Trigonometrical ratios of sub multiple angles: Trigonometry is the branch of geometry that measures the sides and angles. Trigonometry was developed to solve problems and measurements involving triangles. The multiple and sub-multiple angles commonly appear in trigonometric functions. The values of various sub-multiple angles cannot be found directly.

Their values can be computed by expressing each trigonometric function in its expanded form. The concept of angles and compound angles is useful in finding the trigonometric ratios of multiple and sub-multiple angles. In this article, let us learn more about formulas of multiple and sub-multiple angles and some solved examples.

Trigonometric Ratios

These six trigonometric ratios can be defined using this triangle model.

(i) \(\sin \,\theta = \frac{{AB}}{{AC}} = \frac{{{\text{ perpendicular }}}}{{{\text{ hypotenuse }}}}\)
(ii) \(\cos \,\theta = \frac{{BC}}{{AC}} = \frac{{{\text{ base }}}}{{{\text{ hypotenuse }}}}\)
(iii) \(\tan \,\theta = \frac{{AB}}{{BC}} = \frac{{{\text{ perpendicular }}}}{{{\text{ base }}}}\)
(iv) \(\operatorname{cosec} \,\theta = \frac{{AC}}{{AB}} = \frac{{{\text{ hypotenuse }}}}{{{\text{ perpendicular }}}}\)
(v) \(\sec \,\theta = \frac{{AC}}{{BC}} = \frac{{{\text{ hypotenuse }}}}{{{\text{ base }}}}\)
(vi) \(\cot \,\theta = \frac{{BC}}{{AB}} = \frac{{{\text{ base }}}}{{{\text{ perpendicular }}}}\)

Trigonometric Ratios of Special Angles

The trigonometric ratio of some specific angles is the ratio of the sides of a right-angled triangle with respect to any of its acute angles. The trigonometric ratios of some specific angles include \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }\)

For better understanding, we are given the table and the values of the other trigonometric ratios for these angles.

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Compound Angles

A compound angle is an algebraic sum of two or more angles. The sum or difference of functions in trigonometry can be solved by using the compound angle formula.

The formula for trigonometric ratios of compound angles are as follows

\(\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B\)
\(\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B\)
\(\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\)

Trigonometry studies angles and their relationship. When there is a single function or a single angle, the computation is very easy. But there are various formulae for multiples and submultiples of angles too. These multiple and sub-multiple angles formulas should rather be called identities as they hold true for all angles. These formulae are useful in solving complex trigonometric equations.

Multiple Angles

If \(A\) is an angle, then \(2A,3A,4A \ldots \) are called as multiple angles of \(A.\)
We will derive the multiple angle formulas as \(\sin ,\cos ,\) and \(\tan \) using the compound angle formula.
In this section, let us discuss the trigonometric ratios of angle \(A\) in terms of \(2A\) and \(3A.\)

Identities	Proofs
(i) \(\sin 2A = 2\sin A\cos A = \frac{{2\tan A}}{{1 + {{\tan }^2}A}}\)	Proof: Using compound angle formulae \(\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B,\) We can write \(\sin 2A = \sin \left( {A + A} \right) = \sin A\cos A + \cos A\sin A\) \( = 2\sin A\cos A\) \( = 2\frac{{\sin A}}{{\cos A}}{\cos ^2}A\) \( = \frac{{2\tan A}}{{{{\sec }^2}A}}\) \( = \frac{{2\tan A}}{{1 + {{\tan }^2}A}}\)
(ii) \(\cos 2A = {\cos ^2}A – {\sin ^2}A\) \( = 2{\cos ^2}A – 1\) \( = 1 – 2{\sin ^2}A\)	Proof: Using the compound angle formulae \(\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B,\) We can write \(\cos 2A = \cos \left( {A + A} \right) = \cos A\cos A – \sin A\sin A\) \( = {\cos ^2}A – {\sin ^2}A\) \( = {\cos ^2}A – 1 + {\cos ^2}A\) or \(1 – {\sin ^2}A – {\sin ^2}A\) \( = 2{\cos ^2}A – 1\) or \(1 – 2{\sin ^2}A\)
(iii) \(\cos 2A = \frac{{1 – {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)	Proof: We have \(\cos 2A = {\cos ^2}A – {\sin ^2}A\) \( = \frac{{{{\cos }^2}A – {{\sin }^2}A}}{1}\) \( = \frac{{{{\cos }^2}A – {{\sin }^2}A}}{{{{\cos }^2}A + {{\sin }^2}A}}\) \( = \frac{{\frac{{{{\cos }^2}A – {{\sin }^2}A}}{{{{\cos }^2}A}}}}{{\frac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\cos }^2}A}}}},A \ne \left( {2n + 1} \right)\frac{\pi }{2},n \in Z\) \( = \frac{{\frac{{{{\cos }^2}A}}{{{{\cos }^2}A}} – \frac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}{{\frac{{{{\cos }^2}A}}{{{{\cos }^2}A}} + \frac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}\) \( = \frac{{1 – {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\)
(iv) \(\tan 2A = \frac{{2\tan A}}{{1 – {{\tan }^2}A}}\)	Proof: Using the compound angle formulae \(\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\) we can write \(\tan \left( {A + A} \right) = \frac{{\left( {\tan A + \tan A} \right)}}{{\left( {1 – \tan A\tan A} \right)}}\) \( = \frac{{\left( {2\tan A} \right)}}{{\left( {1 – {{\tan }^2}A} \right)}}\)
(v) \(\sin 3A = 3\sin A – 4{\sin ^3}A\)	Proof: Using the compound angle formulae \(\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B,\) we can write \(\sin 3A = \sin \left( {2A + A} \right) = \sin 2A\cos A + \cos 2A\sin A\) \( = 2\sin A{\cos ^2}A + \left( {1 – 2{{\sin }^2}A} \right)\sin A\) [Using \(\sin 2A = 2\sin A\cos A\) and \(\cos 2A = 1 – 2{\sin ^2}A\)] \( = 2\sin A\left( {1 – {{\sin }^2}A} \right) + \left( {1 – 2{{\sin }^2}A} \right)\sin A\) \( = 3\sin A – 4{\sin ^3}A\)
(vi) \(\cos 3A = 4{\cos ^3}A – 3\cos A\)	Proof: Using the compound formulae \(\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B,\) we can write \(\cos 3A = \cos \left( {2A + A} \right) = \cos 2A\cos A – \sin 2A\sin A\) \( = \left( {2{{\cos }^2}A – 1} \right)\cos A – 2\sin A\cos A\sin A\) [Using \(\sin 2A = 2\sin A\cos A\) and \(\cos 2A = 2{\cos ^2}A – 1\)] \( = 2{\cos ^3}A – \cos A – 2\cos A\left( {1 – {{\cos }^2}A} \right)\) \( = 2{\cos ^3}A – \cos A – 2\cos A + 2\cos^3 A\) \( = 4{\cos ^3}A – 3\cos A\)
(vii) \(\tan 3A = \frac{{3\tan A – {{\tan }^3}A}}{{1 – 3{{\tan }^2}A}}\)	Proof: Using the compound angle formulae \(\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}},\) we can write \(\tan 3A = \tan \left( {2A + A} \right) = \frac{{\tan 2A + \tan A}}{{1 – \tan 2A \cdot \tan A}}\) \( = \frac{{\left( {\frac{{2\tan A}}{{1 – {{\tan }^2}A}}} \right) + \tan A}}{{1 – \left( {\frac{{2\tan A}}{{1 – {{\tan }^2}A}}} \right) \cdot \tan A}}\) [Using \(\tan 2A = \frac{{2\tan A}}{{1 – {{\tan }^2}A}}\)] \( = \frac{{2\tan A + \tan A \cdot \left( {1 – {{\tan }^2}A} \right)}}{{1 \cdot \left( {1 – {{\tan }^2}A} \right) – 2{{\tan }^2}A}}\) \( = \frac{{2\tan A + \tan A – {{\tan }^3}A}}{{1 – {{\tan }^2}A – 2{{\tan }^2}A}}\) \( = \frac{{3\tan A – {{\tan }^3}A}}{{1 – 3{{\tan }^2}A}}\)

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Sub Multiple Angles

If \(A\) is an angle \(\frac{A}{2},\frac{A}{3},\frac{A}{4}\) etc. called sub-multiple angles of \(A.\) In this section, let us discuss the trigonometric ratios of angle \(A\) in terms of \(\frac{A}{2}\) and \(\frac{A}{3}.\)

(i) Trigonometric Ratios of Angle \(A\) in terms of \(\frac{A}{2}\)

Identities	Proofs
(a) \(\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}\)	Proof: As we know, \(\sin 2A = 2\sin A\cos A\) Therefore, \(\sin A = \sin \left( {\frac{A}{2} + \frac{A}{2}} \right)\) \( = \sin \left( {2 \cdot \frac{A}{2}} \right)\) \( = 2\sin \frac{A}{2}\cos \frac{A}{2}\)
(b) \(\sin A = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\)	Proof: We have, \(\sin A = 2\sin \frac{A}{2}\cos \frac{A}{2}\) \( = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2}}}{1}\) \( = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2}}}{{{{\cos }^2}\frac{A}{2} + {{\sin }^2}\frac{A}{2}}}\) \( = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\) [Divide numerator and denominator with \({\cos ^2}\frac{A}{2}\)]
(c) \(\sin A = \frac{{2\cot \frac{A}{2}}}{{1 + {{\cot }^2}\frac{A}{2}}}\)	Proof: We have, \(\sin A = \frac{{2\tan \frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}}\) \( = \frac{{\frac{2}{{\cot \frac{A}{2}}}}}{{1 + \frac{1}{{{{\cot }^2}\frac{A}{2}}}}}\) \( = \frac{{2\cot \frac{A}{2}}}{{1 + {{\cot }^2}\frac{A}{2}}}\)
(d) \(\cos \frac{A}{2} = \pm \sqrt {\frac{{1 + \cos A}}{2}} \)	Proof: We have, \(\cos 2A = 2{\cos ^2}A – 1\) Now, \(\cos A = \cos \left( {\frac{A}{2} + \frac{A}{2}} \right)\) \( = \cos \left( {2 \cdot \frac{A}{2}} \right)\) \( = 2{\cos ^2}\frac{A}{2} – 1\) Therefore, \(2{\cos ^2}\frac{A}{2} = 1 + \cos A\) \( \Rightarrow {\cos ^2}\frac{A}{2} = \frac{{1 + \cos A}}{2}\) \( \Rightarrow \cos \frac{A}{2} = \pm \sqrt {\frac{{1 + \cos A}}{2}} \) Also, the sign on the right-hand side depends on the quadrant in which the angle \(\frac{A}{2}\) lies.
(e) \(\sin \frac{A}{2} = \pm \frac{{\sqrt {1 – \cos A} }}{2}\)	Proof: We have, \(\cos 2A = 1 – 2{\sin ^2}A\) Now, \(\cos A = \cos \left( {\frac{A}{2} + \frac{A}{2}} \right)\) \( = \cos \left( {2 \cdot \frac{A}{2}} \right)\) \( = 1 – 2{\sin ^2}\frac{A}{2}\) Therefore, \(2{\sin ^2}\frac{A}{2} = 1 – \cos A\) \( \Rightarrow {\sin ^2}\frac{A}{2} = \frac{{1 – \cos A}}{2}\) \( \Rightarrow \sin \frac{A}{2} = \pm \sqrt {\frac{{1 – \cos A}}{2}} \) Also, the sign on the right-hand side depends on the quadrant in which the angle \(\frac{A}{2}\) lies.
(f) \(\tan \frac{A}{2} = \pm \sqrt {\frac{{1 – \cos A}}{{1 + \cos A}}} \)	Proof: We have, \(\sin \frac{A}{2} = \pm \sqrt {\frac{{1 – \cos A}}{2}} \) and \(\cos \frac{A}{2} = \pm \sqrt {\frac{{1 + \cos A}}{2}} \) Therefore, \(\tan \frac{A}{2} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{A}{2}}}\) \( = \pm \frac{{\sqrt {\frac{{1 – \cos A}}{2}} }}{{\sqrt {\frac{{1 + \cos A}}{2}} }}\) \( = \pm \sqrt {\frac{{1 – \cos A}}{{1 + \cos A}}} \) Also, the sign on the right-hand side depends on the quadrant in which the angle \(\frac{A}{2}\) lies.

(ii) Trigonometric Ratios of Angle \(A\) in terms of \(\frac{A}{3}\)

Identities	Proofs
(a) \(\cos A = 3\cos \frac{A}{3} – 4{\cos ^3}\frac{A}{3}\)	Proof: As we have, \(\cos 3A = 4{\cos ^3}A – 3\cos A\) Therefore, \(\cos A = \cos \left( {\frac{A}{3} + \frac{A}{3} + \frac{A}{3}} \right)\) \( = \cos \left( {3 \cdot \frac{A}{3}} \right)\) \( = 3\cos \frac{A}{3} – 4{\cos ^3}\frac{A}{3}\)
(b) \(\sin A = 3\sin \frac{A}{3} – 4{\sin ^3}\frac{A}{3}\)	Proof: As we have, \(\sin 3A = 3\sin A – 4{\sin ^3}A\) Therefore, \(\sin A = \sin \left( {\frac{A}{3} + \frac{A}{3} + \frac{A}{3}} \right)\) \( = \sin \left( {3 \cdot \frac{A}{3}} \right)\) \( = 3\sin \frac{A}{3} – 4{\sin ^3}\frac{A}{3}\)
(c) \(\tan A = \frac{{3\tan \frac{A}{3} – {{\tan }^3}\frac{A}{3}}}{{1 – 3{{\tan }^2}\frac{A}{3}}}\)	Proof: As we have, \(\tan 3A = \frac{{3\tan A – {{\tan }^3}A}}{{1 – 3{{\tan }^2}A}}\) Therefore, \(\tan A = \tan \left( {\frac{A}{3} + \frac{A}{3} + \frac{A}{3}} \right)\) \( = \tan \left( {3 \cdot \frac{A}{3}} \right)\) \( = \frac{{3\tan \frac{A}{3} – {{\tan }^3}\frac{A}{3}}}{{1 – 3{{\tan }^2}\frac{A}{3}}}\)

Solved Examples – Trigonometrical Ratios of Sub-Multiple Angles

A few important solved examples to elaborate Trigonometrical Ratios of Sub Multiple Angles are given below:

Q.1. If \(\sin A = \frac{1}{2},\) then find the value of \(\cos 2A.\)
Ans: Given that we have \(\sin A = \frac{1}{2}\)
As we know, \(\cos 2A = 1 – 2{\sin ^2}A\)
Therefore, \(\cos 2A = 1 – 2 \times \frac{1}{4}\)
\( = 1 – \frac{1}{2}\)
\( = \frac{1}{2}\)
Hence, the value of \(\cos 2A\) is \(\frac{1}{2}.\)

Q.2. Find the value of \(2\sin {30^ \circ }\cos {30^ \circ }.\)
Ans: Given that we have \(2\sin {30^ \circ }\cos {30^ \circ }\)
As we know, \(\sin 2A = 2\sin A\cos A\)
Therefore, \(2\sin {30^ \circ }\cos {30^ \circ } = \sin 2\left( {{{30}^ \circ }} \right)\)
\( = \sin {60^ \circ }\)
\( = \frac{{\sqrt 3 }}{2}\)
Hence, the value of \(2\sin {30^ \circ }\cos {30^ \circ }\) is \(\frac{{\sqrt 3 }}{2}.\)

Q.3. Prove that \(\frac{{\sin 2x}}{{1 + \cos 2x}} = \tan x.\)
Ans: We need to prove \(\frac{{\sin 2x}}{{1 + \cos 2x}} = \tan x\)
Consider, \({\text{L}}{\text{.H}}{\text{.S}}. = \frac{{\sin 2x}}{{1 + \cos 2x}}\)
\( = \frac{{2\sin x\cos x}}{{2{{\cos }^2}x}}\) [Using \(\sin 2x = 2\sin x\cos x\) and \(1 + \cos 2x = 2{\cos ^2}x\)]
\( = \tan x\)
\( = {\text{R}}{\text{.H}}{\text{.S}}.\)

Q.4. Prove that \(\frac{{1 + \sin 2x + \cos 2x}}{{1 + \sin 2x – \cos 2x}} = \cot x.\)
Ans: We need to prove \(\frac{{1 + \sin 2x + \cos 2x}}{{1 + \sin 2x – \cos 2x}} = \cot x\)
Consider, \({\text{LHS}} = \frac{{1 + \sin 2x + \cos 2x}}{{1 + \sin 2x – \cos 2x}}\)
\( = \frac{{\left( {1 + \cos 2x} \right) + \sin 2x}}{{\left( {1 – \cos 2x} \right) + \sin 2x}}\)
\( = \frac{{2{{\cos }^2}x + 2\sin x\cos x}}{{2{{\sin }^2}x + 2\sin x\cos x}}\) [Using \(\sin 2x = 2\sin x\cos x\) and \(1 + \cos 2x = 2{\cos ^2}x\)]
\( = \frac{{2\cos x\left( {\cos x + \sin x} \right)}}{{2\sin x\left( {\cos x + \sin x} \right)}}\)
\( = \frac{{\cos x}}{{\sin x}}\)
\( = \cot x\)
Hence, proved.

Q.5. Prove that \(\frac{{1 + \sin x – \cos x}}{{1 + \sin x + \cos x}} = \tan \frac{x}{2}.\)
Ans: We need to prove \(\frac{{1 + \sin x – \cos x}}{{1 + \sin x + \cos x}} = \tan \frac{x}{2}\)
Consider, \({\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \frac{{1 + \sin x – \cos x}}{{1 + \sin x + \cos x}}\)
\( = \frac{{\left( {1 – \cos x} \right) + \sin x}}{{\left( {1 + \cos x} \right) + \sin x}}\)
\( = \frac{{2{{\sin }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}\) [Using \(\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2},\cos x = 1 – 2{\sin ^2}\frac{x}{2},\cos x = 2{\cos ^2}\frac{x}{2} – 1\)]
\( = \frac{{2\sin \frac{x}{2}\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}}{{2\cos \frac{x}{2}\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}}\)
\( = \tan \frac{x}{2}\)
\( = {\text{R}}{\text{.H}}{\text{.S}}\)
Hence, proved.

Q.6. Prove that \(\frac{{\sin 2A}}{{1 – \cos 2A}} = \cot A.\)
Ans: We need to prove \(\frac{{\sin 2A}}{{1 – \cos 2A}} = \cot A\)
Consider, \({\text{LHS}} = \frac{{\sin 2A}}{{1 – \cos 2A}}\)
\( = \frac{{2\sin A\cos A}}{{2{{\sin }^2}A}}\) [Using \(\sin 2x = 2\sin x\cos x\) and \(1 + \cos 2x = 2{\cos ^2}x\)]
\( = \frac{{\cos A}}{{\sin A}}\)
\( = \cot A\)
\( = {\text{R}}{\text{.H}}{\text{.S}}\)
Hence,proved.

Q.7. Prove that \(\frac{{1 – \cos 2A + \sin 2A}}{{1 + \cos 2A + \sin 2A}} = \tan A.\)
Ans: We need to prove \(\frac{{1 – \cos 2A + \sin 2A}}{{1 + \cos 2A + \sin 2A}} = \tan A\)
Consider, \({\text{LHS}} = \frac{{1 – \cos 2A + \sin 2A}}{{1 + \cos 2A + \sin 2A}}\)
\( = \frac{{1 – \left( {1 – 2{{\sin }^2}A} \right) + 2\sin A\cos A}}{{1 + 2{{\cos }^2}A – 1 + 2\sin A\cos A}}\) [Using \(\sin 2x = 2\sin x\cos x\) and \(1 + \cos 2x = 2{\cos ^2}x\)]
\( = \frac{{2\sin A\left( {\sin A + \cos A} \right)}}{{2\cos A\left( {\cos A + \sin A} \right)}}\)
\( = \frac{{\sin A}}{{\cos A}}\)
\( = \tan A\)
\( = {\text{R}}{\text{.H}}{\text{.S}}\)
Hence, proved.

Applications of Trigonometry

Summary

If \(A\) is an angle, then \(2A,3A,4A \ldots \) are called as multiple angles of \(A,\) if \(A\) is an angle, then \(\frac{A}{2},\frac{A}{3},\frac{A}{4}\) etc. are called sub-multiple angles of \(A.\) The trigonometric function of sub-multiple angles is also known as the sub-multiple angle formula. The values of multiple and sub-multiple angles cannot be found directly, but their values can be calculated by expressing each trigonometric function in its expanded form. The double and triple angles formula are used under the multiple angle formulas. Sine, tangent, and cosine are the common functions of the multiple and sub-multiples angle formulas. Some identities that can be used to solve multiple and sub-multiple angle problems are listed and proved.

Frequently Asked Questions (FAQs)

The most commonly raised questions on Trigonometrical Ratios of Sub Multiple Angles are answered below:

Q.1. What are sub-multiple angles?
Ans: If \(x\) be a given angle, then \(\frac{x}{2},\frac{x}{3},\frac{x}{4}\) etc. are called sub-multiple angles of \(x.\) So sine, tangent, and cosine are the common functions that are used for the sub-multiple angle formula.

Q.2. What are the trigonometric ratios of special angles?
Ans: The trigonometric ratios of some specific angle is defined as the ratio of the sides of a right-angled triangle with respect to any of its acute angles. The trigonometric ratios of some specific angles include \({0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }\) and \({90^ \circ }.\)

Q.3. What is a multiple angle? Give example.
Ans: If \(x\) be a given angle, then \(2x,3x,4x\) etc. are called multiple angles of \(x.\) So sine, tangent, and cosine are the common functions used for the multiple angle formula. Multiple angle trigonometry expressions are where the angle measure is some multiple of a variable, for example, \(2x\) or \(3y.\)

Q.4. What is the formula of \(\cos⁡2x\) in terms of \(\tan x\)?
Ans: The formula of \(\cos 2x\) can be written in \(\tan x\) as \(\cos \left( {2x} \right) = \frac{{1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}}\)

Q.5. What is the formula for \(\sin 2x\)?
Ans: The formula of \(\sin 2x\) can be written as
\(\sin 2x = 2\sin x\cos x = \frac{{2\tan x}}{{1 + {{\tan }^2}x}}\)

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