Multiple and Sub-multiple Angles MCQ Quiz - Objective Question with Answer for Multiple and Sub-multiple Angles - Download Free PDF

Given:

In a triangle, the relationship between angles is used where A + B + C = 180°.

Formula used:

\(\cos\left(\frac{A + B}{2}\right) = \sin\left(\frac{C}{2}\right)\), where C = 180° - (A + B).

Explanation:

In a triangle:

A + B + C = 180°

⇒ C = 180° - (A + B)

⇒ \(\frac{C}{2} = \frac{180° - (A + B)}{2}\)

Now, using the complementary angle relationship:

\(\cos\left(\frac{A + B}{2}\right) = \sin\left(\frac{C}{2}\right)\)

∴ Option 4 is the correct answer.

Given:

In a triangle, the relationship between angles is used where A + B + C = 180°.

Formula used:

\(\cos\left(\frac{A + B}{2}\right) = \sin\left(\frac{C}{2}\right)\), where C = 180° - (A + B).

Explanation:

In a triangle:

A + B + C = 180°

⇒ C = 180° - (A + B)

⇒ \(\frac{C}{2} = \frac{180° - (A + B)}{2}\)

Now, using the complementary angle relationship:

\(\cos\left(\frac{A + B}{2}\right) = \sin\left(\frac{C}{2}\right)\)

∴ Option 4 is the correct answer.

\(\Longrightarrow 2b=a+c\)

The value of the semiperimeter of the triangle \(s=\dfrac { 3b }{ 2 } \)

\(\therefore\) the value of \(\cos ^{ 2 }{ \dfrac { A }{ 2 } } =\dfrac { s\left( s-a \right) }{ bc } =\dfrac { 3\left( \dfrac { 3b }{ 2 } -a \right) }{ 2c } \)

Similarly the value of \(\cos ^{ 2 }{ \dfrac { C }{ 2 } } =\dfrac { s\left( s-c \right) }{ ab } =\dfrac { 3\left( \dfrac { 3b }{ 2 } -c \right) }{ 2a } \)

\(a\cos ^{ 2 }{ \dfrac { C }{ 2 } } +c\cos ^{ 2 }{ \dfrac { A }{ 2 } } =\dfrac { 3 }{ 2 } \left( 3b-a-c \right) =\dfrac { 3 }{ 2 } \left( b \right) \)

Calculation

\( { (a-b) }^{ 2 }{ \cos }^{ 2 } \dfrac { C }{ 2 } +{ (a+b) }^{ 2 }{ \sin }^{ 2 } \dfrac { C }{ 2 } \ = { a }^{ 2 }+{ b }^{ 2 }+2ab({ \sin }^{ 2 } \dfrac { C }{ 2 } -{ \cos }^{ 2 } \dfrac { C }{ 2 } ) \ = { a }^{ 2 }+{ b }^{ 2 }+2ab(2{ \sin }^{ 2 } \dfrac { C }{ 2 } -1) \)

But \( \sin \dfrac { C }{ 2 } = \sqrt { \dfrac { (s-a)(s-b) }{ ab } } \)

where \( s = \dfrac { a+b+c }{ 2 } \)

\( \therefore \) \( { a }^{ 2 }+{ b }^{ 2 }+2ab(2{ \sin }^{ 2 } \dfrac { C }{ 2 } -1) \ = { a }^{ 2 }+{ b }^{ 2 }+2ab( \dfrac { (c+b-a)(c+a-b) }{ 2ab } -1) \ = { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-{ (b-a) }^{ 2 }-2ab \ = { c }^{ 2 } \)

Hence option 2 is correct

Concept Used:

Trigonometric identities:

sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)

tan θ = \(\frac{sin θ}{cos θ}\)

Calculation:

⇒ \(\frac{sin~7x+sin~5x}{cos~7x+cos~5x}\)

⇒ \(\frac{2 sin\frac{7x+5x}{2} cos\frac{7x-5x}{2}}{2 cos\frac{7x+5x}{2} cos\frac{7x-5x}{2}}\)

⇒ \(\frac{2 sin~6x~cos~x}{2 cos~6x~cos~x}\)

⇒ \(\frac{sin~6x}{cos~6x}\)

⇒ tan 6x

∴ \(\frac{sin~7x+sin~5x}{cos~7x+cos~5x}\) = tan 6x

Hence option 5 is correct

Concept:

cos 2A = cos² A – sin² A

sin 2A = 2 sin A × cos A

Calculation:

\(\Rightarrow \cot \left( {\frac{A}{2}} \right) - \tan \left( {\frac{A}{2}} \right) = \frac{{\cos \left( {\frac{A}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right)}} - \frac{{\sin \left( {\frac{A}{2}} \right)}}{{\cos \left( {\frac{A}{2}} \right)}} = \frac{{{{\cos }^2}\left( {\frac{A}{2}} \right) - {{\sin }^2}\left( {\frac{A}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right) \times \cos \left( {\frac{A}{2}} \right)}}\)

As we know that, cos 2A = cos² A – sin² A

\(\Rightarrow \cot \left( {\frac{A}{2}} \right) - \tan \left( {\frac{A}{2}} \right) = \frac{{{{\cos }^2}\left( {\frac{A}{2}} \right) - {{\sin }^2}\left( {\frac{A}{2}} \right)}}{{\sin \left( {\frac{A}{2}} \right) \times \cos \left( {\frac{A}{2}} \right)}} = \frac{{\cos A}}{{\frac{1}{2} \times 2 \times \sin \left( {\frac{A}{2}} \right) \times \cos \left( {\frac{A}{2}} \right)\;}}\)

As we know that, sin 2A = 2 sin A × cos A

\(\Rightarrow \cot \left( {\frac{A}{2}} \right) - \tan \left( {\frac{A}{2}} \right) = \frac{{\cos A}}{{\frac{1}{2} \times 2 \times \sin \left( {\frac{A}{2}} \right) \times \cos \left( {\frac{A}{2}} \right)\;}} = \;2\frac{{\cos A}}{{\sin A}} = 2\cot A\)

Concept:

cos 2A = cos² A - sin² A = 2 cos² A - 1

sin 2A = 2 sin A × cos A

Calculation:

\(\Rightarrow \cot A+cosec~A=\frac{\cos A}{\sin A}+\frac{1}{\sin A}\)

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}\)

As we know that,

cos 2A = cos² A - sin² A = 2 cos² A - 1

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{\sin A}\)

As we know that,

sin 2A = 2 sin A × cos A

\(\Rightarrow \cot A+cosec~A=\frac{1+\cos A}{\sin A}\)

\(=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{\sin A}=\frac{2\times {{\cos }^{2}}\left( \frac{A}{2} \right)}{2\times \cos \left( \frac{A}{2} \right)\times \sin \left( \frac{A}{2} \right)}=\cot \left( \frac{A}{2} \right)\)

∴ The value of cot A + cosec A is \(\cot \left( \frac{A}{2} \right)\)

Concept:

sin (x + y) = sin x cos y + cos x sin y

sin (x - y) = sin x cos y - cos x sin y

Calculation:

sin 75°

= sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30° 

= \(\rm \frac{1}{\sqrt 2} \times \frac{\sqrt 3}{2} + \frac{1}{\sqrt 2} \times \frac{1}{2}\)

= \(\frac{\sqrt 3 +1 }{2\sqrt2}\)

Concept:

sin² x + cos² x = 1

sin 2x = 2 sin x cos x

Calculation:

Given: sin α + cos α = p

By squaring both the sides, we get

⇒ sin² α + cos² α + 2 sin α cos α = p²

As we know that, sin² x + cos² x = 1 and sin 2x = 2 sin x cos x

⇒ 1 + sin 2α = p²

⇒ sin 2α = p² – 1

As we can write cos² 2α = 1 – sin² 2α

⇒ cos2 2α = 1 – (p² – 1)²

Concept:

• cos 2A = cos² A – sin² A = 1 – 2 sin² A = 2 cos² A – 1
• \(\sin \frac{A}{2} = \; \pm \sqrt {\frac{{1 - \cos A}}{2}}\)
• \(\cos \frac{A}{2} = \; \pm \sqrt {\frac{{1 + \cos A}}{2}}\)

Calculation:

As we know that, cot x = cos x / sin x

\(⇒ \cot \left( {22{{\frac{1}{2}}^\circ }} \right) = \frac{{{\rm{cos\;}}\left( {22{{\frac{1}{2}}^\circ }} \right)}}{{\sin \left( {22{{\frac{1}{2}}^\circ }} \right)}}\)

As we know that, \(\sin \frac{A}{2} = \; \pm \sqrt {\frac{{1 - \cos A}}{2}} \)

\(⇒ \sin \left( {{{\frac{{45}}{2}}^\circ }} \right) = \; \pm \sqrt {\frac{{1 - \cos \left( {45^\circ } \right)}}{2}}\)

As we know that, 0° < A / 2 < 90° where all trigonometric ratios are positive.

\(⇒ \sin \left( {{{\frac{{45}}{2}}^\circ }} \right) = \sqrt {\frac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} \)      -------(1)

As we know that, \(\cos \frac{A}{2} = \; \pm \sqrt {\frac{{1 + \cos A}}{2}} \)

\(⇒ \cos \left( {{{\frac{{45}}{2}}^\circ }} \right) = \; \pm \sqrt {\frac{{1 + \cos \left( {45^\circ } \right)}}{2}} \)      -------(2)

As we know that, 0° < A / 2 < 90° where all trigonometric ratio’s are positive.

\(⇒ \cos \left( {{{\frac{{45}}{2}}^\circ }} \right) = \sqrt {\frac{{\sqrt 2 \; + \;1}}{{2\sqrt 2 }}} \)

So, from equation (1) and (2), we get

\(⇒ \cot \left( {22{{\frac{1}{2}}^\circ }} \right) = \frac{{\sqrt {\frac{{\sqrt 2 \; + \;1}}{{2\sqrt 2 }}} }}{{\sqrt {\frac{{\sqrt 2 - 1}}{{2\sqrt 2 }}} }} = 1 + \sqrt 2 \)

\(\rm ⇒ \sqrt {\frac {\sqrt 2\ + \ 1}{\sqrt 2\ - \ 1}}\)

Rationalization above the equation

\(\rm ⇒ \sqrt {\frac {\sqrt 2\ + \ 1}{\sqrt 2\ - \ 1} \times \frac {\sqrt 2 \ + \ 1}{\sqrt 2\ + \ 1}}\)

\(\rm \Rightarrow \sqrt {\frac {(\sqrt 2\ + \ 1)^2}{(\sqrt 2)^2 - \ (1)^2}}\)       (∵ a² - b² = (a + b)(a - b)

\(\rm \Rightarrow \sqrt {\frac {(\sqrt 2\ + \ 1)^2}{2 - \ 1}} = 1 + \sqrt 2\)

CONCEPT:

Trigonometric Identities:

• \({\rm{cosec\theta }} = \frac{1}{{{\rm{sin\theta }}}}\)
• \({\rm{sec\theta }} = \frac{1}{{{\rm{cos\theta }}}}\)
• \({\rm{cos\;}}30^\circ = {\rm{sin\;}}60^\circ = \frac{{\sqrt 3 }}{2}\)
• \({\rm{cos\;}}60^\circ = {\rm{sin\;}}30^\circ = \frac{1}{2}\)
• (cos A× cos B) – (sin A × sin B) = cos(A + B)
• 2 × sin A × cos A = sin (2A)
• sin (90 - θ) = cos θ

CALCULATION:

Given that: √3 cosec 20° – sec 20°

As we know that, cosec θ = 1/ sin θ and sec θ = 1/ cos θ

\(\Rightarrow \frac{{\sqrt 3 }}{{\sin 20^\circ }} - \frac{1}{{\cos 20^\circ }}\)

\(\Rightarrow 2\left\{ {\frac{{\frac{{\sqrt 3 }}{2}}}{{\sin 20^\circ }} - \frac{{\frac{1}{2}}}{{\cos 20^\circ }}} \right\}\)

As we know that, cos 30° = √3 / 2, sin 30° = ½

\(\Rightarrow 2\left\{ {\frac{{\cos 30^\circ }}{{\sin 20^\circ }} - \frac{{\sin 30^\circ \;}}{{\cos 20^\circ }}} \right\}\)

\(\Rightarrow 2\left\{ {\frac{{\cos 30^\circ \cos 20^\circ - \sin 30^\circ \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }}} \right\}\)

As we know that, (cos A × cos B) – (sin A × sin B) = cos(A + B)

\(\Rightarrow 2\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}} \right\}\)

\(\Rightarrow 2 \times 2\;\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{2 \times \sin 20^\circ \cos 20^\circ }}} \right\}\)

As we know that, 2 × sin A × cos A = sin (2A)

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 40^\circ }}} \right\}\)

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{{\rm{sin}}\left( {90^\circ - 50^\circ } \right)}}} \right\}\)

As we know that, sin (90 - θ) = cos θ

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\cos 50^\circ }}} \right\}\)

Concept:

a² - b² = (a + b)(a - b)

sin C + sin D = \(\rm 2 \sin \;(\frac {C +D}{2}) \cos \;(\frac {C -D}{2})\)

sin C - sin D = \(\rm 2 \cos \;(\frac {C +D}{2}) \sin \;(\frac {C -D}{2})\)

2 sin x cos x = sin 2x

Calculation:

Consider, sin2 6x – sin2 4x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= [ \(\rm 2 \sin \;(\frac {6x + 4x}{2}) \cos \;(\frac {6x- 4x}{2})\) ] [\(\rm 2 \cos \;(\frac {6x + 4x}{2}) \sin \;(\frac {6x- 4x}{2})\)]

= \(\rm [2 \sin \; 5x \cos \;x ] [2 \cos \; 5x \sin \;x]\)

Rearranging the terms, we get

= \(\rm [2 \sin 5x \cos 5x ] [2 \sin x\cos x]\)

= sin 10x sin 2x

= sin 2x sin 10x

Concept:

cosec (2nπ - θ) = - cosec θ

Calculation:

cosec (-1410°) = -cosec (1410°)

= -cosec (4× 360° - 30°)

Since cosec(2nπ - θ) = - cosecθ, we can write:

- cosec (4×360° - 30°) = - (-cosec (30°))

= cosec 30° 

= 2

Hence, option (3) is correct.

Concept:

cos (90 - x) = sin x

sin A - sin B = 2cos\(\rm (A+B) \over 2\)sin \(\rm (A-B) \over 2\)

cos 45° = \(1 \over \sqrt 2\)

Calculation:

Given,

cos18° - sin18°

= cos (90° - 72°) - sin 18°

= sin 72° - sin 18°

= 2 cos \(\rm (72°+18°) \over 2\)× sin \(\rm (72°-18°) \over 2\)

= 2 cos 45° sin 27°

= 2× \(1 \over \sqrt 2\)× sin 27°

= √2 sin 27° 

Concept:

\(\rm sinA - sin B = 2cos\frac{A+B}{2}sin\frac{A-B}{2}\)

\(\rm cosA+cosB=2cos\frac{A+B}{2}cos\frac{A-B}{2}\)

Calculation:

\(\rm \frac{cosA+cosB}{sinA-sinB}=\)

\(\rm \frac {2cos\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}sin\frac{A-B}{2}}\)

=\(\rm cot\frac{A-B}{2}\)

Hence, option (4) is correct.