2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

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Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae
2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae One Mark Questions and Answers

Question 1.
Find the value of Sin 70°. Cos 20° + Cos 70°. Sin 20°.
Answer:
= Sin (70° + 20°) = Sin 90° = 1

Question 2.
Write the value of Sin 75°
Answer:
Sin (45° + 30°) = Sin 45. Cos 30 + Cos 45. Sin 30
\(=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

Question 3.
If Sin \(A=\frac{3}{5}\). Find Sin 2 A
Answer:
Cos \(A=\frac{4}{5}\)
∴ Sin 2 A = 2 Sin A. Cos A
\(=2 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{24}{25}\)

Question 4.
Write the value of Cos 75° ?
Answer:
Cos (45 + 30°)= Cos 45. Cos 30 – Sin 45. Sin 30
\(=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Question 5.
Express the following as sum or difference of two Trigonometric functions.
(i) Sin 5 A. Cos 3 A
Answer:

(ii) Cos 55°. Cos 25°
Answer:

(iii) \(\cos \frac{5 \theta}{2} \cdot \sin \frac{3 \theta}{2}\)
Answer:

(iv) \(\sin \frac{\pi}{5} \cdot \sin \frac{3 \pi}{2}\)
Answer:

Question 6.
Express each of the following as the product of two trigonometric functions.
(i) Sin 12x + Sin 4x
Answer:
\(=2 \sin \frac{12 x+4 x}{2} \cdot \cos \frac{12 x-4 x}{2}\)
= 2 Sin 8x. Cos 4x.

(ii) Sin 7 A-Sin 3 A
Answer:
\(=2 \cos \frac{7 A+3 A}{2} \cdot \sin \frac{7 A-3 A}{2}\)
= 2 Cos 5 A. Sin 2 A

(iii) Cos 2θ + Cos 4θ
Answer:
\(=2 \cos \frac{2 \theta+4 \theta}{2} \cdot \cos \frac{2 \theta-4 \theta}{2}\)
= 2 Cos 3θ. Cos θ

(iv) Sin 80° – Sin 40°
Answer:
= +2 Cos 120°. Sin 20°.

2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae Two Marks Questions and Answers

Question 1.
P.T Cos (A + B). Cos (A – B) = Cos² A – Sin² B.
Answer:
L.H.S = Cos (A + B). Cos (A – B)
= (Cos A. Cos B – sin A. Sin B) (Cos A. Cos B + Sin A Sin B)
= Cos² A. Cos² B – Sin² A. Sin² B = Cos² A (1 – Sin² B) – Sin² B(1 – Cos² A)
= Cos²A – Cos²A – Sin²B – Sin²B + Cos²A. – Sin²B
= Cos² A – Sin² B = R.H.S.

Question 2.
P.T \(\tan \left(\frac{\pi}{4}-A\right)=\frac{1-\tan A}{1+\tan A}\)
Answer:

Question 3.
P.T. Cot (A – B) = Cot A. Cot B + 1
Answer:

 

Question 4.
\(\frac{\cos 3 A}{\sec A}+\frac{\sin 3 A}{\csc A}=\cos _{2} A\)
Answer:
L.H.S = Cos 3 A. Cos A + Sin 3 A. Sin A
= Cos (3 A – A)
= Cos2 A = R.H.S.

Question 5.
P. T. tan 75° + Cot 75° = 4
Answer:

Question 6.
P. T. Sin 105° + Cos 105° = \(\frac{1}{\sqrt{2}}\)
Answer:

Question 7.
If Sin A \(\frac{3}{5}\), Cos B = \(\frac{4}{5}\) find Sin (A+B) ,Cos(A -B)
Answer:

Question 8.
P.T \(\cos \left(A+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) (CosA – SinA)
Answer:

Question 9.
If tan \(\alpha=\frac{n}{n+1}\) and tan β \(=\frac{1}{2 n+1}\)
S.T. α +β = \(\frac{\pi}{4}\)
Answer:

Question 10.
P.T. \(\cos \left(\frac{\pi}{6}+\mathbf{A}\right) \cdot \cos \left(\frac{\pi}{6}-\mathbf{A}\right)-\sin \left(\frac{\pi}{6}+\mathbf{A}\right) \cdot \sin \left(\frac{\pi}{6}-\mathbf{A}\right)=\frac{\sqrt{3}}{2}\)
Answer:

Question 11.
P. T. Cot 2 A + Tan A = Cosec 2 A
Answer:

Question 12.
P.T. Sin 2A = 2 Sin A. Cos A
Answer:
W.K.T. Sin (A + B) = Sin A. Cos B +Cos A Sin B
Put  B = A
Sin (A + A) = Sin A Cos A + Cos A Sin A
Sin 2A = 2 Sin A. Cos A.

Question 13.
P.T. Cos 2 A = Cos² A – Sin² A = 2 Cos² A – 1 = 1 – 2 Sin² A.
Answer:
(i) W.K.T Cos (A + B) = Cos A. Cos B – Sin A Sin B
Put  B = A
Cos (A + A) = cos A. Cos A – Sin A Sin A
∴ Cos 2 A = Cos² A – Sin² A  ………………………… (1)

(ii) W.K.T. Cos²A = Cos² A – Sin² A
Cos² A = Cos² A – (1 – Cos2 A)
∵  Sin² A = I – Cos² A
= Cos² A – 1 + Cos² A
= 2 Cos² A – 1
∴ Cos² A= 2 Cos² A – 1 ………………………… (2)

(iii) W.K.T. Cos² A = Cos² A – Sin² A
= 1 – Sin² A – Sin² A
∵ Cos² A = 1 – Sin² A
Cos² A = 1 – 2 Sin² A  ………………………… (3)

Question 14.
P.T. \(Tan 2A = \frac{2 \tan A}{1-\tan ^{2} A}\frac{2 \tan A}{1-\tan ^{2} A}\)
Answer:

Question 15.
If Cos A = \(\frac{4}{5}\)find Cos 3 A
Answer:

Question 16.
P.T \(\frac{\cos 2 A}{1+\sin 2 A}=\frac{\cos A-\sin A}{\cos A+\sin A}\)
Answer:

Question 17.
P.T \(\frac{\sin A+\sin 2 A}{1+\cos A+\cos 2 A}=\tan A\)
Answer:

 

Question 18.
S.T \( Tan A = \tan A=\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \cdot \tan B}\)
Answer:
R.H.S – tan (A – B + B) = tan A = R.H.S
Using tan (A + B) formula. Where A = (A – B).

Question 19.
P.T \(\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\tan \theta}{1-\tan \theta}\)
Answer:

Question 20.
P.T. (Cos⁴ θ – Sin⁴ θ) = 2 Cos² θ -1.
Answer:
L.H.S = Cos⁴ θ – Sin⁴ θ
= (Cos² θ)² – (Sin² θ)² = (Cos²θ – Sin²θ) (Cos² θ + Sin² θ)
= (Cos² θ – Sin² θ) x 1 = Cos² θ – (1 – Cos² θ)
= Cos² θ – 1 + Cos² θ = 2 Cos² θ – 1 = R.H.S.

Question 21.
P.T \(\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A\)
Answer:

Question 22.
P.T \(\frac{\cos 2 A-\cos 12 A}{\sin 12 A-\sin 2 A}=\tan 7 A\)
Answer:

Question 23.
P.T \(\frac{\sin x-\sin y}{\sin x+\sin y}=\tan \left(\frac{x-y}{2}\right) \cot \left(\frac{x+y}{2}\right)\)
Answer:

Question 24.
If A + B + C = π P.T. tan A + tan B + tan C = tan A. tan B. tan C.
Answer:
Given A + B + C = π
A + B = π – C
tan (A + B) = tan (π – C)
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=\frac{-\tan C}{1}\)
tan A + tan B = – tan C (1 – tan A tan B)
= – tan C + tan A. tan B. tan C.
tan A + tan B + tan C = tan A tan B tan C.

Question 25.
P.T \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}=\tan ^{2} A\)
Answer:

Question 26.
P.T. Sin 65 + Cos 65 = \(\sqrt{2} \)Cos 20°.
Answer:

Question 27.
If A + B + C = \(\frac{\pi}{2}\) P.T. tan A. tan B + tan B . tan C + tan C tan A = 1
Answer:

Question 28.
P.T \(\sin \left(\frac{\pi}{3}+\mathbf{A}\right)-\sin \left(\frac{\pi}{3}-\mathbf{A}\right)=\sin \mathbf{A}\)
Answer:

Question 29.
P.T. Cos A + Cos (120 – A) + Cos (120 + A) + Cos (120 + A) = 0
Answer:

Question 30.
P.T \(\frac{\sin 2 \alpha+\sin 3 \alpha}{\cos 2 \alpha-\cos 3 \alpha}=\cot \left(\frac{\alpha}{2}\right)\)
Answer:

2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae Five Marks Questions and Answers

Question 1.
P. T Sin 3θ = 3 Sin θ – 4 Sin3 θ
Answer:
L.H.S. = Sin3 θ = Sin (θ + 20)
= Sin θ. Cos 2θ + Cos θ Sin 2θ
= Sin θ (1 – 2 Sin² θ) + Cos θ. (2 Sin θ. Cos θ)
= Sin θ – 2 Sin³ θ + 2 Sin θ. Cos² θ
= Sin θ – 2 Sin³ θ + 2 Sin θ (1 – Sin² θ)
= Sin θ – 2 Sin³ θ + 2 Sinθ -2 Sin³ θ
= 3 Sin θ – 4 Sin^3θ
∴ Sin3θ = 3Sinθ – 4Sin³θ

Question 2.
P. T. Cos 3θ = 4 Cos³ θ – 3 Cos θ.
Answer:
L.H.S. Cos 3θ = Cos (θ + 2θ)
= Cos θ. Cos 2θ – Sin θ. Sin 2 θ
= Cos θ (2 Cos² θ – 1) – Sin θ. 2 Sin θ . Cos θ
= 2 Cos³ θ – Cos θ – 2 Cos θ Sin² θ
= 2 Cos³ θ – Cos θ – 2 Cos θ (1 – Cos² θ)
= 2 Cos³ θ – Cos θ – 2 Cos θ + 2 Cos³ θ
= 4 Cos³ θ – 3 Cos θ
∴ Cos3θ = 4Cos³ θ- 3Cosθ .

Question 3.
P.T \(\tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
Answer:

Question 4.
P.T \(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2\)
Answer:

Question 5.
P.T \(\frac{\cos 3 A}{2 \cos 2 A-1}=\cos A\) and Hence find cos 15°
Answer:

Question 6.
P.T  \(\frac{1+\cos 2 A+\sin 2 A}{1-\cos 2 A+\sin 2 A}=C o t A\)
Answer:

Question 7.
P.T. Cos⁶A + Sin⁶A = 1- \(\frac{3}{4}\) Sin² (2A)
Answer:

Question 8.
It tan α \(\frac{1}{3}\) , tan β = \(\frac{1}{7}\) P.T tan (2α + β) – 45°
Answer:

Question 9.
If \(\sin A=\frac{5}{13} \cos B=\frac{-4}{5}\), A is in II^nd Quadrant and B is in III^rd Quadrant. Find (i) Sin (A + B) (ii) Cos (A + B) (iii) tan (A + B), (iv) Sin (A – B)
Answer:

Question 10.
P.T. Cos (120°+A) + Cos (120°-A) = -Cos A.
Answer:

Question 11.
P.T\(\frac{\sin (A+B)}{\sin (A-B)}=\frac{\tan A+\tan B}{\tan A-\tan B}\)
Answer:

Question 12.
P.T. tan 2θ – tan θ = tan θ. Sec 2θ.
Answer:

Question 13.
P.T \(\frac{\cos 7 x+\cos 3 x-\cos 5 x+\cos x}{\sin 7 x-\sin 3 x-\sin 5 x+\sin x}\)
Answer:

Question 14.
If A + B + C = 180° P.T. Sin 2A + Sin 2 B + Sin 2 C = 4 Sin A. Sin B Sin C.
Answer:
GivenA + B + C= 180
L.H.S. = Sin 2 A + Sin 2 B + Sin 2 C
\(=2 \sin \frac{2 A+2 B}{2} \cdot \cos \frac{2 A-2 B}{2}+2 \sin C \cdot \cos C\)
∵ Sin 2 C = 2 Sin C. Cos C
= 2 Sin (A + B). Cos (A – B) + 2 Sin (C). Cos C
= 2 Sin C. Cos (A – B) + 2 Sin C. Cos C                                 Sin (A + B) = Sin C
= 2 sin C [Cos (A – B) + Cos C]                                             Cos(A + B) = -CosC
= 2 Sin C [Cos (A – B) – Cos (A + B)]

= -4 Sin C. Sin A. -Sin B
= 4 Sin A. Sin B. Sin C = R.H.S.

Question 15.
P.T. Cos 2 A+ Cos 2 B + Cos 2 C = – 1 – 4 Cos A. Cos B. Cos C.
Answer:
L.H.S. = Cos 2 A + Cos 2 B + Cos 2 C
\(=2 \cos \frac{2 A+2 B}{2} \cdot \cos \frac{2 A-2 B}{2}+2 \cos ^{2} C-1\)
= 2 Cos (A + B). Cos (A – B) + 2 Cos²C – 1    ∵ Cos (A + B) = – 2 Cos C
= – 1 – 2Cos CCos (A + B) + 2 Cos² C
= – 1 – 2 Cos C [cos (A + B) – Cos C]

Question 16.
P. T. Sin² A + Sin² B – Sin² C = 1 – 4 Sin A. Sin B. Cos C.
Answer:

Question 17.
P.T. Cos² A + Cos² B – Cos² C = 1 – 2 Sin A Sin B Cos C.
Answer:


= 1-Sin A [Sin (B + C) + sin (B – C)
= 1 – Sin A (2 Sin B. Cos C)
= 1 – 2 Sin A. Sin B. Sin C.

Question 18.
P.T. tan2 A + tan 2 B + tan 2 C = tan 2A. tan 2 B. tan 2 C.
Answer:
Given A + B + C = 180°
2A + 2B + 2C = 360°
2A + 2B = 360 -2C
Taking tan both sides
tan (2 A + 2B) = tan (360 – 2C)

tan2A + tan2B = – tan2C (1 – tan 2 A tan 2B)
= – tan 2 A + tan 2 B = – tan 2 C
= – tan 2C + tan 2 A. tan 2 B . tan 2 C

Question 19.
S.T \(\Sigma \frac{\sin (\mathbf{A}-\mathbf{B})}{\cos \mathbf{A} \cdot \mathbf{C} \mathbf{o s} \mathbf{B}}=\mathbf{0}\)
Answer:

Question 20.
P.T tan A. tan 3 A. tan 4 A = tan 4 A – tan 3 A – tan A
Answer:
Consider tan 4 A = tan (A + 3 A).
\(\frac{\tan 4 A}{1}=\frac{\tan A+\tan 3 A}{1-\tan A \cdot \tan 3 A}\)
tan A +tan 3 A = tan 4 A (1 – tan A. tan 3A)
tan A + tan 3 A = tan 4 A – tan A. tan 3 A . tan 4 A
∴ tan A. tan 3 A. tan 4 A = tan 4 A-tan 3 A – tan A.

Question 21.
T. Sin 105° + Cos 105° = 2 Cos 15°.
Answer:
L.H.S. = Sin 105° + Cos (180 – 75°)
= Sin 105° + Sin 75°
= 2 sin \(\left(\frac{105+75}{2}\right) \cos \left(\frac{105-75}{2}\right)\)
= 2 Sin 90°.Cos 15°
= 2.1.Cos 15°
= 2 Cos 15° = R.H.S

Question 22.
P.T \(\cos \left(\frac{\pi}{4}-A\right)-\sin \left(\frac{\pi}{4}+A\right)=0\)
Answer:

Question 23.
P.T \(\sin \left(\frac{\pi}{3}-A\right) \cdot \cos \left(\frac{\pi}{6}+A\right)+\cos \left(\frac{\pi}{3}-A\right) \cdot \sin \left(\frac{\pi}{6}+A\right)=1\)
Answer:
L.H.S. = Sin A. Cos B + Cos A. Sin B

Question 24.
P. T. Cos 20°. Cos 40°. Cos 80° = \(\frac{1}{8}\)
Answer:

Question 25.
P.T. Cos 10° . Cos 30° . Cos 50° Cos 70° \(=\frac{3}{16}\)
Answer:

Question 26.
P.T. Cos² A + Cos² (60 + A) + Cos² (60 – A) \(=\frac{3}{16}\)
Answer: