Unit 5 - Trigonometry | Sub Multiple Angles

Sub Multiple Angles is the second chapter in Trigonometry for Class 10 students who are studying Optional Mathematics subject.

Any angle that can be expressed as a smaller multiple of a reference angle is said to be multiple angles.

For example: let the reference angle be A then A/2, A/3, A/4, ... are said to be the multiple angles of A.

As always, we will have some major formulae in this chapter as well. We have to remember the formulae to express sin, cos, tan and cot in terms of multiple angles of A/2 and A/3.

Exercise 2

1. If $cos \frac{ \theta }{2}= \dfrac{4}{5}$, find the value of:

Solution:
Here,
$cos \theta = 2 cos^2 \frac{\theta}{2} -1$

$= 2 \left ( \dfrac{4}{5} \right )^2 - 1$

$= 2 × \dfrac{16}{25} -1$
$= \dfrac{32 - 25}{25}$
$= \dfrac{7}{25}$
Now,
a) $sin \theta$
Solution:

$= \sqrt{ 1 - cos^2 \theta}$

$= \sqrt{1 -  \left (  \dfrac{7}{25} \right )^2}$

$= \sqrt{\dfrac{25^2 - 49}{25^2}}$

$= \sqrt{\dfrac{625-49}{625}}$

$= \sqrt{\dfrac{576}{625}}$

$= \sqrt{ \left ( \dfrac{24}{25} \right )^2 }$

$= \dfrac{24}{25}$
b) $cos \theta$

Already mentioned above
c) $tan \theta$
Solution:
$= \dfrac{ sin \theta}{cos \theta}$

$= \dfrac{ \dfrac{24}{25} }{\dfrac{7}{25}}$

$= \dfrac{24}{25} × \dfrac{25}{7}$

$= \dfrac{24}{7}$

4. If $cos \theta = - \dfrac{3}{5}$ and $\theta$ lies between 180° and 270°, find the value of:

a) $sin \frac{\theta}{2}$
Solution:
$sin \frac{\theta}{2} = \sqrt{ \dfrac{1 - cos \theta}{2}}$

$= \sqrt{ \dfrac{1 - \left ( - \dfrac{3}{5} \right ) }{2}}$

$= \sqrt{ \dfrac{1 + \dfrac{3}{5} }{2} }$

$= \sqrt{ \dfrac{\dfrac{5+3}{5} } {2}}$

$= \sqrt{ \dfrac{8}{5} × \dfrac{1}{2}}$

$= \sqrt{\dfrac{4}{5}}$
$= \dfrac{2}{√5}$
Since value of $\theta$ lies between 180° and 270°, value of $\frac{\theta}{2}$ should be below 180° so, the required value of $sin \theta$ is $\dfrac{2}{√5}$.


5. If $cos 30° = \dfrac{√3}{2}$, prove that:

a) $sin15° = \dfrac{√3-1}[2√2}$
Solution:

We know,
$sinA = \sqrt{ \dfrac{1 - cos2A}{2} }$

Let, $15° = A and 30° = 2A$
Taking LHS

$sin15° = \sqrt{\dfrac{1 - cos30°}{2}}$

$= \sqrt{\dfrac{1 - \dfrac{√3}{2}}{2}}$

$= \sqrt{ \dfrac{\dfrac{2-√3}{2}}{2}}$

$= \sqrt{ \dfrac{2-√3}{4}}$
$= \sqrt{ \dfrac{2-√3}{4} × \dfrac{2}{2}}$

$= \sqrt{ \dfrac{2(2-√3)}{8}}$

$= \sqrt{ \dfrac{4-2√3}{8}}$

$= \sqrt{ \dfrac{3 - 2√3 + 2}{2^2 × 2}}$

$= \sqrt{ \dfrac{(√3-1)^2}{2^2×2}}$

$= \dfrac{√3-1}{2√2}$
= RHS


8. Prove the following simple identities:

a) $\left ( sin \frac{A}{2} - cos \frac{A}{2} \right )^2 = 1 - sin A$ Answer

b) $cos^4 \frac{A}{2} - sin^4 \frac{A}{2} = cosA$ Answer

c) Answer

e) Answer


9. Prove that:
a) Answer


$b) \dfrac{1 + sec\alpha}{tan \alpha} = cot \frac{\alpha}{2}$

Solution:

LHS

$= \dfrac{1 + sec\alpha}{tan \alpha}$

$= \dfrac{1 + \dfrac{1}{cos \alpha}}{ \dfrac{sin\alpha}{cos \alpha} }$

$= \dfrac{ \dfrac{cos \alpha +1}{cos \alpha} }{ \dfrac{sin \alpha}{cos \alpha} }$

$= \dfrac{cos \alpha +1}{cos \alpha} × \dfrac{cos \alpha}{sin \alpha}$

$= \dfrac{cos \alpha +1}{sin \alpha}$

[Using formula of cos2A = 2cos²A - 1]

$= \dfrac{(2cos^2 \frac{\alpha}{2} -1) +1}{sin \alpha}$

[Using formula of sin2A = 2sinAcosA]

$= \dfrac{2cos^2 \frac{\alpha}{2} }{2sin\frac{\alpha}{2} cos \frac{\alpha}{2}}$

$= \dfrac{cos \frac{\alpha}{2} }{ sin\frac{\alpha}{2}}$

$= cot \frac{ \alpha}{2}$

RHS


c) Answer


$d) \dfrac{ sin \theta + sin \frac{\theta}{2} }{1 + cos \theta + cos\frac{\theta}{2} } = tan \frac{ \theta}{2}$

Solution:

LHS

$= \dfrac{ sin\theta + sin \frac{\theta}{2} }{( 1 + cos \theta )+ cos \frac{ \theta}{2}}$

[ Using formula: 1 + cos 2A = 2cos²A ]

$= \dfrac{sin \theta + sin \frac{\theta}{2}}{2 cos^2 \frac{ \theta}{2} + cos \frac{\theta}{2}}$

$= \dfrac{ 2 sin \frac{ \theta}{2} cos \frac{ \theta}{2} + sin\frac{ \theta}{2} }{cos \frac{ \theta}{2} ( 2cos \frac{ \theta}{2} + 1) }$

$= \dfrac{ sin \frac{ \theta }{2} (2 cos\frac{ \theta}{2} + 1) }{cos \frac{ \theta}{2} ( 2 cos\frac{ \theta}{2} + 1)} $

$= \dfrac{ sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} }$

$= tan \frac{ \theta}{2}$

RHS


e) Answer

g) Answer

k) Answer


10. Show that:

a) $\dfrac{ 1 - tan^2 (45° - \frac{ \theta}{2} )}{ 1 + tan^2(45°- \frac{ \theta}{2})} = sin \theta$

Solution:

LHS

$= \dfrac{ 1 - tan^2(45° - \frac{ \theta}{2} ) }{ 1 + tan^2 (45° - \frac{ \theta}{2})}$

[Using $\dfrac{1 - tan²A}{1+tan²A} = cos2A$]

$= cos \left [2 × \left ( 45° - \frac{\theta}{2} \right ) \right ]$

$= cos \left (45°×2 - \frac{\theta }{2} × 2 \right )$

$= cos (90° - \theta)$

$= sin \theta$

RHS
 

$b) \dfrac{ 2 tan \left ( \frac{\pi}{4} - \frac{A}{2} \right ) } { 1 + tan^2 \left( \frac{\pi}{4} - \frac{A}{2} \right )  } = cos A$

Solution:

LHS

$= \dfrac{ 2 tan ( \frac{\pi}{4} - \frac{A}{2} ) }{1 + tan^2 ( \frac{\pi}{4} - \frac{A}{2} )}$

[Using $\dfrac{2tanA}{1 + tan²A} = sin2A$]

$= sin \left [ 2 × \left ( \frac{\pi}{4} - \frac{A}{2} \right ) \right ]$

$= sin \left ( \frac{\pi}{4} × 2 - \frac{A}{2} × 2 \right ) $

$= sin (\frac{\pi}{2} - A)$

$= sin(90° - A)$

$= cosA$

RHS


c) Answer


d) Similar to above-mentioned solutions of 10 a,b,c. Use the formula of $2cos²A-1 = cos2A$.
11 a) Answer

11 b) If $cos \frac{\theta}{3} = \dfrac{1}{2} \left ( m + \dfrac{1}{m} \right )$, prove that: $cos \theta = \dfrac{1}{2} \left ( m^3 + \dfrac{1}{m^3} \right )$.

Solution:
Here,

$cos \frac{\theta}{3} = \dfrac{1}{2} \left ( m + \dfrac{1}{m} \right )$

Taking LHS of the question,

$= cos \theta$

$= 4cos^3 \frac{ \theta}{3} - 3 cos \frac{\theta}{3}$

$= cos \frac{\theta}{3} \left ( 4cos^2 \frac{\theta}{3} - 3 \right )$

$= cos \frac{\theta}{3} \left [ 4 \left ( \dfrac{1}{2} (m + \frac{1}{m} ) \right) ^2 - 3 \right ]$

$= cos \frac{\theta}{3} \left [ 4 × \dfrac{1}{4} × \left (m^2 + 2 + \dfrac{1}{m^2} \right ) - 3 \right ]$

$= cos \frac{\theta}{3} \left [ m^2 + 2 + \dfrac{1}{m^2} - 3 \right ]$

$= \dfrac{1}{2} \left ( m + \dfrac{1}{m} \right ) \left ( m^2 - 1 + \dfrac{1}{m^2} \right )$

[Using formula (a+b)(a²-ab+b²) = (a³+b³)]

$= \dfrac{1}{2} \left ( m^3 + \dfrac{1}{m^3} \right )$

RHS

12. Prove that:

a) Answer


$b) \dfrac{1}{4} [(cos \theta + cos \beta)^2 + (sin \theta - sin \beta)^2] = cos^2 \frac{ \theta + \beta}{2}$

Solution:

LHS

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

$= \dfrac{1}{4} [ cos^2 \theta + cos^2 \beta +2 cos \theta cos \beta + sin^2 \theta + sin^2 \beta - 2sin\thetasin\beta]$

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

[Using cos(A+B) = cosAcosB - sinAsinB]

$= \dfrac{1}{4} [ 1+1+2cos( \alpha + \beta)]$

$= \dfrac{1}{4} [ 2\{1 + cos (\alpha + \beta)\} ]$

$= \dfrac{2}{4} [1 + cos (\alpha + \beta)]$

[Using 1+cos2A = 2cos²A]

$= \dfrac{1}{2} × 2cos^2 \frac{\alpha + \beta}{2} ]$

$= cos^2 \frac{ \alpha + \beta}{2}$

RHS

13 a) $tan \left ( \dfrac{\pi}{4} - \dfrac{A}{2} \right ) = \dfrac{ cos \frac{A}{2} - sin\frac{A}{2} } {cos \frac{A}{2} + sin \frac{A}{2}} = \dfrac{cosA}{1+sinA}$

Solution:

Left Side

$= tan ( \frac{\pi}{4} - \frac{A}{2} )$

$= tan (\frac{180°}{4} - \frac{A}{2} )$

$= tan (45° - \frac{A}{2})$
$= \dfrac{tan45° - tan \frac{A}{2}}{tan45° + tan \frac{A}{2}}$

$= \dfrac{1 - tan\frac{A}{2} } {1+ tan \frac{A}{2}}$

$= \dfrac{ 1 - \dfrac{sin \frac{A}{2}}{cos \frac{A}{2}}}{1 + \dfrac{sin\frac{A}{2}}{cos \frac{A}{2}}}$

$= \dfrac{ \dfrac{cos \frac{A}{2} - sin\frac}A}{2} }{cos \frac{A}{2}} }{\dfrac{cos \frac{A}{2} + sin\frac{A}{2}}{cos \frac{A}{2}}}$

$= \dfrac{ cos\frac{A}{2} - sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}}$

Middle Side

$= \dfrac{cos \frac{A}{2} - sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}} × \dfrac{cos \frac{A}{2} + sin\frac{A}{2}}{cos \frac{A}{2} + sin\frac{A}{2}}$

$= \dfrac{(cos^2 \frac{A}{2} - sin^2 \frac{A}{2} )}{(cos^2 \frac{A}{2} + sin^2 \frac{A}{2}) +( 2sin \frac{A}{2} cos \frac{A}{2})}$

Contracting to formulae, we get,

$= \dfrac{cosA}{1 + sinA}$
Right Side


e) $cot ( 45° + \frac{ \theta}{2} ) + tan ( 45° - \frac{\theta}{2*) = \dfrac{2 cos \theta}{1 + sin \theta}$

Solution:

LHS

$= cot (45° + \frac{\theta}{2} ) + tan (45° - \frac{\theta}{2})$

$= \dfrac{1}{tan (45° + \frac{\theta}{2} ) } + \dfrac{tan45° - tan\frac{\theta}{2} }{tan45° + tan \frac{\theta}{2} }$

$= \dfrac{1}{ \dfrac{tan45° + tan\frac{\theta}{2} }{tan45° - tan\frac{\theta}{2} } + \dfrac{1 - tan\frac{\theta}{2} }{1 + \frac{\theta}{2} }$

$= \dfrac{1}{ \dfrac{1 + tan\frac{\theta}{2} }{1 - tan\frac{\theta}{2} }} + \dfrac{1 - tan\frac{\theta}{2} }{1 + tan\frac{\theta}{2}}$

$= \dfrac{1 - tan \frac{\theta}{2} }{1 + tan \frac{\theta}{2} } + \dfrac{1 - tan\frac{\theta}{2} }{1 + tan \frac{\theta}{2}}$

$= \dfrac{1 - tan\frac{\theta}{2} + 1 - tan \frac{\theta}{2} }{1 + tan\frac{\theta}{2}}$


About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada
Editor: I. R. Simkhada
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About this page:
Class 10 - Sub Multiple Angles - Trigonometry Solved Exercises | Readmore Optional Mathematics is a collection of the solutions related to sub multiple angles of Trigonometry for Nepal's Secondary Education Examination (SEE) appearing students.

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