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 $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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°) = $\frac{\sqrt{3}}{2}$

Calculation:

Given:

$\frac{1}{\mathrm{c}\mathrm{o}\mathrm{s}{330}^{\circ }}-\frac{1}{\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}{240}^{\circ }}$

= $\frac{1}{\mathrm{c}\mathrm{o}\mathrm{s}({270}^{\circ }+{60}^{\circ })}-\frac{1}{\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}({180}^{\circ }+{60}^{\circ })}$

= $\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}({60}^{\circ })}+\frac{1}{\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}({60}^{\circ })}$

= $\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}\sqrt{3}}$

= $\frac{2}{3}(1+\sqrt{3})$

The value of $\frac{1}{\mathrm{c}\mathrm{o}\mathrm{s}{330}^{\circ }}-\frac{1}{\sqrt{3}\mathrm{s}\mathrm{i}\mathrm{n}{240}^{\circ }}$ is $\frac{2}{3}(1+\sqrt{3})$

Additional Information

Trigonometrical Ratios 

Concept:

Trigonometry Rule

$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{D}=2\sin ({\displaystyle \frac{\mathrm{C}-\mathrm{D}}{2}})\cos ({\displaystyle \frac{\mathrm{C}+\mathrm{D}}{2}})$

$\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{C}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{D}=2\cos ({\displaystyle \frac{\mathrm{C}+\mathrm{D}}{2}})\mathrm{c}\mathrm{o}\mathrm{s}({\displaystyle \frac{\mathrm{C}-\mathrm{D}}{2}})$

Calculations:

Given expression is ${\displaystyle \frac{\sin 4\mathrm{x}-\sin 2\mathrm{x}}{\cos 4\mathrm{x}+\cos 2\mathrm{x}}}$

= ${\displaystyle \frac{2\mathrm{s}\mathrm{i}\mathrm{n}({\displaystyle \frac{4\mathrm{x}-2\mathrm{x}}{2}})\mathrm{c}\mathrm{o}\mathrm{s}({\displaystyle \frac{4\mathrm{x}+2\mathrm{x}}{2}})}{2\mathrm{c}\mathrm{o}\mathrm{s}({\displaystyle \frac{4\mathrm{x}+2\mathrm{x}}{2}})\mathrm{c}\mathrm{o}\mathrm{s}({\displaystyle \frac{4\mathrm{x}-2\mathrm{x}}{2}})}}$

= ${\displaystyle \frac{\sin \mathrm{x}}{\cos \mathrm{x}}}$

= tan x

Hence, the expression ${\displaystyle \frac{\sin 4\mathrm{x}-\sin 2\mathrm{x}}{\cos 4\mathrm{x}+\cos 2\mathrm{x}}}$ 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.
• $\sin [(2\mathrm{n}+1)\frac{\pi }{2}+\theta ]$ = (-1)ⁿ cos θ.
• $\cos [(2\mathrm{n}+1)\frac{\pi }{2}+\theta ]$ = (-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 = $\frac{{\mathrm{p}}^2+{\mathrm{q}}^2}{2}$

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

= $\frac{\mathrm{s}\mathrm{i}\mathrm{n}2\alpha +\mathrm{s}\mathrm{i}\mathrm{n}2\beta }{2}$ + 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$$\frac{{\mathrm{p}}^2+{\mathrm{q}}^2}{2}$

sin (α + β) = $\frac{\mathrm{p}\mathrm{q}}{\frac{{\mathrm{p}}^2+{\mathrm{q}}^2}{2}}$

sin (α + β) = $\frac{2\mathrm{p}\mathrm{q}}{{\mathrm{p}}^2+{\mathrm{q}}^2}$

$\frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{n}(\alpha +\beta )$ = $\frac{\mathrm{p}\mathrm{q}}{{\mathrm{p}}^2+{\mathrm{q}}^2}$

Concept: 

• sin A + sin B = $2\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$  
• cos A + cos B = $2\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$  

Calculation: 

We have ,  $\frac{\cos 7\mathrm{x}+\cos 6\mathrm{x}+\cos 5\mathrm{x}}{\sin 7\mathrm{x}+\sin 6\mathrm{x}+\sin 5\mathrm{x}}$ 

We know that, 

sin A + sin B = $2\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$   

cos A + cos B = $2\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)$   

using the above formula , we get

= $\frac{(\cos 7\mathrm{x}+\cos 5\mathrm{x})+\cos 6\mathrm{x}}{(\sin 7\mathrm{x}+\sin 5\mathrm{x})+\sin 6\mathrm{x}}$ 

= $\frac{2\mathrm{c}\mathrm{o}\mathrm{s}\left(\frac{7\mathrm{x}+5\mathrm{x}}{2}\right)\cos \left(\frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\cos 6\mathrm{x}}{2\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{7\mathrm{x}+5\mathrm{x}}{2}\right)\cos \left(\frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\sin 6\mathrm{x}}$ 

= $\frac{2\cos 6\mathrm{x}\cos \mathrm{x}+\cos 6\mathrm{x}}{2\sin 6\mathrm{x}\cos \mathrm{x}+\sin 6\mathrm{x}}$ 

= $\frac{\cos 6\mathrm{x}\times (2\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}+1)}{\sin 6\mathrm{x}\times (2\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}+1)}$ 

= 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.