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

Last updated on Mar 19, 2025

നേടുക Multiple and Sub-multiple Angles ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Multiple and Sub-multiple Angles MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Multiple and Sub-multiple Angles MCQ Objective Questions

Top Multiple and Sub-multiple Angles MCQ Objective Questions

Multiple and Sub-multiple Angles Question 1:

Suppose sin 2θ = cos 3θ, here 0 < θ < π/2 then what is the value of cos 2θ?

  1. 1+54
  2. 1258
  3. 1+54
  4. 154

Answer (Detailed Solution Below)

Option 1 : 1+54

Multiple and Sub-multiple Angles Question 1 Detailed Solution

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:

T – Ratio’s

Quadrant I

Quadrant II

Quadrant III

Quadrant IV

Sin

+

+

-

-

Cos

+

-

-

+

Cosec

+

+

-

-

Sec

+

-

-

+

Tan

+

-

+

-

Cot

+

-

+

-

 

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±b24ac2a

we get,

sinθ=2±224(4)(1)2(4)=2±4+168=2±208=1±54

Thus, sinθ=1±54.

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

sinθ=1+54

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

cos2θ=12(1+54)2=1+54

Multiple and Sub-multiple Angles Question 2:

What is the value of 1cos33013sin240=?

  1. 23(13)
  2. 23(1+3)
  3. 13(1+3)
  4. 13(13)

Answer (Detailed Solution Below)

Option 2 : 23(1+3)

Multiple and Sub-multiple Angles Question 2 Detailed Solution

Concept:

cos (270° + θ) = sin θ

sin (180° + θ) = - sin θ

sin (60°) = 32

Calculation:

Given:

1cos33013sin240

1cos(270+60)13sin(180+60)

1sin(60)+13sin(60)

23+233

23(1+3)

The value of 1cos33013sin240 is 23(1+3)

Additional Information

Trigonometrical Ratios 

sin (90° + θ) = cos θ

cos (90° + θ) = - sin θ

tan (90° + θ) = - cot θ

cosec (90° + θ) = sec θ

sec ( 90° + θ) = - cosec θ

cot ( 90° + θ) = - tan θ

sin (180° + θ) = - sin θ

cos (180° + θ) = - cos θ

tan (180° + θ) = tan θ

cosec (180° + θ) = -cosec θ

sec (180° + θ) = - sec θ

cot (180° + θ) = cot θ

sin (270° + θ) = - cos θ

cos (270° + θ) = sin θ

tan (270° + θ) = - cot θ

cosec (270° + θ) = - sec θ

sec (270° + θ) = cosec θ

cot (270° + θ) = - tan θ

Multiple and Sub-multiple Angles Question 3:

The expression sin4xsin2xcos4x+cos2x is equal to

  1. - cot x
  2. tan x
  3. cot x
  4. None of these

Answer (Detailed Solution Below)

Option 2 : tan x

Multiple and Sub-multiple Angles Question 3 Detailed Solution

Concept:

Trigonometry Rule

sinCsinD=2sin(CD2)cos(C+D2)

cosC+cosD=2cos(C+D2)cos(CD2)

Calculations:

Given expression is sin4xsin2xcos4x+cos2x

2sin(4x2x2)cos(4x+2x2)2cos(4x+2x2)cos(4x2x2)

sinxcosx

= tan x

Hence, the expression sin4xsin2xcos4x+cos2x is equal to tan x

Multiple and Sub-multiple Angles Question 4:

cos 20 + cos 40 + cos 140 + cos 160 = ?

  1. 0
  2. 1
  3. 2
  4. None of these.

Answer (Detailed Solution Below)

Option 1 : 0

Multiple and Sub-multiple Angles Question 4 Detailed Solution

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.
  • sin[(2n+1)π2+θ] = (-1)n cos θ.
  • cos[(2n+1)π2+θ] = (-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

Multiple and Sub-multiple Angles Question 5:

If sin α + sin β = p and cos α + cos β = q, then find the value of 12 sin(α+β).

  1. pqp2+q2
  2. 2pqp2q2
  3. 2pqp2+q2
  4. None of these

Answer (Detailed Solution Below)

Option 1 : pqp2+q2

Multiple and Sub-multiple Angles Question 5 Detailed Solution

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 = p2+q22

1 + cos (α - β) = p2+q22      ....(iii)

Now 

× q = (sin α + sin β) × (cos α + cos β)

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

sin2α+sin2β2 + 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) ×p2+q22

sin (α + β) = pqp2+q22

sin (α + β) = 2pqp2+q2

12 sin(α+β) = pqp2+q2

Multiple and Sub-multiple Angles Question 6:

Evaluate , cos7x + cos6x + cos5xsin7x + sin6x + sin5x .

  1. tan 6x
  2. tan 2x
  3.  cot 4x
  4. cot 6x

Answer (Detailed Solution Below)

Option 4 : cot 6x

Multiple and Sub-multiple Angles Question 6 Detailed Solution

Concept: 

  • sin A + sin B = 2sin(A+B2)cos(AB2)  
  • cos A + cos B = 2cos(A+B2)cos(AB2)  

Calculation: 

We have ,  cos7x + cos6x + cos5xsin7x + sin6x + sin5x 

We know that, 

sin A + sin B = 2sin(A+B2)cos(AB2)   

cos A + cos B = 2cos(A+B2)cos(AB2)   

using the above formula , we get

(cos7x + cos5x )+ cos6x(sin7x + sin5x )+ sin6x 

2cos(7x+5x2)cos(7x5x2)+cos6x2sin(7x+5x2)cos(7x5x2)+sin6x 

2cos6x cosx+cos6x2sin6x cosx+sin6x 

cos6x ×(2 cosx+1)sin6x ×(2 cosx+1) 

= cot 6x .

The correct option is 4

Multiple and Sub-multiple Angles Question 7:

Find the value of sin 450° 

  1. 1
  2. -1
  3. 0
  4. 12

Answer (Detailed Solution Below)

Option 1 : 1

Multiple and Sub-multiple Angles Question 7 Detailed Solution

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

Multiple and Sub-multiple Angles Question 8:

Find the value of expression: cos (π / 15) cos (2π / 15) cos (4π / 15) cos (7π / 15)?

  1. 1 / 16
  2. 1 / 32
  3. 1 / 64
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 1 / 16

Multiple and Sub-multiple Angles Question 8 Detailed Solution

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)

=2sin(π15)2sin(π15)×cos(π15)×cos(2π15)×cos(4π15)×cos(7π15)

=2sin(π15)×cos(π15)2sin(π15)×cos(4π15)×cos(7π15)

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.

Multiple and Sub-multiple Angles Question 9:

If tan A = x +1 and tan B = x – 1, then x2 tan (A - B) = ?

  1. 1
  2. x
  3. 0
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Multiple and Sub-multiple Angles Question 9 Detailed Solution

Concept:

I. tan(AB)=tanAtanB1+tanAtanB

Calculation:

Given: tan A = x +1 and tan B = x – 1

As we know that, tan(AB)=tanAtanB1+tanAtanB

tan(AB)=(x+1)(x1)1+(x+1)(x1)

tan(AB)=2x2

⇒ x2 tan (A - B) = 2

Multiple and Sub-multiple Angles Question 10:

If cos x = 1/2 then find the value of cos 3x ?

  1. 1
  2. -1
  3. 3/4
  4. None of these

Answer (Detailed Solution Below)

Option 2 : -1

Multiple and Sub-multiple Angles Question 10 Detailed Solution

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.

Get Free Access Now
Hot Links: teen patti sweet teen patti apk download teen patti flush teen patti baaz