2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3

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Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3

Part – A

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 One Marks Questions and Answers

Question 1.
Express each of the following a sum or difference of two trigonometric functions.
(i) sin 5A · cos3A
(ii) cos 4A sin 2A
(iii) cos \(\left(\frac{5 \theta}{2}\right)\) . sin \(\left(\frac{\theta}{2}\right)\)
(iv) 2 cos 70° cos 10°
Answer:
(i) Using transformation formulae
sin5A . cos3A = \(\frac { 1 }{ 2 }\)[sin(5A +3A) + sin(5A – 3A)]
= \(\frac { 1 }{ 2 }\)(sin 8A + sin 2A)

(ii) cos4A – sin2A = \(\frac { 1 }{ 2 }\) [sin 6A – sin 2A]

(iii) cos \(\left(\frac{5 \theta}{2}\right)\) . sin \(\left(\frac{\theta}{2}\right)\) = \(\frac { 1 }{ 2 }\)[sin 3θ – sin 2θ]

(iv) 2cos 70° . cos10° = 2 x \(\frac { 1 }{ 2 }\) [cos80° + cos60°]
= cos80° + \(\frac { 1 }{ 2 }\)

Question 2.
Express each of the following as the product of two trigonometric reactions.
(i) sin 12x + sin 4x
(ii) sin 7A – sin 3A
(iii) cos 2θ + cos 6θ
(iv) sin 80° – sin 40°
Answer:
(i) sin12x + sin4x = 2.sin \(\frac{12 x+4 x}{2}\) . cos \(\frac{12 x-4 x}{2}\)

(ii) sin7A – sin3A = 2 cos 5A.sin2A

(iii) cos6θ + cos2θ = 2 cos4θ.cos2θ

(iv) sin 80° – sin 40° = 2 cos60°. sin 20° = 2.\(\frac { 1 }{ 2 }\)sin 20° = sin 20°

Part – B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3Two Marks Questions and Answers

Prove the following

Question 1.
\(\frac{\cos 2 A-\cos 12 A}{\sin 12 A-\sin 2 A}=\tan 7 A\)
Answer:

Question 2.
\(\frac{\sin x-\sin y}{\sin x \sin y}=\tan \left(\frac{x-y}{2}\right) \cdot \cot \left(\frac{x+y}{2}\right)\)
Answer:

Question 3.
\(\frac{\sin 2 \alpha+\sin 3 \alpha}{\cos 2 \alpha-\cos 3 \alpha}=\cot \left(\frac{\alpha}{2}\right)\)
Answer:

Question 4.
\(\sin \left(\frac{\pi}{3}+A\right)-\sin \left(\frac{\pi}{3}-A\right)=\sin A\)
Answer:

Question 5.
cos A + cos (120° – A) + cos(120° + A) = 0
Answer:
L.H.S. = cosA + cos A(120° – A) + cos(120° + A)
= cosA + 2cos 120°. cosA = cosA + 2cosA (-cos60°)
= cos A – 2 cos A . \(\frac { 1 }{ 2 }\) = 0 = R.H.S.

Question 6.
sin 65° + cos 65° = \(\sqrt{2} \cos 20^{\circ}\)
Answer:
L.H.S. sin65° + cos65°
= sin(45° + 20°) + cos(45° + 20°)

= 2. \(\frac{1}{\sqrt{2}}\). cos20°= \(\sqrt{2}\) cos20° = R.H.S.

Question 7.
If A + B + C = π, Prove that tan A + tan B + tan C = tan A . tanB · tan C
Answer:
Given A + B + C = π
⇒ A + B = π – C
tan(A + B) = tan(180 – C)
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=-\frac{\tan C}{1}\)
tanA + tanB = – tanc – tanB tanB · tanc
tanA + tanB + tanC = tanA tanB tanC

Question 8.
Prove that \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}=\tan 2 A\)
Answer:

Question 9.
If A + B + C = 180°. Prove that cot B. cot C + cot C · cot A + cot A · cot B = 1
Answer:
Given A + B + C = 180°; A + B = 180° – C; cot(A + B) = cot (180° – C)
\(\frac{\cot (\mathrm{A}) \cdot \cot (\mathrm{B})-1}{\cot \mathrm{A}+\cot \mathrm{B}}=-\cot \mathrm{C}\)
cotA . cotB – 1 = -cotA · cotC – cot B . cotC.
∴ cotA . cotB + cotB cotC + cotC . cotA = 1.

Question 10.
If A + B + C = \(\frac{\pi}{2}\) . P.T tanA · tanB + tan B . tan C + tan C · tan A = 1
Answer:
Given A+B+C = \(\frac{\pi}{2}\); A + B = 90° – C; tan(A + B) = tan(90° – C)
\(\frac{\tan A+\tan B}{1-\tan A \tan B}=\cot C=\frac{1}{\tan C}\)
⇒ tanA tanC + tanB tanC = 1 – tanA · tanB
⇒ tan A tanB + tanB tanC + tanC tanA = 1.

Part – C

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 Five Marks Questions and Answers

Prove the following

Question 1
\(\frac{\cos 7 x+\cos 3 x-\cos 5 x-\cos x}{\sin 7 x-\sin 3 x-\sin 5 x+\sin x}=\cot 2 x\)
Answer:

Question 2.
cos10°.cos30°. cos50°.cos 70° = \(\frac { 3 }{ 16 }\)
Answer:

Question 3.
cos 20° · cos40° · cos60° . cos80° = \(\frac { 1 }{ 16 }\)
Answer:
L.H.S. = cos 20°.cos40°.cos60°.cos80°
= cos60°.cos80°.cos40°.c0s60°

Question 4.
If A + B + C= 180°C. Prove that sin²a + sin² \(\left(\frac{2 \pi}{3}+A\right)\) + sin \(\left(\frac{2 \pi}{3}-A\right)\) = \(\frac { 3 }{ 2 }\)
Answer:
Given A + B + C = 180° \(\left[\because \frac{2 \pi}{3}=120^{\circ}\right]\)
sin²A + sin²(120° + A) + sin²(120° – A)

Question 5.
sin 2A + sin 2B + sin 2C = 4 sin A · sinB · sin C
Answer:
L.H.S. = sin2A + sin2B + sin2C
= 2sin(A + B) · cos(A – B) + 2sinC cosC
= 2sinC.cos(A – B) + 2 sinC cosC = 2sinC [cos(A – B) – cos(A + B)]
= 2 sinC[-2 sinA · sin(-B)]
= 4sinA . sinB . sinC ∵ sin(-B)
= R.H.S. = SinB.

Question 6.
tan 2A + tan 2B + tan 2C = tan 2A . tan 2B . tan 2C
Answer:
Given
A + B + C = 180°
⇒ 2A + 2B + 2C = 360°
2A + 2B = 360° – 2C
tan(2A + 2B) = tan(360° – 2C)
\(\frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \cdot \tan 2 B}=\frac{-\tan 2 C}{1}\)
tan2A + tan2B = – tan2C + tan2A tan2B tan2C
tan2A + tan2B + tan2C = tan 2A tan2B · tan2C

Question 7.
sin 4A + sin 4B + sin 4C = -4sin2A . sin2B · sin2C
Answer:
L.H.S = sin4A + sin4B + sin4C [A+B+C = 180]
= 2 sin(2A + 2B) · cos(2A – 2B) + 2 sinc.cos2C [2A + 2B + 2C = 360]
= (-sin2C) · cos(2A – 2B) + (2 sin2C · cos2C) [2A + 2B = 360 – 20]
= -2 sin 2C [cos(2A – 2B) – cos2C] [sin(2A + 2B) = – sin2C]
= – 2 sin 2C [cos(2A – 2B) – cos(2A + 2B)] [cos(2A + 2B) = cos 2C]
= -2 sin2C[-2 sin2A · sin(-2B)]
= – 4 sin2A · sin2B · sin2C
= R.H.S.

Question 8.
cos²A + cos²B – cos²C = 1 – 2 sin A . sin B · sin C
Answer:
L.H.S. = cos²A + cos²B – cos²C

= \(\frac { 1 }{ 2 }\)[1+2cos(A+B) cos(A – B) – (2cos²C – 1)]
= \(\frac { 1 }{ 2 }\)[1+2cos(A+B).cos(A – B) – 2cos²C +1]
= \(\frac { 1 }{ 2 }\)[2 + 2(-cosc) cos(A – B) 2cos²C]
= \(\frac { 1 }{ 2 }\)[2 – 2 cos C[cos (A – B) + cos C]
= \(\frac { 1 }{ 2 }\)[2 – 2cosC[cos(A – B) – cos(A+B)]]
= 1 – cosC[-2sin A . sin(-B)]
=1 – cosC[2sin A sin B]
= 1 – 2sinA sinB cosC = R.H.S.

Question 9.
\(\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin 2 A+\sin 2 B-\sin 2 C}=\tan A \cdot \tan B\)
Answer:

Question 10.
cos2A + cos2B + cos2C = -1 – 4 cosA cosB cosC
Answer:
L.H.S. = cos2A + cos2B + cos2C
= 2cos(A + B) · cos(A – B) + 2 cos²C – 1
= -2cosC . cos(A – B) + 2 cos²C – 1
= -2 cosC[cos (A – B) – cos C] – 1
= -2 cosC{cos (A – B) + cos(A + B)} – 1
= -2 cosC (2 cosA · cosB) – 1
= – 1 – 4 cos A cosB cosC = R.H.S.

Question 11.
sin²A + sin²B – sin²C = 2sinA · sin B. cosC
Answer:
L.H.S. = sin²A + sin²B – sin²C

= \(\frac { 1 }{ 2 }\)[1-(cos 2A + cos2B –cos2C)]
= \(\frac { 1 }{ 2 }\)[1-{{2cos(A+B). Cos(A – B) – 2 cos²C+1}]

= \(\frac { 1 }{ 2 }\)[+2cosC . cos(A – B) +2cos²c]
= \(\frac { 1 }{ 2 }\)[2cosC(cos(A – B)+cosC]
= \(\frac { 1 }{ 2 }\)[2cosC(cos(A – B) – cos(A + B))]
= \(\frac { 1 }{ 2 }\)[2cosC[-2sin A.sin(-B)]]
= 2 sinA sinB sinc

Question 12.
If A + B + C = π. Prove that:
cos 2A + cos 2B – cos2C = 1 – 4 sin A . sin B · cos C
Answer:
L.H.S. = cos2A + cos2B – cos2C
= 2cos(A + B) cos(A – B) – (2 cos²C – 1)
= -2cosC Cos(A – B) – 2 cos²C + 1 [∵ cosC=-cos(A+B)]
= 1 – 2cos C[cos(A – B) – cos (A + B)]
= 1 – 2cos C[-2 sinA . sin(- B)]
= 1 – 4 sin A sin B cos C = R.H.S.

Question 13.
If A + B + C = 180°. Prove that:
sin 2A – sin 2B + sin 2C = 4 cos A · sin B · cos C
Answer:
Given A + B + C = 180°
2A + 2B + 2C = 360°
2A + 2B = 360° – 2C
sin(2A + 2B) = sin(360° – 20) = – sin2C
cos(2A + 2B) = cos(360° – 2C) = cos 2C
L.H.S = sin2A – sin2B + sin2C
= 2cos(A + B) · sin(A – B) + 2sinC. cosC
= – 2cosC.sin(A – B) + 2 sinc.cosC
= 2 cosC [sinC – sin (A – B)]
= 2 cosC [sin(A + B) – sin(A – B)]
= 2 cos C [2cosA . sinB]
= 4 cosA sinB.cosC = R.H.S.

Question 14.
cos 2A – cos 2B + cos 2C = 1 – 4 sin A · cos B · sin C
Answer:
L.H.S. = cos 2A – cos2B + cos2C
= -2 sin(A + B) . sin(A – B) + 1 – 2sin²C
= 1 – 2 sinC sin (A – B) – 2 sin²C [∵ sin(A+B)-sinc]
= 1 – 2 sinC [sin(A – B) + sinC]
= 1 – 2 sinC [sin(A + B) + sin(A – B)]
= 1 – 2 sinC [2sinA . cosB]
= 1 – 4 sinA cosB sinC = R.H.S.