Multiple and Sub-multiple Angles MCQ Quiz in मल्याळम - Objective Question with Answer for Multiple and Sub-multiple Angles - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Used formulas:

• sin 2θ = 2 sin θ cos θ.
• Cos 2θ = cos² θ – sin² θ = 1 – 2 sin² θ
• cos 3θ = 4cos³ θ – 3cos θ
• sin² θ + cos² θ = 1

The table below shows the sign of trigonometric functions in different quadrants:

Calculation:

Given: sin 2θ = cos 3θ, 0 < θ < π/2.

As we know that, sin 2θ = 2 sin θ cos θ and cos 3θ = 4cos³ θ – 3cos θ.

⟹ 2 sin θ cos θ = 4cos³ θ – 3cos θ

⟹ 2 sin θ = 4cos² θ – 3

⟹ 2 sin θ = 4(1 – sin² θ) – 3 = 4 - 4 sin² θ – 3

⟹ 4 sin² θ + 2 sin θ – 1 = 0

Comparing the above equation with quadratic equation ax² + bx + c = 0, a = 4, b = 2 and c = – 1.

Now substituting the values in the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

we get,

\(sin\;\theta \; = \;\frac{{ - 2 \pm \sqrt { {2^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2\left( 4 \right)}} = \frac{{ - 2 \pm \sqrt {4 + 16} }}{8} = \frac{{ - 2 \pm \sqrt {20} }}{8} = \frac{{ - 1 \pm \sqrt 5 }}{4}\)

Thus, \(sin\;\theta = \frac{{ - 1 \pm \sqrt 5 }}{4}\).

Since, 0 < θ < π/2 ⟹ θ lies between 0° to 90°⟹ all ratios are positive.

\(\Rightarrow sin\;\theta = \frac{{ - 1 + \sqrt 5 }}{4}\)

As we know that, cos 2θ = cos² θ – sin² θ = 1 – 2 sin² θ

\( \Rightarrow \cos 2\theta = \;1 - 2{\left( {\frac{{ - 1 + \sqrt 5 }}{4}} \right)^2} = \frac{{1 + \sqrt 5 }}{4}\)

Concept:

cos (270° + θ) = sin θ

sin (180° + θ) = - sin θ

sin (60°) = \(\rm \frac{\sqrt{3}}{2}\)

Calculation:

Given:

\(\rm \frac{1}{cos \: 330^{\circ}} - \frac{1}{\sqrt{3} \: sin \: 240^{\circ} } \)

= \(\rm \frac{1}{cos \: (270^{\circ} + 60^{\circ})} - \frac{1}{\sqrt{3} \: sin \: (180^{\circ} + 60^{\circ}) }\)

= \(\rm \frac{1}{sin \: (60^{\circ})} + \frac{1}{\sqrt{3} \: sin \: ( 60^{\circ}) }\)

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

= \(\rm \frac{2}{3} (1 + \sqrt{3})\)

The value of \(\rm \frac{1}{cos \: 330^{\circ}} - \frac{1}{\sqrt{3} \: sin \: 240^{\circ} } \) is \(\rm \frac{2}{3} (1 + \sqrt{3})\)

Additional Information

Trigonometrical Ratios 

Concept:

Trigonometry Rule

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

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

Calculations:

Given expression is \(\rm \dfrac {\sin 4x - \sin 2x}{\cos 4x + \cos 2x} \)

= \(\rm \dfrac {2sin\; (\dfrac {4x-2x}{2})\;cos\; (\dfrac {4x+2x}{2})}{2cos\; (\dfrac {4x+2x}{2})\;cos\; (\dfrac {4x-2x}{2})} \)

= \(\rm \dfrac {\sin x}{\cos x}\)

= tan x

Hence, the expression \(\rm \dfrac {\sin 4x - \sin 2x}{\cos 4x + \cos 2x} \) is equal to tan x

Concept:

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (2nπ + θ) = sin θ.
• cos (2nπ + θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
• \(\rm \sin \left [(2n+1)\frac{\pi}{2}+\theta \right ]\) = (-1)ⁿ cos θ.
• \(\rm \cos \left [(2n+1)\frac{\pi}{2}+\theta \right ]\) = (-1)ⁿ (-sin θ).

Calculation:

We know that cos (π - θ) = -cos θ.

∴ cos 20∘ + cos 40∘ + cos 140∘ + cos 160∘

= cos 20∘ + cos 40∘ + cos (180^∘ - 40∘) + cos (180∘ - 20∘)

= cos 20∘ + cos 40∘ -  cos 40∘ - cos 20∘

= 0

Concept:

(a + b)² = a² + b² + 2ab

cos (a - b) = cos a. cos b + sin a. sin b

sin 2a = 2sin a .cos a

sin (a + b) = sin a. cos b + sin b. cos a

sin²a + cos²a = 1

sin a + sin b = 2 sin [(a + b)/2] ⋅ cos [(a - b)/2]

Calculation:

sin α + sin β = p

squaring both the sides,

(sin α + sin β)² = p²

sin² α + sin² β + 2sin α.sin β = p²      ....(i)

cos α + cos β = q

squaring both the sides,

(cos α + cos β)² = q²

cos2 α + cos2 β + 2cos α.cos β = q2      ....(ii)

Adding equ (i) and (ii)

sin2 α + sin2 β + cos2 α + cos2 β + 2sin α.sin β + 2cos α.cos β = p2 + q2

1 + 1 + 2(sin α.sin β + cos α.cos β) = p2 + q2

2 + 2(sin α.sin β + cos α.cos β) = p2 + q2

2[1 + sin α.sin β + cos α.cos β] = p2 + q2

1 + sin α.sin β + cos α.cos β = \(\rm \frac{p^2 + q^2}{2}\)

1 + cos (α - β) = \(\rm \frac{p^2 + q^2}{2}\)      ....(iii)

Now 

p \(\times\) q = (sin α + sin β) \(\times\) (cos α + cos β)

= sin α.cos α + sin β.cos β + sin α.cos β + sin β.cos α

= \(\rm \frac{sin 2\alpha + sin 2\beta }{2}\) + sin (α + β)

= sin (α + β) . cos (α - β) + sin (α + β)

= sin (α + β)[cos (α - β) + 1]

From equ (iii) we can write 

= sin (α + β) \(\times\)\(\rm \frac{p^2 + q^2}{2}\)

sin (α + β) = \(\rm \frac{pq}{\frac{p^2 + q^2}{2}}\)

sin (α + β) = \(\rm \frac{2pq}{p^{2} + q^{2}}\)

\(\rm {1\over 2}\ sin(\alpha + \beta) \) = \(\rm \frac{pq}{p^{2} + q^{2}}\)

Concept: 

• sin A + sin B = \(\rm 2sin\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)  
• cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)  

Calculation: 

We have ,  \(\rm \frac{\cos 7x\ +\ \cos 6x\ +\ \cos5x }{\sin 7x\ +\ \sin 6x\ +\ \sin 5x}\) 

We know that, 

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

cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)   

using the above formula , we get

= \(\rm \frac{(\cos 7x\ +\ \cos 5x\ ) +\ \cos6x }{(\sin 7x\ +\ \sin 5x\ )+\ \sin 6x}\) 

= \(\rm \frac{ 2cos\left ( \frac{7x+5x}{2} \right )\cos \left ( \frac{7x-5x}{2} \right )+\cos 6x}{2sin\left ( \frac{7x+5x}{2} \right )\cos \left ( \frac{7x-5x}{2} \right )+\sin 6x}\) 

= \(\rm \frac{2 \cos 6x\ \cos x +\cos 6x }{2 \sin 6x\ \cos x +\sin6x}\) 

= \(\rm \frac{\cos6x\ \times (2 \ cosx + 1)}{\sin6x\ \times(2 \ cosx + 1)}\) 

= cot 6x .

The correct option is 4

Concept:

sin (2nπ + θ) = sin θ, n ∈ Integer

sin 90° = 1

Calculation:

We have to find the value of sin 450°

sin 450° = sin (360° + 90°)                    

= sin (2π + 90°)                                 (∵ π = 180°)

= sin 90°                                            (∵ sin (2nπ + θ) = sin θ)

= 1

Concept:

• Sin 2x = 2 sin x cos x
• 2 sin x cos y = sin (x + y) + sin (x - y)

Calculation:

Given: cos (π / 15) cos (2π / 15) cos (4π / 15) cos (7π / 15)

By dividing and multiplying the given expression with 2 sin (π / 15), we get

⇒ cos (π / 15) cos (2π / 15) cos (4π / 15) cos (7π / 15)

\(= \frac{{2\sin \left( {\frac{\pi }{{15}}} \right)}}{{2\sin \left( {\frac{\pi }{{15}}} \right)}} \times \cos \left( {\frac{\pi }{{15}}} \right) \times \;\cos \left( {\frac{{2\pi }}{{15}}} \right) \times \cos \left( {\frac{{4\pi }}{{15}}} \right) \times \cos \left( {\frac{{7\pi }}{{15}}} \right)\)

\(= \frac{{2\sin \left( {\frac{\pi }{{15}}} \right) \times \;\cos \left( {\frac{\pi }{{15}}} \right)}}{{2\sin \left( {\frac{\pi }{{15}}} \right)}} \times \cos \left( {\frac{{4\pi }}{{15}}} \right) \times \cos \left( {\frac{{7\pi }}{{15}}} \right)\)

As we know that, sin 2x = 2 sin x cos x

⇒ cos (π / 15) cos (2π / 15) cos (4π / 15) cos (7π / 15)

 = [sin (2π / 15) / 2 sin (π / 15)] × cos (2π / 15) cos (4π / 15) cos (7π / 15)

 = [2 sin (2π / 15) × cos (2π / 15) / 2 × 2 sin (π / 15)] cos (4π / 15) cos (7π / 15)

As we know that, sin 2x = 2 sin x cos x

= [sin (4π / 15) / 4 sin (π / 15)] × cos (4π / 15) cos (7π / 15)

= [2 × sin (4π / 15) × cos (4π / 15) / 2 × 4 sin (π / 15)] × cos (7π / 15)

As we know that, sin 2x = 2 sin x cos x

= [sin (8π / 15) / 8 × sin (π / 15)] × cos (7π / 15)

= [2 × sin(8π / 15) × cos (7π / 15)] / 2 × 8 × sin (π / 15)

As we know that, 2 sin x cos y = sin (x + y) + sin (x - y)

= [sin π + sin (π / 15)] / 16 sin (π / 15)

As we know that, sin π = 0

⇒ cos (π / 15) cos (2π / 15) cos (4π / 15) cos (7π / 15) = sin (π / 15) / 16 sin (π / 15) = 1 / 16.

Concept:

I. \(\tan \left( {A - B} \right) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\)

Calculation:

Given: tan A = x +1 and tan B = x – 1

As we know that, \(\tan \left( {A - B} \right) = \frac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\)

\(\Rightarrow \tan \left( {A - B} \right) = \frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{1 + \left( {x + 1} \right)\;\left( {x - 1} \right)}}\)

\(\Rightarrow \tan \left( {A - B} \right) = \frac{2}{{{x^2}}}\)

CONCEPT:

cos 3x = 4 cos³ x - 3 cos x

CALCULATION:

Given: cos x = 1/2

As we know that, cos 3x = 4 cos3 x - 3 cos x

⇒ cos 3x = (1/2) - (3/2) =  -1

Hence, the correct option is 2.