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Submultiple Angle Formula Class 10 Exercise 7.3 Solutions
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1. Submultiple Angles: Class 10 Optional Mathematics Trigonometry Overview and Key Formulas

Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.3 Submultiple Angles

A submultiple angle of \(A\) is an angle obtained by dividing \(A\) by \(2,3,4,\dots\), such as \(\frac A2\) and \(\frac A3\). This page gives the submultiple angle formula, a short submultiple angle proof, submultiple angle examples and step by step solutions to every question of Exercise 7.3 (multiple and submultiple angles in trigonometry, Class 10 Optional Mathematics, SEE, Nepal), with the Nepali question in brackets.

This is part of the unit prescribed by the Curriculum Development Centre (CDC), Nepal. Multiple choice solutions also have Nepali in brackets, and the other solutions are in English because the steps and symbols are the same in both languages.

  • Submultiple angles (Exercise 7.3): replace \(A\) by \(\frac A2\) in the double angle formulas (or by \(\frac\theta3\) in the triple angle formulas) to express \(\sin A\), \(\cos A\), \(\tan A\) through \(\frac A2\), and use \(\cos A=1-2\sin^2\frac A2=2\cos^2\frac A2-1\) to find \(\sin\frac A2\) or \(\cos\frac A2\).

Key formulas to remember

Submultiple angles: \(\sin A=2\sin\frac A2\cos\frac A2=\dfrac{2\tan\frac A2}{1+\tan^2\frac A2}\), \(\cos A=\cos^2\frac A2-\sin^2\frac A2=\dfrac{1-\tan^2\frac A2}{1+\tan^2\frac A2}\), \(\tan A=\dfrac{2\tan\frac A2}{1-\tan^2\frac A2}\). Also \(\sin\theta=3\sin\frac\theta3-4\sin^3\frac\theta3\) and \(\cos\theta=4\cos^3\frac\theta3-3\cos\frac\theta3\).
Half angle (from \(\cos A=1-2\sin^2\frac A2=2\cos^2\frac A2-1\)): \(\sin\frac A2=\pm\sqrt{\dfrac{1-\cos A}{2}}\), \(\cos\frac A2=\pm\sqrt{\dfrac{1+\cos A}{2}}\). Proof idea: replace \(A\) by \(\frac A2\) in the double angle formulas.

For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about trigonometric identities for extra understanding.

2. Exercise 7.3 Submultiple Angles: Class 10 Optional Mathematics Trigonometry

Exercise 7.3 Submultiple Angles: Multiple Choice Questions (बहुवैकल्पिक प्रश्न)

Q.1 Tick (✓) the correct option for the given questions:

(दिइएका प्रश्नहरूको उपयुक्त विकल्पमा ठिक चिह्न (✓) लगाउनुहोस् : )

A. Which of the following is equal to \(\sin A\)? (तलका मध्ये \(\sin A\) कुनसँग बराबर हुन्छ ?)

  • a) \(\cos^2\frac A2+\sin^2\frac A2\)
  • b) \(\cos^2\frac A2-\sin^2\frac A2\)
  • c) \(1-2\cos^2\frac A2\)
  • d) \(2\sin\frac A2\cos\frac A2\)

Correct Answer: d) \(2\sin\frac A2\cos\frac A2\)

Solution:

  • Put \(X=\frac A2\) in the double angle formula \(\sin2X=2\sin X\cos X\).
\[\sin A=\sin\left(2\cdot\frac A2\right)=2\sin\frac A2\cos\frac A2\]

(\(\sin2X=2\sin X\cos X\) मा \(X=\frac A2\) राख्दा \(\sin A=2\sin\frac A2\cos\frac A2\) हुन्छ ।)

B. Which of the following is equal to \(\cos\theta\)? (तलका मध्ये कुन चाहिँ \(\cos\theta\) सँग बराबर हुन्छ ?)

  • a) \(4\sin\frac\theta3-3\sin^3\frac\theta3\)
  • b) \(4\cos^3\frac\theta3-3\cos\frac\theta3\)
  • c) \(3\cos\frac\theta3-4\cos^3\frac\theta3\)
  • d) \(3\sin\frac\theta3-4\sin^4\frac\theta3\)

Correct Answer: b) \(4\cos^3\frac\theta3-3\cos\frac\theta3\)

Solution:

  • Put \(X=\frac\theta3\) in the triple angle formula \(\cos3X=4\cos^3X-3\cos X\).
\[\cos\theta=\cos\left(3\cdot\frac\theta3\right)=4\cos^3\frac\theta3-3\cos\frac\theta3\]

(\(\cos3X=4\cos^3X-3\cos X\) मा \(X=\frac\theta3\) राख्दा \(\cos\theta=4\cos^3\frac\theta3-3\cos\frac\theta3\) हुन्छ ।)

C. If \(A=60^\circ\), what is the value of \(\sin\frac A2\)? (यदि \(A=60^\circ\) भए \(\sin\frac A2\) को मान कति हुन्छ ?)

  • a) \(0\)
  • b) \(1\)
  • c) \(\frac12\)
  • d) \(\frac{\sqrt3}{2}\)

Correct Answer: c) \(\frac12\)

Solution:

  • Here, \(\dfrac A2=\dfrac{60^\circ}{2}=30^\circ\).
\[\sin\frac A2=\sin30^\circ=\frac12\]

(\(\frac A2=30^\circ\) भएकाले \(\sin30^\circ=\frac12\) हुन्छ ।)

D. Which is a submultiple angle of \(60^\circ\)? (\(60^\circ\) को अपवर्तक कोण कुन हो ?)

  • a) \(15^\circ\)
  • b) \(25^\circ\)
  • c) \(35^\circ\)
  • d) \(45^\circ\)

Correct Answer: a) \(15^\circ\)

Solution:

  • A submultiple angle of \(A\) is \(\frac An\) where \(n=2,3,4,\dots\)
  • \(\frac{60^\circ}{2}=30^\circ\), \(\frac{60^\circ}{3}=20^\circ\), \(\frac{60^\circ}{4}=15^\circ\).
\[\frac{60^\circ}{4}=15^\circ\]

(\(60^\circ\) लाई \(4\) ले भाग गर्दा \(15^\circ\) आउँछ, त्यसैले \(15^\circ\) अपवर्तक कोण हो ।)

E. Which is equal to \(\tan A\)? (\(\tan A\) सँग बराबर कुन हुन्छ ?)

  • a) \(\dfrac{2\tan\frac A2}{1+\tan^2\frac A2}\)
  • b) \(\dfrac{1-\tan^2\frac A2}{1+\tan^2\frac A2}\)
  • c) \(\dfrac{2\tan\frac A2}{1-\tan^2\frac A2}\)
  • d) \(\dfrac{1+\tan^2\frac A2}{1-\tan^2\frac A2}\)

Correct Answer: c) \(\dfrac{2\tan\frac A2}{1-\tan^2\frac A2}\)

Solution:

  • Put \(X=\frac A2\) in \(\tan2X=\dfrac{2\tan X}{1-\tan^2X}\).
\[\tan A=\tan\left(2\cdot\frac A2\right)=\frac{2\tan\frac A2}{1-\tan^2\frac A2}\]

(\(\tan2X\) को सूत्रमा \(X=\frac A2\) राख्दा \(\tan A=\frac{2\tan\frac A2}{1-\tan^2\frac A2}\) हुन्छ ।)

Exercise 7.3 Submultiple Angles: Written Questions (लेख्ने प्रश्न)

Q.2
a. Define submultiple angle with an example. (अपवर्तक कोणलाई उदाहरणसहित परिभाषित गर्नुहोस् ।)

An angle obtained by dividing a given angle \(A\) by the integers \(2,3,4,\dots\) is called a submultiple angle of \(A\). Its submultiple angles are \(\dfrac A2,\ \dfrac A3,\ \dfrac A4,\dots\)

Example: if \(A=60^\circ\), then

\[\begin{aligned}\frac{60^\circ}{2}&=30^\circ \\ \frac{60^\circ}{3}&=20^\circ \\ \frac{60^\circ}{4}&=15^\circ\end{aligned}\]
So \(30^\circ,\ 20^\circ,\ 15^\circ\) are submultiple angles of \(60^\circ\).
b. Express \(\sin\theta\) in terms of \(\tan\frac\theta2\). (\(\sin\theta\) लाई \(\tan\frac\theta2\) मा व्यक्त गर्नुहोस् ।)

Use \(\sin2X=\dfrac{2\tan X}{1+\tan^2X}\) with \(X=\dfrac\theta2\).

\[\begin{aligned}\sin\theta&=\sin\left(2\cdot\frac\theta2\right) \\ &=\frac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}\end{aligned}\]
\(\sin\theta=\dfrac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}\)
c. Express \(\cos\theta\) in terms of \(\cos\frac\theta2\). (\(\cos\theta\) लाई \(\cos\frac\theta2\) मा व्यक्त गर्नुहोस् ।)

Use \(\cos2X=2\cos^2X-1\) with \(X=\dfrac\theta2\).

\[\begin{aligned}\cos\theta&=\cos\left(2\cdot\frac\theta2\right) \\ &=2\cos^2\frac\theta2-1\end{aligned}\]
\(\cos\theta=2\cos^2\dfrac\theta2-1\)
d. Express \(\cos\theta\) in terms of \(\tan\frac\theta2\). (\(\cos\theta\) लाई \(\tan\frac\theta2\) मा व्यक्त गर्नुहोस् ।)

Use \(\cos2X=\dfrac{1-\tan^2X}{1+\tan^2X}\) with \(X=\dfrac\theta2\).

\[\begin{aligned}\cos\theta&=\cos\left(2\cdot\frac\theta2\right) \\ &=\frac{1-\tan^2\frac\theta2}{1+\tan^2\frac\theta2}\end{aligned}\]
\(\cos\theta=\dfrac{1-\tan^2\frac\theta2}{1+\tan^2\frac\theta2}\)
e. Express \(\sin\theta\) in terms of \(\sin\frac\theta3\). (\(\sin\theta\) लाई \(\sin\frac\theta3\) मा व्यक्त गर्नुहोस् ।)

Use \(\sin3X=3\sin X-4\sin^3X\) with \(X=\dfrac\theta3\).

\[\begin{aligned}\sin\theta&=\sin\left(3\cdot\frac\theta3\right) \\ &=3\sin\frac\theta3-4\sin^3\frac\theta3\end{aligned}\]
\(\sin\theta=3\sin\dfrac\theta3-4\sin^3\dfrac\theta3\)
Q.3
a. If \(\sin\frac A2=\frac45\), find the values of \(\sin A\), \(\cos A\) and \(\tan A\). (यदि \(\sin\frac A2=\frac45\) भए \(\sin A\), \(\cos A\) र \(\tan A\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\sin\frac A2=\frac45\) (taking \(\frac A2\) acute).

\[\begin{aligned}\cos\frac A2&=\sqrt{1-\sin^2\frac A2} \\ &=\sqrt{1-\frac{16}{25}} \\ &=\frac35\end{aligned}\]\[\begin{aligned}\sin A&=2\sin\frac A2\cos\frac A2 \\ &=2\cdot\frac45\cdot\frac35 \\ &=\frac{24}{25}\end{aligned}\]\[\begin{aligned}\cos A&=\cos^2\frac A2-\sin^2\frac A2 \\ &=\frac{9}{25}-\frac{16}{25} \\ &=-\frac{7}{25}\end{aligned}\]\[\begin{aligned}\tan A&=\frac{\sin A}{\cos A} \\ &=\frac{24/25}{-7/25} \\ &=-\frac{24}{7}\end{aligned}\]
\(\sin A=\dfrac{24}{25},\ \cos A=-\dfrac{7}{25},\ \tan A=-\dfrac{24}{7}\)
b. If \(\tan\frac A2=\frac34\), find the values of \(\sin A\), \(\cos A\) and \(\tan A\). (यदि \(\tan\frac A2=\frac34\) भए \(\sin A\), \(\cos A\) र \(\tan A\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\tan\frac A2=\frac34\), so \(\tan^2\frac A2=\frac9{16}\).

\[\begin{aligned}\sin A&=\frac{2\tan\frac A2}{1+\tan^2\frac A2} \\ &=\frac{2\cdot\frac34}{1+\frac9{16}} \\ &=\frac{3/2}{25/16} \\ &=\frac{24}{25}\end{aligned}\]\[\begin{aligned}\cos A&=\frac{1-\tan^2\frac A2}{1+\tan^2\frac A2} \\ &=\frac{1-\frac9{16}}{1+\frac9{16}} \\ &=\frac{7/16}{25/16} \\ &=\frac{7}{25}\end{aligned}\]\[\begin{aligned}\tan A&=\frac{2\tan\frac A2}{1-\tan^2\frac A2} \\ &=\frac{3/2}{7/16} \\ &=\frac{24}{7}\end{aligned}\]
\(\sin A=\dfrac{24}{25},\ \cos A=\dfrac{7}{25},\ \tan A=\dfrac{24}{7}\)
c. If \(\cos\frac A2=\frac5{13}\), find the values of \(\sin A\), \(\cos A\) and \(\tan A\). (यदि \(\cos\frac A2=\frac5{13}\) भए \(\sin A\), \(\cos A\) र \(\tan A\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\cos\frac A2=\frac5{13}\) (taking \(\frac A2\) acute).

\[\begin{aligned}\sin\frac A2&=\sqrt{1-\cos^2\frac A2} \\ &=\sqrt{1-\frac{25}{169}} \\ &=\frac{12}{13}\end{aligned}\]\[\begin{aligned}\sin A&=2\sin\frac A2\cos\frac A2 \\ &=2\cdot\frac{12}{13}\cdot\frac5{13} \\ &=\frac{120}{169}\end{aligned}\]\[\begin{aligned}\cos A&=2\cos^2\frac A2-1 \\ &=2\cdot\frac{25}{169}-1 \\ &=-\frac{119}{169}\end{aligned}\]\[\begin{aligned}\tan A&=\frac{120/169}{-119/169} \\ &=-\frac{120}{119}\end{aligned}\]
\(\sin A=\dfrac{120}{169},\ \cos A=-\dfrac{119}{169},\ \tan A=-\dfrac{120}{119}\)
d. If \(\sin\frac\theta3=\frac12\), find the values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\). (यदि \(\sin\frac\theta3=\frac12\) भए \(\sin\theta\), \(\cos\theta\) र \(\tan\theta\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\sin\frac\theta3=\frac12\) (taking \(\frac\theta3\) acute).

\[\begin{aligned}\cos\frac\theta3&=\sqrt{1-\frac14} \\ &=\frac{\sqrt3}{2}\end{aligned}\]\[\begin{aligned}\sin\theta&=3\sin\frac\theta3-4\sin^3\frac\theta3 \\ &=\frac32-4\cdot\frac18 \\ &=1\end{aligned}\]\[\begin{aligned}\cos\theta&=4\cos^3\frac\theta3-3\cos\frac\theta3 \\ &=4\cdot\frac{3\sqrt3}{8}-\frac{3\sqrt3}{2} \\ &=0\end{aligned}\]

Since \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac10\), \(\tan\theta\) is not defined (\(\theta=90^\circ\)).

\(\sin\theta=1,\ \cos\theta=0,\ \tan\theta\) is undefined \((\infty)\)
e. If \(\cos\frac\theta3=\frac45\), find the values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\). (यदि \(\cos\frac\theta3=\frac45\) भए \(\sin\theta\), \(\cos\theta\) र \(\tan\theta\) को मान पत्ता लगाउनुहोस् ।)

Here, \(\cos\frac\theta3=\frac45\) (taking \(\frac\theta3\) acute).

\[\begin{aligned}\sin\frac\theta3&=\sqrt{1-\frac{16}{25}} \\ &=\frac35\end{aligned}\]\[\begin{aligned}\sin\theta&=3\sin\frac\theta3-4\sin^3\frac\theta3 \\ &=\frac95-4\cdot\frac{27}{125} \\ &=\frac{225-108}{125} \\ &=\frac{117}{125}\end{aligned}\]\[\begin{aligned}\cos\theta&=4\cos^3\frac\theta3-3\cos\frac\theta3 \\ &=4\cdot\frac{64}{125}-\frac{12}{5} \\ &=\frac{256-300}{125} \\ &=-\frac{44}{125}\end{aligned}\]\[\begin{aligned}\tan\theta&=\frac{117/125}{-44/125} \\ &=-\frac{117}{44}\end{aligned}\]
\(\sin\theta=\dfrac{117}{125},\ \cos\theta=-\dfrac{44}{125},\ \tan\theta=-\dfrac{117}{44}\)
f. If \(\tan\frac\theta3=\frac12\), find the value of \(\tan\theta\). (यदि \(\tan\frac\theta3=\frac12\) भए \(\tan\theta\) को मान पत्ता लगाउनुहोस् ।)

Use \(\tan3X=\dfrac{3\tan X-\tan^3X}{1-3\tan^2X}\) with \(X=\dfrac\theta3\).

\[\begin{aligned}\tan\theta&=\frac{3\cdot\frac12-\left(\frac12\right)^3}{1-3\left(\frac12\right)^2} \\ &=\frac{\frac32-\frac18}{1-\frac34} \\ &=\frac{11/8}{1/4} \\ &=\frac{11}{2}\end{aligned}\]
\(\tan\theta=\dfrac{11}{2}\)
Q.4
a. If \(\cos A=\frac7{25}\), find the value of \(\sin\frac A2\). (यदि \(\cos A=\frac7{25}\) भए \(\sin\frac A2\) को मान पत्ता लगाउनुहोस् ।)

Use \(\cos A=1-2\sin^2\frac A2\) (taking \(\frac A2\) acute).

\[\begin{aligned}&\frac7{25}=1-2\sin^2\frac A2 \\ &\implies 2\sin^2\frac A2=\frac{18}{25} \\ &\implies \sin^2\frac A2=\frac9{25} \\ &\implies \sin\frac A2=\frac35\end{aligned}\]
\(\sin\dfrac A2=\dfrac35\)
b. If \(\cos A=\frac{13}{36}\), find the value of \(\cos\frac A2\). (यदि \(\cos A=\frac{13}{36}\) भए \(\cos\frac A2\) को मान पत्ता लगाउनुहोस् ।)

Use \(\cos A=2\cos^2\frac A2-1\) (taking \(\frac A2\) acute).

\[\begin{aligned}&\frac{13}{36}=2\cos^2\frac A2-1 \\ &\implies 2\cos^2\frac A2=\frac{49}{36} \\ &\implies \cos^2\frac A2=\frac{49}{72} \\ &\implies \cos\frac A2=\frac{7}{6\sqrt2}\end{aligned}\]
\(\cos\dfrac A2=\dfrac{7}{6\sqrt2}\)
c. If \(\tan A=\frac43\), find the value of \(\tan\frac A2\). (यदि \(\tan A=\frac43\) भए \(\tan\frac A2\) को मान पत्ता लगाउनुहोस् ।)

Let \(t=\tan\frac A2\). Use \(\tan A=\dfrac{2t}{1-t^2}\).

\[\begin{aligned}&\frac43=\frac{2t}{1-t^2} \\ &\implies 4-4t^2=6t \\ &\implies 2t^2+3t-2=0 \\ &\implies 2t^2+4t-t-2=0 \\ &\implies (2t-1)(t+2)=0\end{aligned}\]

So \(t=\frac12\) or \(t=-2\).

\(\tan\dfrac A2=\dfrac12\) or \(\tan\dfrac A2=-2\)

Q.5 If \(\cos30^\circ=\frac{\sqrt3}{2}\), prove that:

  • i. \(\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}\)
  • ii. \(\cos15^\circ=\dfrac{\sqrt3+1}{2\sqrt2}\)
  • iii. \(\tan15^\circ=2-\sqrt3\)

(यदि \(\cos30^\circ=\frac{\sqrt3}{2}\) भए प्रमाणित गर्नुहोस् : )

i. \(\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}\)

Given \(\cos30^\circ=\dfrac{\sqrt3}{2}\).

Use \(\cos2X=1-2\sin^2X\) with \(X=15^\circ\).

\[\begin{aligned}\cos30^\circ&=1-2\sin^215^\circ \\ \frac{\sqrt3}{2}&=1-2\sin^215^\circ \\ 2\sin^215^\circ&=\frac{2-\sqrt3}{2} \\ \sin^215^\circ&=\frac{2-\sqrt3}{4}=\frac{4-2\sqrt3}{8} \\ &=\frac{(\sqrt3-1)^2}{(2\sqrt2)^2}\end{aligned}\]

Since \(15^\circ\) is acute, \(\sin15^\circ>0\). Hence

\[\begin{aligned}\sin15^\circ&=\frac{\sqrt3-1}{2\sqrt2}\end{aligned}\]
Hence proved.
ii. \(\cos15^\circ=\dfrac{\sqrt3+1}{2\sqrt2}\)

Given \(\cos30^\circ=\dfrac{\sqrt3}{2}\).

Use \(\cos2X=2\cos^2X-1\) with \(X=15^\circ\).

\[\begin{aligned}\cos30^\circ&=2\cos^215^\circ-1 \\ \frac{\sqrt3}{2}&=2\cos^215^\circ-1 \\ 2\cos^215^\circ&=\frac{2+\sqrt3}{2} \\ \cos^215^\circ&=\frac{2+\sqrt3}{4}=\frac{4+2\sqrt3}{8} \\ &=\frac{(\sqrt3+1)^2}{(2\sqrt2)^2}\end{aligned}\]

Since \(15^\circ\) is acute, \(\cos15^\circ>0\). Hence

\[\begin{aligned}\cos15^\circ&=\frac{\sqrt3+1}{2\sqrt2}\end{aligned}\]
Hence proved.
iii. \(\tan15^\circ=2-\sqrt3\)

Using parts (i) and (ii):

\[\begin{aligned}\tan15^\circ&=\frac{\sin15^\circ}{\cos15^\circ} \\ &=\frac{\sqrt3-1}{\sqrt3+1} \\ &=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)} \\ &=\frac{4-2\sqrt3}{2} \\ &=2-\sqrt3\end{aligned}\]
Hence proved.
Answer
1. A. d   B. b   C. c   D. a   E. c
3. a. \(\frac{24}{25},-\frac{7}{25},-\frac{24}{7}\)   b. \(\frac{24}{25},\frac{7}{25},\frac{24}{7}\)   c. \(\frac{120}{169},-\frac{119}{169},-\frac{120}{119}\)   d. \(1,0,\) undefined   e. \(\frac{117}{125},-\frac{44}{125},-\frac{117}{44}\)   f. \(\frac{11}{2}\)
4. a. \(\frac35\)   b. \(\frac{7}{6\sqrt2}\)   c. \(\frac12\) or \(-2\)
2, 5. Proved or expressed as shown

3. Frequently Asked Questions on Submultiple Angle Formula (Class 10 Optional Mathematics)

What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.3 Submultiple Angles? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.3 Submultiple Angles मा के के समावेश छ ?)

This page solves all 22 questions of Exercise 7.3 Submultiple Angles, in textbook order, with the Nepali question in brackets and an Answer box at the end. Submultiple angle formula, proof and examples with all Exercise 7.3 questions solved. Multiple and submultiple angles for Class 10 Optional Math SEE.

What is a submultiple angle? (अपवर्तक कोण भनेको के हो ?)

An angle obtained by dividing a given angle \(A\) by \(2,3,4,\dots\) is a submultiple angle of \(A\), for example \(\dfrac A2\) and \(\dfrac A3\). For \(A=60^\circ\) they are \(30^\circ\), \(20^\circ\) and \(15^\circ\) (Question 2a).

What is the submultiple angle formula? (अपवर्तक कोणका सूत्रहरू के के हुन् ?)
  • \(\sin A=2\sin\frac A2\cos\frac A2=\dfrac{2\tan\frac A2}{1+\tan^2\frac A2}\)
  • \(\cos A=\cos^2\frac A2-\sin^2\frac A2=\dfrac{1-\tan^2\frac A2}{1+\tan^2\frac A2}\)
  • \(\tan A=\dfrac{2\tan\frac A2}{1-\tan^2\frac A2}\)
  • \(\sin\theta=3\sin\frac\theta3-4\sin^3\frac\theta3\) and \(\cos\theta=4\cos^3\frac\theta3-3\cos\frac\theta3\)
What is the proof of the submultiple angle formula? (अपवर्तक कोणको सूत्रको प्रमाण कसरी गर्ने ?)

Start from the double angle formula \(\cos2X=1-2\sin^2X\) and put \(X=\frac A2\). Then \(\cos A=1-2\sin^2\frac A2\), so \(\sin^2\frac A2=\dfrac{1-\cos A}{2}\). In the same way \(\sin2X=2\sin X\cos X\) with \(X=\frac A2\) gives \(\sin A=2\sin\frac A2\cos\frac A2\).

Can you give a submultiple angle example? (अपवर्तक कोणको उदाहरण दिनुहोस् ।)

Example: if \(\cos30^\circ=\frac{\sqrt3}{2}\), then \(15^\circ\) is a submultiple angle of \(30^\circ\) and \(\sin^215^\circ=\dfrac{1-\cos30^\circ}{2}=\dfrac{2-\sqrt3}{4}\), so \(\sin15^\circ=\dfrac{\sqrt3-1}{2\sqrt2}\) (Question 5).

How do I find \(\sin A\), \(\cos A\) and \(\tan A\) when \(\tan\frac A2\) is given? (\(\tan\frac A2\) दिइएको छ भने \(\sin A\), \(\cos A\) र \(\tan A\) कसरी निकाल्ने ?)

Use the three formulas with \(\tan\frac A2\). For \(\tan\frac A2=\frac34\) these give \(\frac{24}{25}\), \(\frac7{25}\) and \(\frac{24}7\) (Question 3b).

Are multiple and submultiple angles studied in Class 11 as well? (के बहुगुणित र अपवर्तक कोण कक्षा ११ मा पनि पढाइन्छ ?)

The same multiple and submultiple angle formulas appear again in higher classes (for example Class 11 trigonometry), so the practice in Exercise 7.3 is useful beyond SEE.

Is there a PDF of the multiple and submultiple angles formulas? (बहुगुणित र अपवर्तक कोणका सूत्रको सूची कहाँ पाइन्छ ?)

The formula boxes at the top of this page list the submultiple angle formulas, and the multiple angle formulas are on the Exercise 7.2 page. Each solution here shows which formula is used.

Are these Exercise 7.3 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)

Yes. Exercise 7.3 covers expressing one angle in terms of its half or one-third angle and finding values from a given submultiple ratio, which are standard SEE-style question types.

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