# Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Ex 6(f)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(f) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(f)

Question 1.
If A, B, C are angles in a triangle, then prove that
(i) sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C
Solution:
∵ A, B, C are angles in a triangle
⇒ A + B + C = 180° ……….(1)
LHS = sin 2A – sin 2B + sin 2C
= sin 2A + sin 2C – sin 2B
= 2 sin ($$\frac{2 A+2 C}{2}$$) . cos($$\frac{2 A-2 C}{2}$$) – sin 2B
= 2 sin (A + C) cos (A – C) – sin B
= 2 sin (180° – B) cos (A – C) – 2 sin B cos B
= 2 sin B cos (A – C) – 2 sin B cos B
= 2 sin B [cos (A – C) – cos B]
= 2 sin B [cos (A – C) – cos (180° – (A + C)]
= 2 sin B [cos (A – C) + cos (A + C)]
= 2 sin B (2 cos A cos C)
= 4 cos A sin B cos C
∴ sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C

(ii) cos 2A – cos 2B + cos 2C = 1 – 4 sin A cos B Sin C
Solution:
L.H.S. = -(cos 2B – cos 2A) + cos 2C
= -2 sin (A + B) sin (A – B) + cos 2C
= -2 sin (180° – C) sin (A – B) + cos 2C
= -2 sin C sin (A – B) + 1 – 2 sin2C
= 1 – 2 sin C (sin (A – B) + sin C)
= 1 – 2 sin C sin (A – B) + sin (180° – $$\overline{\mathrm{A}+\mathrm{B}}$$)
= 1 – 2 sin C (sin (A – B) + sin (A + B))
= 1 – 2 sin C (2 sin A cos B)
= 1 – 4 sin A cos B sin C
= R.H.S.

Question 2.
If A, B, C are angles in a triangle, then prove that
(i) sin A + sin B – sin C = 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ cos $$\frac{C}{2}$$
Solution:
L.H.S. = (sin A + sin B) – sin C
= 2 sin ($$\frac{A+B}{2}$$) cos ($$\frac{A-B}{2}$$) – sin C

(ii) cos A + cos B – cos C = -1 + 4 cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ sin $$\frac{C}{2}$$
Solution:
A, B, C are angles in a triangle
A + B + C = 180° ………(1)
LHS = cos A + cos B – cos C

Question 3.
If A, B, C are angles in a triangle, then prove that
(i) sin2A + sin2B – sin2C = 2 sin A sin B cos C
Solution:
Given A + B + C = 180°
L.H.S. = sin2A + [sin2B – sin2C]
= sin2A + sin (B + C) sin (B – C)
= sin2A + sin (180° – A) . sin (B – C)
= sin2A + sin A . sin (B – C)
= sin A (sin A + sin (B – C))
= sin A [sin (180° – $$\overline{\mathrm{B}+\mathrm{C}}$$) + sin (B – C)]
= sin A [sin (B + C) + sin (B – C)]
= sin A [2 sin B cos C]
= 2 sin A sin B cos C
= R.H.S

(ii) cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C
Solution:
A, B, C are angles in a triangle
⇒ A + B + C = 180° ……..(1)
L.H.S = cos2A + cos2B – cos2C
= cos2A + cos2B – cos2C

= 1 + cos (A + B) cos (A – B) – cos2C
= 1 + cos (180° – C) cos (A – B) – cos2C [By (1)]
= 1 – cos C cos (A – B) – cos2C
= 1 – cos C [cos (A – B) + cos C]
= 1 – cos C [cos (A – B) + cos(180° – $$\overline{A+B}$$] [By eq. (1)]
= 1 – cos C [cos (A – B) – cos (A + B)]
= 1 – cos C [2 sin A sin B]
= 1 – 2 sin A sin B cos C
∴ cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C

Question 4.
If A + B + C = π, then prove that
(i) $$\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}$$ = $$2\left[1+\sin \frac{A}{2}+\sin \frac{B}{2} \sin \frac{C}{2}\right]$$
Solution:

(ii) $$\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}-\cos ^{2} \frac{C}{2}=2 \cos \frac{A}{2}$$ $$\cos \frac{B}{2} \sin \frac{C}{2}$$
Solution:

Question 5.
In triangle ABC, prove that
(i) $$\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2}$$ = $$4 \cos \frac{\pi-A}{4} \cos \frac{\pi-B}{4} \cos \frac{\pi-C}{4}$$
Solution:

(ii) $$\cos \frac{A}{2}+\cos \frac{B}{2}-\cos \frac{C}{2}$$ = $$4 \cos \frac{\pi+A}{4} \cdot \cos \frac{\pi+B}{4} \cos \frac{\pi-C}{4}$$
Solution:

(iii) $$\sin \frac{A}{2}+\sin \frac{B}{2}-\sin \frac{C}{2}$$ = $$-1+4 \cos \frac{\pi-A}{4} \cos \frac{\pi-B}{4} \sin \frac{\pi-C}{4}$$
Solution:

Question 6.
If A + B + C = π/2, then prove that cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Solution:
A + B + C = π/2 ………(1)
LHS = cos 2A + cos 2B + cos 2C
= 2 cos ($$\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}$$) cos ($$\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}$$) + cos 2C
= 2 cos (A + B) . cos (A – B) + cos 2C
= 2 cos (90° – C) cos (A – B) + cos 2C
= 2 sin C cos (A – B) + (1 – 2 sin2C)
= 1 + 2 sin C [cos (A – B) – sin C]
= 1 + 2 sin C [cos (A – B)- sin (90° – $$\overline{A+B}$$)]
= 1 + 2 sin C [cos (A – B) – cos (A +B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C
= RHS
∴ cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C

Question 7.
If A + B + C = 3π/2, then prove that
(i) cos2A + cos2B – cos2C = -2 cos A cos B sin C
Solution:
A + B + C = 3π/2 ……..(1)
L.H.S. = cos2A + cos2B – cos2C
= cos2A + (1 – sin2B) – cos2C
= (cos2A – sin2B) + (1 – cos2C)
= cos (A + B) cos (A – B) + sin2C
= cos (270° – C) cos(A – B) + sin2C
= -sin C cos (A – B) + sin2C
= sin C [sin C – cos (A – B)]
= sin C [sin (270°- $$\overline{A+B}$$) – cos (A – B)]
= sin C [-cos (A + B) – cos (A – B)]
= -sin C [cos (A + B) + cos (A – B)]
= -sin C [2 cos A cos B]
= -2 cos A cos B sin C
= RHS
∴ cos2A + cos2B – cos2C = -2 cos A cos B sin C

(ii) sin 2A + sin 2B – sin 2C = -4 sin A sin B cos C
Solution:
Here A + B + C = 270° ………(1)
LHS = sin 2A + sin 2B – sin 2C
= 2 sin ($$\frac{2 A+2 B}{2}$$) cos ($$\frac{2 A-2 B}{2}$$) – sin 2C
= 2 sin (A + B) . cos (A – B) – 2 sin C cos 2C
= 2 sin (270° – C) cos (A – B) – 2 sin C cos C
= -2 cos C cos (A – B) – 2 sin C cos C
= -2 cos C [cos (A – B) + sin C]
= -2 cos C [cos (A – B) + sin (270° – $$\overline{A+B}$$)]
= -2 cos C [cos (A – B) – cos (A + B)]
= -2 cos C [2 sin A sin B]
= -4 sin A sin B cos C
= RHS
∴ sin 2A + sin 2B – sin 2C = -4 sin A sin B cos C

Question 8.
If A + B + C = 0°, then prove that
(i) sin 2A + sin 2B + sin 2C = -4 sin A sin B sin C
Solution:
Here A + B + C = 0 ………(1)
LHS = sin 2A + sin 2B + sin 2C
= 2 sin ($$\frac{2 A+2 B}{2}$$) cos ($$\frac{2 A-2 B}{2}$$) + sin 2C
= 2 sin (A + B) . cos (A – B) + 2 sin C cos 2C
= 2 sin (-C) cos (A – B) + 2 sin C cos C
= -2 sin C cos (A – B) + 2 sin C cos C
= -2 sin C [cos (A – B) – cos C]
= -2 sin C [cos (A – B) – cos (-A – B)]
= -2 sin C [cos (A – B) – cos (A + B)]
= -2 sin C [2 sin A sin B]
= -4 sin A sin B sin C
= RHS
∴ sin 2A + sin 2B + sin 2C = -4 sin A sin B sin C

(ii) sin A + sin B – sin C = – 4 cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ sin $$\frac{C}{2}$$
Solution:

Question 9.
If A + B + C + D = 2π then prove that
(i) sin A – sin B + sin C – sin D = $$-4 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A+D}{2}\right)$$
Solution:

(ii) cos 2A + cos 2B + cos 2C + cos 2D = 4 cos (A + B) cos (A + C) cos (A + D)
Solution:

Question 10.
If A + B + C = 2S, then prove that
(i) sin (S – A) + sin (S – B) + sin C = $$4 \cos \left(\frac{S-A}{2}\right) \cos \left(\frac{S-B}{2}\right) \sin \frac{C}{2}$$
Solution:

(ii) cos (S – A) + cos (S – B) + cos C = $$-1+4 \cos \left(\frac{S-A}{2}\right) \cos \left(\frac{S-B}{2}\right) \cdot \cos \frac{C}{2}$$
Solution: