Trigonometric Ratio of Submultiple of an Angle

The trigonometric ratios of a given angle are the ratios of a right-angled triangle's sides with regard to any one of its acute angles. The six trigonometric ratios are sine (sin) , cosine (cos), tangent(tan), cotangent(cot), secant(sec), cosecant( $\mathrm{cosec})$. The trignometric ratio of a submultiple of an angle involves the conversion of a trignometric function of an angle into the trignometric expression of a submultiple of angles. Sometimes it is difficult to find the value of the trignometric ratio of an angle but its submultiple is known. So. with the help of submutiple we can find the value of an angle.

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In this article, we will cover the concept of the Trigonometric Ratio of Submultiple of an Angle. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Trigonometric Ratio of Submultiple of an Angle

If A is the angle then $A/2,A/3,$ and $A/4$ are called the submultiple angles. The Trigonometric Ratio of the Submultiple of an Angle is used to transform the trigonometric ratios of an Angles into the trigonometric ratios of its submultiple angles. We can use double-angle or triple-angle formulas to derive the trigonometric ratio of the submultiple of an angle.

1. Trigonometric Ratio of θ in terms of θ/2

With the help of double-angle formulas of $\sin ,\cos$, and $\tan$ functions we can derive the Trigonometric Ratio of $\theta$ in terms of $\theta /2$

$\begin{array}{rl}\sin (\theta )& =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\\ & =\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\\ \cos (\theta )& ={\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\\ & =1-2{\sin }^2\frac{\theta }{2}\\ & =2{\cos }^2\frac{\theta }{2}-1\\ & =\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\\ \tan (\theta )& =\frac{2{\tan }^{\frac{\theta }{2}}}{1-{\tan }^2\frac{\theta }{2}}\end{array}$

All the above trigonometric ratios can be derived by replacing ' $\theta$ ' with ' $\theta /2$ ' in the double-angle formulas.

2. Trigonometric Ratio of θ in terms of θ/3

The above trigonometric ratio of angle ‘$A^{\prime }$ in terms of ‘$A/3$’ can be derived by replacing ‘$A$’ with ‘$A/3$’ in the triple angle formulas

With the help of Triple-angle formulas of $\sin ,\cos$, and $\tan$ functions we can derive the Trigonometric Ratio of $\theta$ in terms of $\theta /3$
1. $\sin \mathrm{A}=3\sin \frac{\mathrm{A}}{3}-4{\sin }^3\frac{\mathrm{A}}{3}$
2. $\cos \mathrm{A}=4{\cos }^3\frac{\mathrm{A}}{3}-3\cos \frac{A}{3}$
3. $\tan \mathrm{A}=\frac{3\tan \frac{\mathrm{A}}{3}-{\tan }^3\frac{\mathrm{A}}{3}}{1-3{\tan }^2\frac{\mathrm{A}}{3}}$

The above trigonometric ratio of angle ' $A$ ' in terms of ' $A/3$ ' can be derived by replacing ' $A$ ' with ' $A/3$ ' in the triple angle formulas

3. Trigonometric Ratio of θ/2 in terms of θ

With the help of reduction formulas, we can derive the Trigonometric Ratio of $\theta /2$ in terms of $\theta$ and we can use it when we have an angle that is half the size of a special angle.
1. $\sin \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \theta }{2}}$
2. $\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}$
3. $\tan \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$

Summary

The submultiple of angles is a versatile tool for solving problems across diverse fields. Mastery of these concepts enhances mathematical proficiency and enables individuals to apply trigonometric principles effectively in real-world scenarios. Understanding submultiples helps us to solve and analyze real-life problems.

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Solved Example Based on Trigonometric Ratio of Submultiple of an Angle

Example 1: The value of $\cot \frac{\pi }{24}$ is
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Solution

$\frac{\pi }{24}=\frac{{180}^{\circ }}{24}={\left(\frac{15}{2}\right)}^o$

we knowthat $\cot \theta =\frac{1+\cos 2\theta }{\sin 2\theta }$

$\begin{array}{rl}& \Rightarrow \cot {\left(\frac{15}{2}\right)}^0=\frac{1+\cos {15}^{\circ }}{\sin {15}^{\circ }}--(i)\\ & \cos {15}^{\circ }=\cos (45-30)=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{3}+1}{2+\sqrt{2}}\end{array}$

Similarly $\sin {15}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$
From (i)

$\begin{array}{rl}& \cot {\left(\frac{15}{2}\right)}^o=\frac{1+\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}{\frac{\sqrt{3}-1}{2\sqrt{2}}}\\ & =\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\cdot \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}\\ & =\sqrt{2}+\sqrt{3}+2+\sqrt{6}\end{array}$

Hence, the answer is $\sqrt{2}+\sqrt{3}+2+\sqrt{6}$

Example 2: What is the value of $\cot \frac{A}{2}$ if $\sin A=\frac{3}{5}$ ?
Solution

$\tan \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}=\frac{1-\cos A}{\sin A}$

$\begin{array}{rl}& \tan \frac{A}{2}=\frac{\sin A}{1+\cos A}=\frac{\frac{3}{5}}{1+\frac{4}{5}}=\frac{1}{3}\\ & \therefore \cot \frac{A}{2}=3\end{array}$

Hence, the answer is 3

Example 3: If $\sin x+\cos x=\frac{\sqrt{5}}{2}$. Find the value of $\tan \left(\frac{x}{2}\right)$.
Solution

$\begin{array}{rl}& \sin x+\cos x=\frac{\sqrt{5}}{2}\\ & \frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{\sqrt{5}}{2}\\ & \frac{2\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=\frac{\sqrt{5}}{2}\\ & (\sqrt{5}+2){\tan }^2\frac{x}{2}-4\tan \frac{x}{2}+\sqrt{5}-2=0\\ & \tan \frac{x}{2}=\frac{4\pm \sqrt{16-4\ast (\sqrt{5}+2)\ast (\sqrt{5}-2)}}{2(\sqrt{5}+2)}\\ & =\frac{4\pm \sqrt{12}}{2(\sqrt{5}+2)}\end{array}$

Hence, the answer is $\frac{4\pm \sqrt{12}}{2(\sqrt{5}+2)}$

Example 4: What is the range of $\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }$ ?

Solution

$\begin{array}{rl}& \frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=\frac{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}{1-\frac{{\sin }^2\theta }{{\cos }^2\theta }}\\ =& \frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta -{\sin }^2\theta }\\ =& \frac{1}{\cos 2\theta }=\sec 2\theta \end{array}$

Since the range of $\sec (2\theta )$ is $(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$.
So the range of the given function is also $(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$.
Hence, the answer is $(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$