b\)
Ans: \(2\sin a\cos b = \sin (a + b) + \sin (a – b)\)

Q.5: What is \(\sin 3x\) formula?
Ans: \(\sin 3x = 3\sin x – 4{\sin ^3}x\)

We hope you find this detailed article on sin cos formulas helpful. Stay tuned to Embibe for such informative articles. Happy learning!