/ Class 10 Optional Math, SEE
Transformation of Product and Sum Exercise 7.4 Solutions
Importantedunotes.com
Back to Optional Mathematics Notes

1. Transformation of Products and Sums: Class 10 Optional Mathematics Trigonometry Overview and Key Formulas

Unit 7: Trigonometry (Optional Mathematics, Grade 10, SEE, Nepal) • Exercise 7.4 Transformation of Products and Sums

The transformation of a product into a sum or difference, and of a sum or difference into a product, is the topic of Exercise 7.4. This page gives the transformation of product and sum formulas, the sum-to-product formula proof, transformation of product and sum examples and step by step solutions to every question of Exercise 7.4 (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.

  • Transformation of products and sums (Exercise 7.4): add or subtract the compound angle formulas to turn a product such as \(2\sin A\cos B\) into a sum, and put \(A=\frac{C+D}{2}\), \(B=\frac{C-D}{2}\) to turn a sum such as \(\sin C+\sin D\) into a product.

Key formulas to remember

Product to sum: \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\), \(2\cos A\sin B=\sin(A+B)-\sin(A-B)\), \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\), \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\).
Sum to product: \(\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2}\), \(\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\), \(\cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}\), \(\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\).
Proof idea for the sum to product formula: in \(2\sin X\cos Y=\sin(X+Y)+\sin(X-Y)\) put \(X=\frac{C+D}{2}\) and \(Y=\frac{C-D}{2}\). Then \(\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\).

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

2. Exercise 7.4 Transformation of Products and Sums: Class 10 Optional Mathematics Trigonometry

Exercise 7.4 Transformation of Products and Sums: 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 trigonometric expressions is equal to \(2\sin A\cos B\)? (तलका मध्ये \(2\sin A\cos B\) सँग बराबर हुने त्रिकोणमितीय अभिव्यञ्जक कुन हो ?)

  • a) \(\sin(A+B)-\sin(A-B)\)
  • b) \(\sin(A+B)+\sin(A-B)\)
  • c) \(\cos(A+B)+\cos(A-B)\)
  • d) \(\cos(A+B)-\sin(A-B)\)

Correct Answer: b) \(\sin(A+B)+\sin(A-B)\)

Solution:

  • Add \(\sin(A+B)=\sin A\cos B+\cos A\sin B\) and \(\sin(A-B)=\sin A\cos B-\cos A\sin B\).
\[\sin(A+B)+\sin(A-B)=2\sin A\cos B\]

(दुवै सूत्र जोड्दा \(\sin(A+B)+\sin(A-B)=2\sin A\cos B\) हुन्छ ।)

B. Which of the trigonometric expressions is equal to \(\sin A-\sin B\)? (तलका मध्ये \(\sin A-\sin B\) सँग बराबर हुने त्रिकोणमितीय अभिव्यञ्जक कुन हो ?)

  • a) \(2\sin\frac{A+B}{2}\cos\frac{A-B}{2}\)
  • b) \(-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\)
  • c) \(2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\)
  • d) \(2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\)

Correct Answer: c) \(2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\)

Solution:

  • Use the formula \(\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\).
\[\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\]

(\(\sin C-\sin D\) को सूत्र अनुसार \(\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}\) हुन्छ ।)

C. Which of the following trigonometric expressions is equal to \(\cos C-\cos D\)? (तलका मध्ये \(\cos C-\cos D\) सँग बराबर हुने त्रिकोणमितीय अभिव्यञ्जक कुन हो ?)

  • a) \(2\sin\frac{C+D}{2}\cos\frac{C-D}{2}\)
  • b) \(2\cos\frac{C+D}{2}\cos\frac{C-D}{2}\)
  • c) \(-2\sin\frac{C+D}{2}\sin\frac{D-C}{2}\)
  • d) \(-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\)

Correct Answer: d) \(-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\)

Solution:

  • Use the formula \(\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\).
  • Option (c) is not equal to it: \(\sin\frac{D-C}{2}=-\sin\frac{C-D}{2}\), so option (c) equals \(+2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\).
\[\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\]

(\(\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}\) हुन्छ, जुन विकल्प (घ) हो ।)

D. What is obtained when \(\cos6A+\cos4A\) is transformed into product form? (\(\cos6A+\cos4A\) लाई गुणनफलमा रूपान्तरण गर्दा के हुन्छ ?)

  • a) \(2\sin6A\cos4A\)
  • b) \(2\sin5A\cos A\)
  • c) \(2\cos5A\sin A\)
  • d) \(2\cos5A\cos A\)

Correct Answer: d) \(2\cos5A\cos A\)

Solution:

  • Use \(\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\) with \(C=6A\), \(D=4A\).
\[\begin{aligned}\cos6A+\cos4A&=2\cos\frac{10A}{2}\cos\frac{2A}{2} \\ &=2\cos5A\cos A\end{aligned}\]

(\(\cos C+\cos D\) को सूत्रमा \(C=6A,\ D=4A\) राख्दा \(2\cos5A\cos A\) हुन्छ ।)

E. Which of the following is equal to \(\sin60^\circ+\sin40^\circ\)? (\(\sin60^\circ+\sin40^\circ\) कुनसँग बराबर हुन्छ ?)

  • a) \(2\sin50^\circ\cos10^\circ\)
  • b) \(2\sin50^\circ\sin10^\circ\)
  • c) \(2\cos50^\circ\sin10^\circ\)
  • d) \(2\cos50^\circ\cos5^\circ\)

Correct Answer: a) \(2\sin50^\circ\cos10^\circ\)

Solution:

  • Use \(\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\) with \(C=60^\circ\), \(D=40^\circ\).
\[\begin{aligned}\sin60^\circ+\sin40^\circ&=2\sin\frac{100^\circ}{2}\cos\frac{20^\circ}{2} \\ &=2\sin50^\circ\cos10^\circ\end{aligned}\]

(\(\sin C+\sin D\) को सूत्र प्रयोग गर्दा \(2\sin50^\circ\cos10^\circ\) हुन्छ ।)

Exercise 7.4 Transformation of Products and Sums: Written Questions (लेख्ने प्रश्न)

Q.2 Transform the following trigonometric expressions in product form into sum or difference form:

  • a. \(\sin50^\circ\cos30^\circ\)
  • b. \(\cos25^\circ\cos15^\circ\)
  • c. \(\cos10A\sin6A\)
  • d. \(\sin27^\circ\sin32^\circ\)

(दिइएका गुणनफल रूपका त्रिकोणमितीय व्यञ्जकहरूलाई योग वा अन्तरको रूपमा रूपान्तरण गर्नुहोस् : )

a. \(\sin50^\circ\cos30^\circ\)

Use \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\).

\[\begin{aligned}\sin50^\circ\cos30^\circ&=\frac12\left[2\sin50^\circ\cos30^\circ\right] \\ &=\frac12\left[\sin80^\circ+\sin20^\circ\right]\end{aligned}\]
\(\dfrac{\sin80^\circ+\sin20^\circ}{2}\)
b. \(\cos25^\circ\cos15^\circ\)

Use \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\).

\[\begin{aligned}\cos25^\circ\cos15^\circ&=\frac12\left[2\cos25^\circ\cos15^\circ\right] \\ &=\frac12\left[\cos40^\circ+\cos10^\circ\right]\end{aligned}\]
\(\dfrac{\cos40^\circ+\cos10^\circ}{2}\)
c. \(\cos10A\sin6A\)

Use \(2\cos A\sin B=\sin(A+B)-\sin(A-B)\).

\[\begin{aligned}\cos10A\sin6A&=\frac12\left[2\cos10A\sin6A\right] \\ &=\frac12\left[\sin16A-\sin4A\right]\end{aligned}\]
\(\dfrac{\sin16A-\sin4A}{2}\)
d. \(\sin27^\circ\sin32^\circ\)

Use \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\) with \(A=32^\circ\), \(B=27^\circ\).

\[\begin{aligned}\sin27^\circ\sin32^\circ&=\frac12\left[2\sin32^\circ\sin27^\circ\right] \\ &=\frac12\left[\cos5^\circ-\cos59^\circ\right]\end{aligned}\]
\(\dfrac{\cos5^\circ-\cos59^\circ}{2}\)

Q.3 Transform the following trigonometric expressions in sum or difference form into product form:

  • a. \(\sin50^\circ+\sin30^\circ\)
  • b. \(\cos25^\circ+\cos15^\circ\)
  • c. \(\cos7A-\cos5A\)
  • d. \(\sin67^\circ-\sin43^\circ\)

(दिइएका योग वा अन्तर रूपका त्रिकोणमितीय व्यञ्जकहरूलाई गुणनफल रूपमा रूपान्तरण गर्नुहोस् : )

a. \(\sin50^\circ+\sin30^\circ\)

Use \(\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\).

\[\begin{aligned}\sin50^\circ+\sin30^\circ&=2\sin\frac{80^\circ}{2}\cos\frac{20^\circ}{2} \\ &=2\sin40^\circ\cos10^\circ\end{aligned}\]
\(2\sin40^\circ\cos10^\circ\)
b. \(\cos25^\circ+\cos15^\circ\)

Use \(\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\).

\[\begin{aligned}\cos25^\circ+\cos15^\circ&=2\cos\frac{40^\circ}{2}\cos\frac{10^\circ}{2} \\ &=2\cos20^\circ\cos5^\circ\end{aligned}\]
\(2\cos20^\circ\cos5^\circ\)
c. \(\cos7A-\cos5A\)

Use \(\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\).

\[\begin{aligned}\cos7A-\cos5A&=-2\sin\frac{12A}{2}\sin\frac{2A}{2} \\ &=-2\sin6A\sin A\end{aligned}\]
\(-2\sin6A\sin A\)
d. \(\sin67^\circ-\sin43^\circ\)

Use \(\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\).

\[\begin{aligned}\sin67^\circ-\sin43^\circ&=2\cos\frac{110^\circ}{2}\sin\frac{24^\circ}{2} \\ &=2\cos55^\circ\sin12^\circ\end{aligned}\]
\(2\cos55^\circ\sin12^\circ\)

Q.4 Prove that:

  • a. \(\cos50^\circ+\cos40^\circ=\sqrt2\cos5^\circ\)
  • b. \(\sin50^\circ-\sin70^\circ=-\sin10^\circ\)
  • c. \(\sin10^\circ+\sin70^\circ=\sqrt3\sin40^\circ\)
  • d. \(\cos10^\circ+\cos70^\circ=\sqrt3\cos40^\circ\)
  • e. \(\cos40^\circ-\sin40^\circ=\sqrt2\sin5^\circ\)
  • f. \(\sin37^\circ-\cos67^\circ=\sqrt3\sin7^\circ\)

(प्रमाणित गर्नुहोस् : )

a. \(\cos50^\circ+\cos40^\circ=\sqrt2\cos5^\circ\)
\[\begin{aligned}\text{L.H.S.}&=\cos50^\circ+\cos40^\circ \\ &=2\cos\frac{50^\circ+40^\circ}{2}\cos\frac{50^\circ-40^\circ}{2} \\ &=2\cos45^\circ\cos5^\circ \\ &=2\cdot\frac1{\sqrt2}\cos5^\circ \\ &=\sqrt2\cos5^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\sin50^\circ-\sin70^\circ=-\sin10^\circ\)
\[\begin{aligned}\text{L.H.S.}&=\sin50^\circ-\sin70^\circ \\ &=2\cos\frac{50^\circ+70^\circ}{2}\sin\frac{50^\circ-70^\circ}{2} \\ &=2\cos60^\circ\sin(-10^\circ) \\ &=2\cdot\frac12\cdot(-\sin10^\circ) \\ &=-\sin10^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\sin10^\circ+\sin70^\circ=\sqrt3\sin40^\circ\)
\[\begin{aligned}\text{L.H.S.}&=\sin70^\circ+\sin10^\circ \\ &=2\sin\frac{70^\circ+10^\circ}{2}\cos\frac{70^\circ-10^\circ}{2} \\ &=2\sin40^\circ\cos30^\circ \\ &=2\sin40^\circ\cdot\frac{\sqrt3}{2} \\ &=\sqrt3\sin40^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\cos10^\circ+\cos70^\circ=\sqrt3\cos40^\circ\)
\[\begin{aligned}\text{L.H.S.}&=\cos70^\circ+\cos10^\circ \\ &=2\cos\frac{70^\circ+10^\circ}{2}\cos\frac{70^\circ-10^\circ}{2} \\ &=2\cos40^\circ\cos30^\circ \\ &=\sqrt3\cos40^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\cos40^\circ-\sin40^\circ=\sqrt2\sin5^\circ\)

Here \(\cos40^\circ=\sin(90^\circ-40^\circ)=\sin50^\circ\).

\[\begin{aligned}\text{L.H.S.}&=\sin50^\circ-\sin40^\circ \\ &=2\cos\frac{50^\circ+40^\circ}{2}\sin\frac{50^\circ-40^\circ}{2} \\ &=2\cos45^\circ\sin5^\circ \\ &=\sqrt2\sin5^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\sin37^\circ-\cos67^\circ=\sqrt3\sin7^\circ\)

Here \(\cos67^\circ=\sin(90^\circ-67^\circ)=\sin23^\circ\).

\[\begin{aligned}\text{L.H.S.}&=\sin37^\circ-\sin23^\circ \\ &=2\cos\frac{37^\circ+23^\circ}{2}\sin\frac{37^\circ-23^\circ}{2} \\ &=2\cos30^\circ\sin7^\circ \\ &=\sqrt3\sin7^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.5 Prove that:

  • a. \(\dfrac{\cos10^\circ-\sin10^\circ}{\cos10^\circ+\sin10^\circ}=\tan35^\circ\)
  • b. \(\dfrac{\cos9^\circ-\sin9^\circ}{\cos9^\circ+\sin9^\circ}=\cot54^\circ\)
  • c. \(\dfrac{\cos12^\circ-\sin12^\circ}{\cos12^\circ+\sin12^\circ}=\tan33^\circ\)
  • d. \(\dfrac{\sin A-\sin B}{\sin A+\sin B}=\cot\dfrac{A+B}{2}\tan\dfrac{A-B}{2}\)
  • e. \(\dfrac{\cos B-\cos A}{\sin A+\sin B}=\tan\dfrac{A-B}{2}\)
  • f. \(\dfrac{\sin(60^\circ+A)+\sin(60^\circ-A)}{\cos(60^\circ+A)+\cos(60^\circ-A)}=\sqrt3\)
  • g. \(2\sin(45^\circ+\phi)\cos(45^\circ-\phi)=1+\sin2\phi\)
  • h. \(2\cos(45^\circ+\theta)\cos(45^\circ-\theta)=\cos2\theta\)

(प्रमाणित गर्नुहोस् : )

a. \(\dfrac{\cos10^\circ-\sin10^\circ}{\cos10^\circ+\sin10^\circ}=\tan35^\circ\)

Here \(\cos10^\circ=\sin80^\circ\).

\[\begin{aligned}\text{L.H.S.}&=\frac{\sin80^\circ-\sin10^\circ}{\sin80^\circ+\sin10^\circ} \\ &=\frac{2\cos45^\circ\sin35^\circ}{2\sin45^\circ\cos35^\circ} \\ &=\frac{\sin35^\circ}{\cos35^\circ}\quad(\because\sin45^\circ=\cos45^\circ) \\ &=\tan35^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\dfrac{\cos9^\circ-\sin9^\circ}{\cos9^\circ+\sin9^\circ}=\cot54^\circ\)

Divide the numerator and denominator by \(\cos9^\circ\):

\[\begin{aligned}\text{L.H.S.}&=\frac{1-\tan9^\circ}{1+\tan9^\circ} \\ &=\frac{\tan45^\circ-\tan9^\circ}{1+\tan45^\circ\tan9^\circ} \\ &=\tan(45^\circ-9^\circ) \\ &=\tan36^\circ \\ &=\cot(90^\circ-36^\circ) \\ &=\cot54^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\dfrac{\cos12^\circ-\sin12^\circ}{\cos12^\circ+\sin12^\circ}=\tan33^\circ\)

Divide the numerator and denominator by \(\cos12^\circ\):

\[\begin{aligned}\text{L.H.S.}&=\frac{1-\tan12^\circ}{1+\tan12^\circ} \\ &=\frac{\tan45^\circ-\tan12^\circ}{1+\tan45^\circ\tan12^\circ} \\ &=\tan(45^\circ-12^\circ) \\ &=\tan33^\circ \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\dfrac{\sin A-\sin B}{\sin A+\sin B}=\cot\dfrac{A+B}{2}\tan\dfrac{A-B}{2}\)
\[\begin{aligned}\text{L.H.S.}&=\frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}} \\ &=\frac{\cos\frac{A+B}{2}}{\sin\frac{A+B}{2}}\cdot\frac{\sin\frac{A-B}{2}}{\cos\frac{A-B}{2}} \\ &=\cot\frac{A+B}{2}\tan\frac{A-B}{2} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\dfrac{\cos B-\cos A}{\sin A+\sin B}=\tan\dfrac{A-B}{2}\)

Here \(\cos B-\cos A=2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\).

\[\begin{aligned}\text{L.H.S.}&=\frac{2\sin\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}} \\ &=\frac{\sin\frac{A-B}{2}}{\cos\frac{A-B}{2}} \\ &=\tan\frac{A-B}{2} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\dfrac{\sin(60^\circ+A)+\sin(60^\circ-A)}{\cos(60^\circ+A)+\cos(60^\circ-A)}=\sqrt3\)

Use \(\sin(X+Y)+\sin(X-Y)=2\sin X\cos Y\) and \(\cos(X+Y)+\cos(X-Y)=2\cos X\cos Y\).

\[\begin{aligned}\text{L.H.S.}&=\frac{2\sin60^\circ\cos A}{2\cos60^\circ\cos A} \\ &=\tan60^\circ \\ &=\sqrt3 \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
g. \(2\sin(45^\circ+\phi)\cos(45^\circ-\phi)=1+\sin2\phi\)

Use \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\).

\[\begin{aligned}\text{L.H.S.}&=\sin\left[(45^\circ+\phi)+(45^\circ-\phi)\right]+\sin\left[(45^\circ+\phi)-(45^\circ-\phi)\right] \\ &=\sin90^\circ+\sin2\phi \\ &=1+\sin2\phi \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
h. \(2\cos(45^\circ+\theta)\cos(45^\circ-\theta)=\cos2\theta\)

Use \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\).

\[\begin{aligned}\text{L.H.S.}&=\cos90^\circ+\cos2\theta \\ &=0+\cos2\theta \\ &=\cos2\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.6 Prove that:

  • a. \(\dfrac{\sin\theta+\sin3\theta-\sin2\theta}{\cos\theta-\cos2\theta+\cos3\theta}=\tan2\theta\)
  • b. \(\dfrac{\sin2A+\sin3A+\sin5A+\sin6A}{\cos2A+\cos3A+\cos5A+\cos6A}=\tan4A\)
  • c. \(\dfrac{\sin A-\sin3A+\sin5A-\sin7A}{\cos A-\cos3A-\cos5A+\cos7A}=\cot3A\)
  • d. \(\dfrac{\cos3A\cos2A-\sin5A\sin7A}{\sin4A\sin3A-\sin2A\sin5A}=\cot A\)

(प्रमाणित गर्नुहोस् : )

a. \(\dfrac{\sin\theta+\sin3\theta-\sin2\theta}{\cos\theta-\cos2\theta+\cos3\theta}=\tan2\theta\)

Group \(\sin3\theta+\sin\theta=2\sin2\theta\cos\theta\) and \(\cos3\theta+\cos\theta=2\cos2\theta\cos\theta\).

\[\begin{aligned}\text{L.H.S.}&=\frac{(\sin3\theta+\sin\theta)-\sin2\theta}{(\cos3\theta+\cos\theta)-\cos2\theta} \\ &=\frac{2\sin2\theta\cos\theta-\sin2\theta}{2\cos2\theta\cos\theta-\cos2\theta} \\ &=\frac{\sin2\theta(2\cos\theta-1)}{\cos2\theta(2\cos\theta-1)} \\ &=\tan2\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\dfrac{\sin2A+\sin3A+\sin5A+\sin6A}{\cos2A+\cos3A+\cos5A+\cos6A}=\tan4A\)

Pair the angles whose sum is \(8A\): \(6A,2A\) and \(5A,3A\).

\[\begin{aligned}\text{L.H.S.}&=\frac{(\sin6A+\sin2A)+(\sin5A+\sin3A)}{(\cos6A+\cos2A)+(\cos5A+\cos3A)} \\ &=\frac{2\sin4A\cos2A+2\sin4A\cos A}{2\cos4A\cos2A+2\cos4A\cos A} \\ &=\frac{2\sin4A(\cos2A+\cos A)}{2\cos4A(\cos2A+\cos A)} \\ &=\tan4A \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\dfrac{\sin A-\sin3A+\sin5A-\sin7A}{\cos A-\cos3A-\cos5A+\cos7A}=\cot3A\)

Note on the printed question

As printed, the right side should be \(\cot2A\), not \(\cot3A\). For \(A=10^\circ\) the left side is \(\approx2.7475=\cot20^\circ\), while \(\cot30^\circ\approx1.7321\). We therefore prove the corrected relation.

Proof of the corrected relation

\[\begin{aligned}\text{Numerator}&=(\sin A-\sin7A)+(\sin5A-\sin3A) \\ &=-2\cos4A\sin3A+2\cos4A\sin A \\ &=-2\cos4A(\sin3A-\sin A) \\ &=-4\cos4A\cos2A\sin A\end{aligned}\]\[\begin{aligned}\text{Denominator}&=(\cos A-\cos3A)+(\cos7A-\cos5A) \\ &=2\sin2A\sin A-2\sin6A\sin A \\ &=2\sin A(\sin2A-\sin6A) \\ &=-4\sin A\cos4A\sin2A\end{aligned}\]\[\begin{aligned}\text{L.H.S.}&=\frac{-4\cos4A\cos2A\sin A}{-4\sin A\cos4A\sin2A} \\ &=\frac{\cos2A}{\sin2A} \\ &=\cot2A\end{aligned}\]
The expression equals \(\cot2A\) (the printed \(\cot3A\) does not hold)
d. \(\dfrac{\cos3A\cos2A-\sin5A\sin7A}{\sin4A\sin3A-\sin2A\sin5A}=\cot A\)

Note on the printed question

As printed, this relation is not true. For \(A=10^\circ\), the left side is \(\approx1.582\) but \(\cot10^\circ\approx5.671\). The simplification below shows why: the numerator does not reduce to a multiple of the denominator, so there appears to be a misprint in the question.

What the expression simplifies to

\[\begin{aligned}2\times\text{Numerator}&=2\cos3A\cos2A-2\sin7A\sin5A \\ &=(\cos5A+\cos A)-(\cos2A-\cos12A)\end{aligned}\]\[\begin{aligned}2\times\text{Denominator}&=2\sin4A\sin3A-2\sin5A\sin2A \\ &=(\cos A-\cos7A)-(\cos3A-\cos7A) \\ &=\cos A-\cos3A=2\sin2A\sin A\end{aligned}\]

So the denominator is \(\sin2A\sin A\), and the numerator is \(\tfrac12(\cos12A-\cos2A+\cos5A+\cos A)\), which does not simplify to \(\cos A\sin2A\).

The identity cannot be proved as printed (see the note above)

Q.7 Prove that:

  • a. \(4\cos\theta\cos(60^\circ+\theta)\cos(60^\circ-\theta)=\cos3\theta\)
  • b. \(4\cot\theta\cot(60^\circ+\theta)\cot(60^\circ-\theta)=\cot3\theta\)
  • c. \(\sec\left(45^\circ+\dfrac\theta2\right)\sec\left(45^\circ-\dfrac\theta2\right)=2\sec\theta\)
  • d. \(\tan\left(\dfrac{\pi^c}{4}+\theta\right)-\tan\left(\dfrac{\pi^c}{4}-\theta\right)=2\tan2\theta\)

(प्रमाणित गर्नुहोस् : )

a. \(4\cos\theta\cos(60^\circ+\theta)\cos(60^\circ-\theta)=\cos3\theta\)
\[\begin{aligned}\text{L.H.S.}&=2\cos\theta\left[2\cos(60^\circ+\theta)\cos(60^\circ-\theta)\right] \\ &=2\cos\theta\left[\cos120^\circ+\cos2\theta\right] \\ &=2\cos\theta\left[-\frac12+\cos2\theta\right] \\ &=-\cos\theta+2\cos2\theta\cos\theta \\ &=-\cos\theta+(\cos3\theta+\cos\theta) \\ &=\cos3\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(4\cot\theta\cot(60^\circ+\theta)\cot(60^\circ-\theta)=\cot3\theta\)

First prove \(4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta)=\sin3\theta\):

\[\begin{aligned}4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta)&=2\sin\theta\left[\cos2\theta-\cos120^\circ\right] \\ &=2\sin\theta\cos2\theta+\sin\theta \\ &=(\sin3\theta-\sin\theta)+\sin\theta \\ &=\sin3\theta\end{aligned}\]

By part (a), \(4\cos\theta\cos(60^\circ+\theta)\cos(60^\circ-\theta)=\cos3\theta\). Hence

\[\begin{aligned}\text{L.H.S.}&=\frac{4\cos\theta\cos(60^\circ+\theta)\cos(60^\circ-\theta)}{4\sin\theta\sin(60^\circ+\theta)\sin(60^\circ-\theta)} \\ &=\frac{\cos3\theta}{\sin3\theta} \\ &=\cot3\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\sec\left(45^\circ+\dfrac\theta2\right)\sec\left(45^\circ-\dfrac\theta2\right)=2\sec\theta\)
\[\begin{aligned}\text{L.H.S.}&=\frac{2}{2\cos\left(45^\circ+\frac\theta2\right)\cos\left(45^\circ-\frac\theta2\right)} \\ &=\frac{2}{\cos90^\circ+\cos\theta} \\ &=\frac{2}{\cos\theta} \\ &=2\sec\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\tan\left(\dfrac{\pi^c}{4}+\theta\right)-\tan\left(\dfrac{\pi^c}{4}-\theta\right)=2\tan2\theta\)

Here \(\dfrac{\pi^c}{4}=45^\circ\), \(\tan(45^\circ+\theta)=\dfrac{1+\tan\theta}{1-\tan\theta}\) and \(\tan(45^\circ-\theta)=\dfrac{1-\tan\theta}{1+\tan\theta}\).

\[\begin{aligned}\text{L.H.S.}&=\frac{(1+\tan\theta)^2-(1-\tan\theta)^2}{1-\tan^2\theta} \\ &=\frac{4\tan\theta}{1-\tan^2\theta} \\ &=2\cdot\frac{2\tan\theta}{1-\tan^2\theta} \\ &=2\tan2\theta \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.

Q.8 Prove that:

  • a. \(\sin10^\circ\sin50^\circ\sin70^\circ=\dfrac18\)
  • b. \(\cos10^\circ\cos50^\circ\cos70^\circ=\dfrac{\sqrt3}{8}\)
  • c. \(\sin10^\circ\sin30^\circ\sin50^\circ\sin70^\circ=\dfrac1{16}\)
  • d. \(\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\dfrac{\sqrt3}{16}\)
  • e. \(\cos20^\circ\cos30^\circ\cos40^\circ\cos80^\circ=\dfrac{\sqrt3}{16}\)
  • f. \(\cos20^\circ\cos40^\circ\cos80^\circ=\dfrac18\)

(प्रमाणित गर्नुहोस् : )

a. \(\sin10^\circ\sin50^\circ\sin70^\circ=\dfrac18\)

Using \(\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\frac14\sin3\theta\) (Question 7b) with \(\theta=10^\circ\):

\[\begin{aligned}\text{L.H.S.}&=\sin10^\circ\sin(60^\circ-10^\circ)\sin(60^\circ+10^\circ) \\ &=\frac14\sin30^\circ \\ &=\frac14\cdot\frac12 \\ &=\frac18 \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
b. \(\cos10^\circ\cos50^\circ\cos70^\circ=\dfrac{\sqrt3}{8}\)

Using \(\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)=\frac14\cos3\theta\) (Question 7a) with \(\theta=10^\circ\):

\[\begin{aligned}\text{L.H.S.}&=\frac14\cos30^\circ \\ &=\frac14\cdot\frac{\sqrt3}{2} \\ &=\frac{\sqrt3}{8} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
c. \(\sin10^\circ\sin30^\circ\sin50^\circ\sin70^\circ=\dfrac1{16}\)

Using part (a), \(\sin10^\circ\sin50^\circ\sin70^\circ=\frac18\):

\[\begin{aligned}\text{L.H.S.}&=\sin30^\circ\left(\sin10^\circ\sin50^\circ\sin70^\circ\right) \\ &=\frac12\cdot\frac18 \\ &=\frac1{16} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
d. \(\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\dfrac{\sqrt3}{16}\)

Using part (a):

\[\begin{aligned}\text{L.H.S.}&=\sin60^\circ\left(\sin10^\circ\sin50^\circ\sin70^\circ\right) \\ &=\frac{\sqrt3}{2}\cdot\frac18 \\ &=\frac{\sqrt3}{16} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
e. \(\cos20^\circ\cos30^\circ\cos40^\circ\cos80^\circ=\dfrac{\sqrt3}{16}\)

Here \(40^\circ=60^\circ-20^\circ\) and \(80^\circ=60^\circ+20^\circ\). Using \(\cos\theta\cos(60^\circ-\theta)\cos(60^\circ+\theta)=\frac14\cos3\theta\) with \(\theta=20^\circ\):

\[\begin{aligned}\cos20^\circ\cos40^\circ\cos80^\circ&=\frac14\cos60^\circ \\ &=\frac18\end{aligned}\]\[\begin{aligned}\text{L.H.S.}&=\cos30^\circ\cdot\frac18 \\ &=\frac{\sqrt3}{2}\cdot\frac18 \\ &=\frac{\sqrt3}{16} \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
f. \(\cos20^\circ\cos40^\circ\cos80^\circ=\dfrac18\)

Multiply and divide by \(2\sin20^\circ\), using \(2\sin x\cos x=\sin2x\) repeatedly:

\[\begin{aligned}\text{L.H.S.}&=\frac{2\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ} \\ &=\frac{\sin40^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ} \\ &=\frac{2\sin40^\circ\cos40^\circ\cos80^\circ}{4\sin20^\circ} \\ &=\frac{\sin80^\circ\cos80^\circ}{4\sin20^\circ} \\ &=\frac{2\sin80^\circ\cos80^\circ}{8\sin20^\circ} \\ &=\frac{\sin160^\circ}{8\sin20^\circ} \\ &=\frac{\sin20^\circ}{8\sin20^\circ}\quad(\because\sin160^\circ=\sin20^\circ) \\ &=\frac18 \\ &=\text{R.H.S.}\end{aligned}\]
Therefore, L.H.S. = R.H.S., proved.
Answer
1. A. b   B. c   C. d   D. d   E. a
2. a. \(\frac{\sin80^\circ+\sin20^\circ}{2}\)   b. \(\frac{\cos40^\circ+\cos10^\circ}{2}\)   c. \(\frac{\sin16A-\sin4A}{2}\)   d. \(\frac{\cos5^\circ-\cos59^\circ}{2}\)
3. a. \(2\sin40^\circ\cos10^\circ\)   b. \(2\cos20^\circ\cos5^\circ\)   c. \(-2\sin6A\sin A\)   d. \(2\cos55^\circ\sin12^\circ\)
4–8. Proved (see the notes for Questions 6c and 6d)

3. Frequently Asked Questions on Transformation of Product and Sum (Class 10 Optional Mathematics)

What is covered in Class 10 Optional Mathematics Trigonometry Exercise 7.4 Transformation of Products and Sums? (कक्षा १० ऐच्छिक गणित त्रिकोणमितिको अभ्यास 7.4 Transformation of Products and Sums मा के के समावेश छ ?)

This page solves all 41 questions of Exercise 7.4 Transformation of Products and Sums, in textbook order, with the Nepali question in brackets and an Answer box at the end. Transformation of product and sum: formulas, proof, examples and all Exercise 7.4 questions solved. Sum to product identities, Class 10 Optional Math.

What are the sum to product and product to sum formulas? (योगलाई गुणनफल र गुणनफललाई योगमा रूपान्तरण गर्ने सूत्रहरू के के हुन् ?)

Product to sum:

  • \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\)
  • \(2\cos A\sin B=\sin(A+B)-\sin(A-B)\)
  • \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\)
  • \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\)

Sum to product:

  • \(\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\)
  • \(\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\)
  • \(\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2}\)
  • \(\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}\)
What are the sum to product identities? (योगलाई गुणनफलमा बदल्ने सर्वसमिकाहरू के के हुन् ?)

They are the four formulas \(\sin C\pm\sin D\) and \(\cos C\pm\cos D\) written as products (see the formula boxes above). They are used in Exercise 7.4, Question 3 to 6.

What is the proof of the sum to product formula? (योगलाई गुणनफलमा बदल्ने सूत्रको प्रमाण कसरी गर्ने ?)

Add the formulas \(\sin(X+Y)=\sin X\cos Y+\cos X\sin Y\) and \(\sin(X-Y)=\sin X\cos Y-\cos X\sin Y\) to get \(2\sin X\cos Y=\sin(X+Y)+\sin(X-Y)\). Now put \(X+Y=C\) and \(X-Y=D\), so \(X=\frac{C+D}{2}\) and \(Y=\frac{C-D}{2}\). This gives \(\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2}\). The other three formulas are proved in the same way.

How do I change a product like \(\sin50^\circ\cos30^\circ\) into a sum? (\(\sin50^\circ\cos30^\circ\) जस्तो गुणनफललाई योगमा कसरी बदल्ने ?)

Multiply and divide by \(2\) and use \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\). So \(\sin50^\circ\cos30^\circ=\dfrac{\sin80^\circ+\sin20^\circ}{2}\) (Question 2a).

Can you give transformation of product and sum examples? (गुणनफल र योगको रूपान्तरणका उदाहरण दिनुहोस् ।)

Product to sum: \(\cos25^\circ\cos15^\circ=\dfrac{\cos40^\circ+\cos10^\circ}{2}\) (Question 2b). Sum to product: \(\cos7A-\cos5A=-2\sin6A\sin A\) (Question 3c).

Do I need a calculator for the transformation of product and sum? (के यसका लागि क्याल्कुलेटर चाहिन्छ ?)

No. Every question of Exercise 7.4 is solved by the formulas alone. A calculator is only useful for a quick numerical check, for example to see whether a printed identity is true.

Are there misprints in Exercise 7.4? (के अभ्यास ७.४ मा छापाइका गल्ती छन् ?)

Two relations as printed do not hold: Question 6(c) simplifies to \(\cot2A\) (not \(\cot3A\)), and Question 6(d) cannot be proved as printed. The solutions show the numerical check and the corrected result.

Are these Exercise 7.4 questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू SEE को तयारीका लागि उपयोगी छन् ?)

Yes. Exercise 7.4 covers changing products to sums and sums to products and proving identities with them, which are common SEE-style question types.

Students also search for:

Transformation of product and sum pdfTransformation of product and sum examplesTransformation of product and sum proofTransformation of product and sum formulasTransformation of product and sum calculatorSum to product and product to sum formulasSum to product identitiesSum-to-product formula proofClass 10 optional math trigonometryClass 10 opt math

More Class 10 Optional Mathematics Trigonometry Exercises

Also Read

Related Posts

Scroll to Top