3 - Transformation of Trigonometric Ratios (Trigonometry)

In this post, you can check the solutions to the exercises of Transformation of Trigonometric Ratios of Trigonometry Unit of Readmore Optional Mathematics by DR Simkhada. It consists of complete answers to all odd questions mentioned in the textbook.

Before starting, here is a quick revision of the formulae from this chapter:
  1. $sinC + sinD = 2sin\dfrac{C+D}{2}cos \dfrac{C-D}{2}$
  2. $sinC - sinD = 2cos\dfrac{C+D}{2}sin\dfrac{C-D}{2}$
  3. $cosC + cosD = 2cos\dfrac{C+D}{2}cos\dfrac{C-D}{2}$
  4. $cosC - cosD = 2sin\dfrac{C+D}{2}sin\dfrac{D-C}{2}$
  5. $2sinAcosB = sin(A+B) + sin(A-B)$
  6. $2cosAsinB = sin(A+B) - sin(A-B)$
  7. $2cosAcosB = cos(A+B) + cos(A-B)$
  8. $2sinAsinB = cos(A-B) - cos(A+B)$
Here are more solutions from Algebra:
  1. Multiple Angles
  2. Sub-Multiple Angles
  3. Transformation of Trigonometric Ratios


Now, you can check the solutions below. But, make sure to check the description after the solution to know more about this textbook series, author, and the publisher.

1. Express each of the following as a sum or difference:

$a) 2cos85°cos45°$
Solution:

We know,
$2cosAcosB = cos(A+B)+cos(A-B)$

So,
$2cos85°cos45°$
$= cos(85°+45°)+cos(85°-45°)$
$= cos130° + cos 40°$

$b) 2sin27°cos15°$
Solution:

We know,
$2sinAcosB = sin(A +B) + sin(A-B)$

So,
$2sin27°cos15°$
$= sin(27°+15°) + sin(27°-15°)$

$= sin42° + sin12°$

$c) 2cos7\theta.sin3\theta$
Solution:

We know,
$2cosAsinB = sin(A+B) - sin(A-B)$

So,
$2cos 7\theta sin3\theta$
$= sin(7\theta + 3\theta) - sin(7\theta - 3\theta)$

$= sin10\theta - sin4\theta$

$d) sin16°.cos42°$
Solution:

We know,
$2sinAcosB = sin(A + B) + sin(A - B)$

So,
$sin16°cos42°$
$= sin16°cos42° × \dfrac{2}{2}$

$= \dfrac{2sin16°cos42°}{2}$
$= \dfrac{sin(16°+42°)+sin(16°-42°)}{2}$

$= \dfrac{sin58° - sin26°}{2}$


$e) sin3\theta. sin7\theta$
Solution:

We know,
$2sinAsinB = cos(A-B)-cos(A+B)$

So,
$sin3\theta.sin7\theta$
$= sin3\theta. sin7\theta × \dfrac{2}{2}$

$= \dfrac{cos(3\theta - 7\theta) - cos (3\theta + 7\theta)}{2}$

$= \dfrac{cos(-4\theta) - cos 10\theta}{2}$

$= \dfrac{cos4\theta - cos10\theta}{2}$


$f) sin13°.cos115°$
Solution:
We know,
$2sinAcosB = sin(A +B) + sin(A-B)$

So,
$sin13°.cos115°$
$= sin13°.cos115° × \dfrac{2}{2}$

$= \dfrac{2sin13°cos115°}{2}$
$= \dfrac{sin(13°+115°)+sin(13°-115°)}{2}$

$= \dfrac{sin128° + sin(-102°)}{2}$

$= \dfrac{sin128° - sin102°}{2}$


2. Find the value of:
$a) 2sin75°sin45°$
Solution:
$2sin75°sin45°$
$= cos(75°-45°) - cos(75°+45°)$

$= cos30° - cos120°$
$= cos30° - (-sin30°)$

$= cos30° + sin30°$
$= \dfrac{√3}{2} + \dfrac{1}{2}$

$= \dfrac{√3+1}{2}$

b) 2cos105°cos15°
Solution:
$2cos105°cos15°$

$= cos(105°+15°)+cos(105°-15°)$

$= cos120° + cos90°$
$= - sin30° + 0$
$= - \dfrac{1}{2}$

c) sin15°sin105°
Solution:
$sin15°sin105°$
$= sin15°sin105° × \dfrac{2}{2}$

$= \dfrac{2sin15°sin105°}{2}$

$= \dfrac{cos(15°-105°) - cos(15°+105°)}{2}$

$= \dfrac{cos(-90°) - cos120°}{2}$

$= \dfrac{cos90° - (-sin30°)}{2}$

$= \dfrac{0 + sin30°}{2}$
$= \dfrac{sin30°}{2}$
$= \dfrac{ \dfrac{1}{2}}{2}$
$= \dfrac{1}{4}$

3. Express the following sum or difference into product of trigonometric ratios.
Important Formulae:
  1. $sinC + sinD = 2sin\dfrac{C+D}{2}cos \dfrac{C-D}{2}$
  2. $sinC - sinD = 2cos\dfrac{C+D}{2}sin\dfrac{C-D}{2}$
  3. $cosC + cosD = 2cos\dfrac{C+D}{2}cos\dfrac{C-D}{2}$
  4. $cosC - cosD = 2sin\dfrac{C+D}{2}sin\dfrac{D-C}{2}$

a) sin80° + sin16°

Solution:
$sin80° + sin16°$
$= 2 sin \dfrac{80°+16°}{2} cos \dfrac{80°-16°}{2}$

$= 2 sin\dfrac{96°}{2} cos \dfrac{64°}{2}$

$= 2sin48°cos32°$

c) cos30° - cos80°

Solution:
$cos30° - cos80°$
$= 2 sin\dfrac{30°+80°}{2} sin\dfrac{80°-30°}{2}$

$= 2 sin\dfrac{110°}{2} sin\dfrac{50°}{2}$

$= 2sin55°sin25°$

e) cos80° - cos40°

Solution:
$cos80° - cos40°$
$= 2sin\dfrac{80°+40°}{2} sin\dfrac{40°-80°}{2}$

$= 2sin\dfrac{120°}{2} sin\dfrac{-40°}{2}$

$= 2sin60°sin(-20°)$
$= -2× \dfrac{√3}{2} × sin20°$

$= -√3sin20°$

4. Find the value of:
a) sin75° - sin15°
Solution:
$sin75° - sin15°$
$= 2 cos \dfrac{75°+15°}{2} sin \dfrac{75°-15°}{2}$

$= 2 cos \dfrac{90°}{2} sin\dfrac{60°}{2}$

$= 2 cos45° sin30°$
$= 2 × \dfrac{1}{√2} × \dfrac{1}{2}$

$= \dfrac{1}{√2}$

c) cos15° - cos75°

Solution:
$cos 15° - cos75°$
$= 2 sin \dfrac{15°+75°}{2} sin \dfrac{75°-15°}{2}$

$= 2 sin\dfrac{90°}{2} sin\dfrac{60°}{2}$

$= 2sin45°sin30°$
$= 2 × \dfrac{1}{√2} × \dfrac{1}{2}$

$= \dfrac{1}{√2}$

e) cos130° + cos110° + cos10°
Solution:
$cos130° + cos110° + cos10°$
$= 2 cos \dfrac{130°+110°}{2} cos\dfrac{130°-110°}{2} + cos10°$

$= 2cos\dfrac{240°}{2} cos \dfrac{20°}{2} + cos10°$

$= 2 cos120°cos10° + cos10°$
$= 2 (-sin30°) cos10° + cos10°$

$= - 2 × \dfrac{1}{2} cos10° + cos10°$

$= -cos10° + cos10°$
$= 0$

5. Prove that:
$a) cos15°.sin75° = \dfrac{2+√3}{4}$

Solution:
LHS
$= cos15° sin75°$
$= cos15°sin75° × \dfrac{2}{2}$

$= \dfrac{2cos15°sin75°}{2}$
$= \dfrac{sin(15°+75°) - sin(15°-75°)}{2}$

$= \dfrac{sin90° - sin(-60°)}{2}$

$= \dfrac{1 + sin60°}{2}$
$= \dfrac{1 + \dfrac{√3}{2}}{2}$

$= \dfrac{\dfrac{2+√3}{2}}{2}$

$= \dfrac{2+√3}{4}$
RHS

c) 2 sin(45°+2A) cos(45°-2A) = 1 + sin4A

Solution:
LHS
$= 2sin(45°+2A)cos(45°-2A)$
$= sin[(45°+2A)+(45°-2A)] + sin[(45°+2A)-(45°-2A)]$

$= sin[90°] + sin[45°+2A-45°+2A]$

$= 1 + sin4A$
RHS

$e) \ cos \frac{3 \pi}{13} cos \frac{5\pi}{13} - cos \frac{8 \pi}{13} cos \frac{10 \pi}{13} = 0$

Solution:
LHS
$= cos \frac{3 \pi}{13} cos \frac{5 \pi}{13} - cos \frac{8 \pi}{13} cos \frac{10 \pi}{13}$

$= [cos \frac{3 \pi}{13} cos \frac{5 \pi}{13} - cos \frac{8\pi}{13} cos \frac{10 \pi}{13} ] × \dfrac{2}{2}$

$= \dfrac{1}{2} [2cos \frac{3\pi}{13} cos \frac{5 \pi}{13} - 2cos \frac{8 \pi}{13} cos \frac{10 \pi}{13} ]$

$= \dfrac{1}{2} [cos ( \frac{3 \pi}{13} + \frac{5 \pi}{13} ) + cos ( \frac{3 \pi}{13} - cos \frac{5 \pi}{13} ) - cos (\frac{8 \pi}{13} + \frac{10 \pi}{13}) - cos ( \frac{ 8 \pi}{13} - frac{10 \pi}{13} )]$

$= \dfrac{1}{2} [cos \frac{8 \pi}{13} + cos \frac{-2 \pi}{13} - cos \frac{18 \pi}{13} - cos \frac{-2 \pi}{13}]$

$= \dfrac{1}{2} [cos \frac{8\pi}{13} + cos \frac{2 \pi}{13} - cos \frac{18 \pi}{13} - cos \frac{2 \pi}{13}]$

$= \dfrac{1}{2} [cos \frac{8 \pi}{13} - cos \frac{18 \pi}{13}]$

$= \dfrac{1}{2} [ 2 sin ( \frac{8\pi}{13} + frac{18 \pi}{13}) sin ( \frac{18 \pi}{13} - \frac{8 \pi}{13} )]$

$= sin \frac{8\pi+18\pi}{13} sin \frac{18 \pi - 8\pi}{13}$

$= sin \frac{26 \pi}{13} sin \frac{10 \pi}{13}$

$= sin (2 \pi) sin\frac{10 \pi}{13}$

$= sin360° × sin \frac{10 \pi}{13}$

$= 0 × sin \frac{10 \pi}{13}$
$= 0$
RHS

6. Prove that:
a) tan50° - tan40° = 2 tan10°

Solution:
LHS

$= tan 50° - tan40°$
$= \dfrac{sin50°}{cos 50°} - \dfrac{sin40°}{cos40°}$

$= \dfrac{sin50°cos40° - cos50°sin40°}{cos50°cos40°}$

[Using sinAcosB + cosAsinB = sin(A-B)]

$= \dfrac{sin(50°-40°)}{cos40°cos50°}$

$= \dfrac{sin10°}{cos40°cos50°} × \dfrac{2}{2}$

$= \dfrac{2sin10°}{2cos40°cos50°}$

$= \dfrac{2sin10°}{cos (40°+50° ) + cos (40°-50°)}$

$= \dfrac{2sin10°}{ cos90° + cos(-10°)}$

$= \dfrac{2sin10°}{0 + cos10°}$

$= \dfrac{2sin10°}{cos10°}$
$= 2tan10°$
RHS


c) 2tan40° = tan65° - tan25°

Solution:
Taking RHS
$= tan65° - tan25°$

$= \dfrac{sin65°}{cos65°} - \dfrac{sin25°}{cos25°}$

$= \dfrac{sin65°cos25° - cos65°sin25°}{cos65°cos25°}$

[Using sinAcosB-cosAsinB = sin(A-B)]

$= \dfrac{sin(65°-25°)}{cos65°cos25°} × \dfrac{2}{2}$

$= \dfrac{2sin40°}{2cos25°cos65°}$

$= \dfrac{2sin40°}{cos(25°+65°) + cos(25°-65°)}$

$= \dfrac{2sin40°}{cos90° + cos(-40°)}$

$= \dfrac{2sin40°}{cos40°}$
$= 2 tan 40°$
LHS

7. Prove that:


$a) \dfrac{cos285° + cos345°}{sin435° - sin375°} = √3$

Solution:

LHS
$= \dfrac{cos285° + cos345°}{sin435° - sin375°}$

$= \dfrac{cos285° + cos345°}{sin435° - sin375°}$

$= \dfrac{ 2 cos \frac{285°+345°}{2} cos \frac{285°-345°}{2} }{ 2cos \frac{435°+375°}{2} sin \frac{435°-375°}{2} }$

$= \dfrac{2cos\frac{630°}{2} cos \frac{-60°}{2}}{ 2cos \frac{810°}{2} sin \frac{60°}{2}}$

$= \dfrac{cos 315° cos30°}{cos405°sin30°}$

$= \dfrac{cos(360°-45°)cos30°}{cos(360°+45°)sin30°}$

$= \dfrac{cos45° cos30°}{cos45° sin30°}$

$= \dfrac{cos30°}{sin30°}$
$= \dfrac{ \dfrac{√3}{2}}{\dfrac{1}{2}}$

$= \dfrac{√3}{2} × 2$
$= √3$
RHS

8. Prove that:
$a) sin20° sin40° sin80° = \dfrac{√3}{8}$

Solution:
LHS
$= sin20° sin40° sin80°$
$= \dfrac{2}{2} sin20° sin40° sin80°$

$= \dfrac{1}{2} [2sin20°sin80°]sin40°$

$= \dfrac{1}{2} [cos(20°-80°)-cos(20°+80°)]sin40°$

$= \dfrac{1}{2} [cos60° - cos100°]sin40°$

$= \dfrac{1}{2} [\frac{1}{2} - cos(90°+10°)]sin40°$

$= \dfrac{1}{2} [ \frac{1}{2} + sin10°] sin40°$

$= \dfrac{1}{2} [ \frac{1 + 2sin10°}{2} ] sin40°$

$= \dfrac{1}{4} [ sin40° + 2sin10°sin40°]$

$= \dfrac{1}{4} [sin40° + cos(10°-40°) - cos(10°+40°)]$

$= \dfrac{1}{4} [sin40° + cos30° - cos50°]$

$= \dfrac{1}{4} [sin40° + cos30° - cos(90°-40°)]$

$= \dfrac{1}{4} [sin40° + cos30° - sin40°]$

$= \dfrac{1}{4} × cos30°$
$= \dfrac{1}{4} × \dfrac{√3}{2}$

$= \dfrac{√3}{8}$
RHS


9) Prove that:


$a) cos \theta cos (60° + \theta) cos (60° - \theta) = \dfrac{1}{4} cos 3\theta$

Solution:

LHS

$= cos \theta cos(60°+\theta) cos (60°-\theta)$

$= \dfrac{2}{2} × cos\theta cos (60° + \theta) cos (60° - \theta)$

$= \dfrac{cos \theta × 2cos(60°+\theta)cos(60°-\theta)}{2} $

$= \dfrac{cos \theta × [cos\{60°+\theta+(60°-\theta)\} + cos\{ 60° + \theta - (60° - \theta)\} ]}{2}$

$= \dfrac{cos \theta ×[cos120° + cos2\theta]}{2}$

$= \dfrac{cos \theta × [- \frac{1}{2} + cos 2\theta]}{2}$

$= \dfrac{cos \theta × \dfrac{2cos2\theta - 1}{2}}{2}$

$= \dfrac{1}{4} × [cos \theta (2cos 2\theta -1)]$

$= \dfrac{1}{4} × [2cos 2\theta cos \theta - cos\theta]$

$= \dfrac{1}{4} × [cos(2\theta + cos\theta) + cos(2\theta - cos\theta) - cos\theta]$

$= \dfrac{1}{4} × [cos3\theta + cos\theta - cos\theta]$

$= \dfrac{1}{4} cos3\theta$

RHS



About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada

Readmore Publishers and Distributors
T.U. Road, Kuleshwor - 14, Kathmandu, Nepal
Phone: 4672071, 5187211, 5187226
readmorenepal6@gmail.com

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Transformation of Trigonometric Ratios - Solved Exercises | Class 10 Readmore Optional Mathematics is a collection of the solutions related to exercises of Transformation of Trigonometric Ratios from the Trigonometry chapter for Nepal's Secondary Education Examination (SEE) appearing students.

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