11类RD Sharma解决方案–第8章转换公式–练习8.2 |套装1

问题1.将以下每一项表示为正弦和余弦的乘积：

(i)罪12θ+罪4θ

解决方案：

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 12θ + sin 4θ = 2 sin (12θ + 4θ)/2 cos (12θ – 4θ)/2

= 2 sin 16θ/2 cos 8θ/2

= 2 sin 8θ cos 4θ

(ii)sin 5θ– sinθ

解决方案：

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 5θ – sin θ = 2 cos (5θ + θ)/2 sin (5θ – θ)/2

= 2 cos 6θ/2 sin 4θ/2

= 2 cos 3θ sin 2θ

(iii)cos12θ+ cos8θ

解决方案：

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 12θ + cos 8θ = 2 cos (12θ + 8θ)/2 cos (12θ – 8θ)/2

= 2 cos 20θ/2 cos 4θ/2

= 2 cos 10θ cos 2θ

(iv)cos12θ– cos4θ

解决方案：

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

cos 12θ – cos 4θ = –2 sin (12θ + 4θ)/2 sin (12θ – 4θ)/2

= –2 sin 16θ/2 sin 8θ/2

= –2 sin 8θ sin 4θ

(v)正弦2θ+余弦4θ

解决方案：

sin 2θ + cos 4θ = sin 2θ + sin (90^o – 4θ)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 2θ + sin (90^o – 4θ) = 2 sin (2x + 90^o – 4θ)/2 cos (2θ – 90^o + 4θ)/2

= 2 sin (90^o – 2θ)/2 cos (6θ – 90^o)/2

= 2 sin (45^o – θ) cos (3θ – 45^o)

= 2 sin (45^o – θ) cos [–(45^o – 3θ)]

= 2 sin (45^o – θ) cos (45^o – 3θ)

= 2 sin (π/4 – θ) cos (π/4 – 3θ)

问题2.证明：

(i)正弦38°+正弦22°=正弦82°

解决方案：

Given, L.H.S. = sin 38° + sin 22°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 38° + sin 22° = 2 sin (38^o + 22^o)/2 cos (38^o – 22^o)/2

= 2 sin 60^o/2 cos 16^o/2

= 2 sin 30^o cos 8^o

= 2 × (1/2) × cos 8^o

= cos 8^o

= cos (90° – 82°)

= sin 82°

= R.H.S.

Hence proved.

(ii)cos 100°+ cos 20°= cos 40°

解决方案：

Given, L.H.S. = cos 100° + cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 100° + cos 20° = 2 cos (100^o + 20^o)/2 cos (100^o – 20^o)/2

= 2 cos 120^o/2 cos 80^o/2

= 2 cos 60^o cos 40^o

= 2 × (1/2) × cos 40^o

= cos 40^o

= R.H.S.

Hence Proved.

(iii)正弦50°+正弦10°=余弦20°

解决方案：

Given, L.H.S. = sin 50° + sin 10°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2.

sin 50° + sin 10° = 2 sin (50^o + 10^o)/2 cos (50^o – 10^o)/2

= 2 sin 60^o/2 cos 40^o/2

= 2 sin 30^o cos 20^o

= 2 × (1/2) × cos 20^o

= cos 20^o

= R.H.S

Hence Proved.

(iv)正弦23°+正弦37°=余弦7°

解决方案：

Given L.H.S. = sin 23° + sin 37°.

We know sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 23° + sin 37° = 2 sin (23^o + 37^o)/2 cos (23^o – 37^o)/2

= 2 sin 60^o/2 cos (–14^o/2)

= 2 sin 30^o cos (–7^o)

= 2 × (1/2) × cos 7^o

= cos 7^o

= R.H.S.

Hence Proved.

(v)正弦105°+余弦105°=余弦45°

解决方案：

Given, L.H.S. = sin 105° + cos 105°

sin 105° + cos 105° = sin 105^o + sin (90^o – 105^o)

= sin 105^o + sin (–15^o)

= sin 105^o – sin 15^o

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 105^o – sin 15^o = 2 cos (105^o + 15^o)/2 sin (105^o – 15^o)/2

= 2 cos 120^o/2 sin 90^o/2

= 2 cos 60^o sin 45^o

= 2 × (1/2) × (1/√2)

= 1/√2

= cos 45^o

= R.H.S.

Hence Proved.

(vi)正弦40°+正弦20°=余弦10°

解决方案：

Given, L.H.S. = sin 40° + sin 20°.

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 40° + sin 20° = 2 sin (40^o + 20^o)/2 cos (40^o – 20^o)/2

= 2 sin 60^o/2 cos 20^o/2

= 2 sin 30^o cos 10^o

= 2 × (1/2) × cos 10^o

= cos 10^o

= R.H.S.

Hence Proved.

问题3.证明：

(i)cos 55°+ cos 65°+ cos 175°= 0

解决方案：

Given, L.H.S. = cos 55° + cos 65° + cos 175°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 55° + cos 65° + cos 175° = 2 cos (55^o + 65^o)/2 cos (55^o – 65^o) + cos (180^o – 5^o)

= 2 cos 120^o/2 cos (–10o)/2 – cos 5^o

= 2 cos 60° cos (–5°) – cos 5°

= 2 × (1/2) × cos 5^o – cos 5^o

= cos 5^o – cos 5^o

= 0

= R.H.S.

Hence Proved.

(ii)正弦50°–正弦70°+正弦10°= 0

解决方案：

Given, L.H.S. = sin 50° – sin 70° + sin 10°.

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

sin 50° – sin 70° + sin 10° = 2 cos (50^o + 70^o)/2 sin (50^o – 70^o) + sin 10^o

= 2 cos 120^o/2 sin (–20^o)/2 + sin 10^o

= 2 cos 60^o (–sin 10^o) + sin 10^o

= 2 × (1/2) × (–sin 10^o) + sin 10^o

= 0

= R.H.S.

Hence Proved.

(iii)cos 80°+ cos 40°– cos 20°= 0

解决方案：

Given L.H.S. = cos 80° + cos 40° – cos 20°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 80° + cos 40° – cos 20° = 2 cos (80^o + 40^o)/2 cos (80^o – 40^o) – cos 20^o

= 2 cos 120^o/2 cos 40^o/2 – cos 20^o

= 2 cos 60° cos 20^o – cos 20°

= 2 × (1/2) × cos 20^o – cos 20^o

= 0

= R.H.S.

Hence Proved.

(iv)cos 20°+ cos 100°+ cos 140°= 0

解决方案：

Given, L.H.S. = cos 20° + cos 100° + cos 140°.

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2.

cos 20° + cos 100° + cos 140° = 2 cos (20^o + 100^o)/2 cos (20^o – 100^o) + cos (80^o – 40^o)

= 2 cos 120^o/2 cos (–80^o)/2 – cos 40^o

= 2 cos 60° cos (–40°) – cos 40°

= 2 × (1/2) × cos 40^o – cos 40^o

= 0

= R.H.S.

Hence Proved.

(v)sin5π/ 18 – cos4π/ 9 =√3sinπ/ 9

解决方案：

Given, L.H.S. = sin 5π/18 – cos 4π/9

= sin 5π/18 – sin (π/2 – 4π/9)

= sin 5π/18 – sin (9π – 8π)/18

= sin 5π/18 – sin π/18

We know, sin A – sin B = 2 cos (A+B)/2 sin (A– B)/2

= 2 cos (6π/36) sin (4π/36)

= 2 cos π/6 sin π/9

= 2 cos 30^o sin π/9

= 2 × (√3/2) × sin π/9

= √3 sin π/9

= R.H.S.

Hence Proved.

(vi)cosπ/ 12 – sinπ/ 12 = 1 /√2

解决方案：

Given, cos π/12 – sin π/12 = sin (π/2 – π/12) – sin π/12

= sin (6π – 5π)/12 – sin π/12

= sin 5π/12 – sin π/12

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= 2 cos (6π/24) sin (4π/24)

= 2 cos π/4 sin π/6

= 2 cos 45^o sin 30^o

= 2 × (1/√2) × (1/2)

= 1/√2

= R.H.S.

Hence Proved.

(vii)正弦80°– cos 70°= cos 50°

解决方案：

We have, sin 80° = cos 50° + cos 70^o

Here, R.H.S. = cos 50° + cos 70^o

We know,

cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos 50° + cos 70^o = 2 cos (50^o + 70^o)/2 cos (50^o – 70^o)/2

= 2 cos 120^o/2 cos (–20^o)/2

= 2 cos 60^o cos (–10^o)

= 2 × (1/2) × cos 10^o

= cos 10^o

= cos (90° – 80°)

= sin 80°

= L.H.S.

Hence Proved.

(viii)正弦51°+余弦81°=余弦21°

解决方案：

Given, L.H.S. = sin 51° + cos 81°

= sin 51^o + sin (90^o – 81^o)

= sin 51^o + sin 9^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 51^o + sin 9^o = 2 sin (51^o + 9^o)/2 cos (51^o – 9^o)/2

= 2 sin 60^o/2 cos 42^o/2

= 2 sin 30^o cos 21^o

= 2 × (1/2) × cos 21^o

= cos 21^o

= R.H.S.

Hence Proved.

问题4：证明：

(i)cos(3π/ 4 + x)– cos(3π/ 4 – x)= –√2 sin x

解决方案：

Given, L.H.S. = cos (3π/4 + x) – cos (3π/4 – x)

We know, cos A – cos B = –2 sin (A+B)/2 sin (A–B)/2

cos (3π/4 + x) – cos (3π/4 – x) = –2 sin (3π/4 + x + 3π/4 – x)/2 sin (3π/4 + x – 3π/4 + x)/2

= –2 sin (6π/4)/2 sin 2x/2

= –2 sin 6π/8 sin x

= –2 sin 3π/4 sin x

= –2 sin (π – π/4) sin x

= –2 sin π/4 sin x

= –2 × (1/√2) × sin x

= –√2 sin x

= R.H.S.

Hence proved.

(ii)cos(π/ 4 + x)+ cos(π/ 4 – x)=√2cos x

解决方案：

Given, L.H.S. = cos (π/4 + x) + cos (π/4 – x)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

cos (π/4 + x) + cos (π/4 – x) = 2 cos (π/4 + x + π/4 – x)/2 cos (π/4 + x – π/4 + x)/2

= 2 cos (2π/4)/2 cos 2x/2

= 2 cos 2π/8 cos x

= 2 sin π/4 cos x

= 2 × (1/√2) × cos x

= √2 cos x

= R.H.S.

Hence proved.

问题5.证明：

(i)罪65 ^o + cos 65 ^o =√2cos 20 ^o

解决方案：

Given L.H.S. = sin 65^o + cos 65^o

= sin 65^o + sin (90^o – 65^o)

= sin 65^o + sin 25^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 65^o + sin 25^o = 2 sin (65^o + 25^o)/2 cos (65^o – 25^o)/2

= 2 sin 90^o/2 cos 40^o/2

= 2 sin 45^o cos 20^o

= 2 × (1/√2) × cos 20^o

= √2 cos 20^o

= R.H.S.

Hence proved.

(ii)罪47 ^o + cos 77 ^o = cos 17 ^o

解决方案：

Given, L.H.S. = sin 47^o + cos 77^o

= sin 47^o + sin (90^o – 77^o)

= sin 47^o + sin 13^o

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

sin 47^o + sin 13^o = 2 sin (47^o + 13^o)/2 cos (47^o – 13^o)/2

= 2 sin 60^o/2 cos 34^o/2

= 2 sin 30^o cos 17^o

= 2 × (1/2) × cos 17^o

= cos 17^o

= R.H.S.

Hence proved.

问题6：证明：

(i)cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

解决方案：

Given, L.H.S. = cos 3A + cos 5A + cos 7A + cos 15A

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 5A + cos 3A) + (cos 15A + cos 7A)

= [2 cos (5A+3A)/2 cos (5A–3A)/2] + [2 cos (15A+7A)/2 cos (15A–7A)/2]

= [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2]

= [2 cos 4A cos A] + [2 cos 11A cos 4A]

= 2 cos 4A (cos 11A + cos A)

= 2 cos 4A [2 cos (11A+A)/2 cos (11A-A)/2]

= 2 cos 4A [2 cos 12A/2 cos 10A/2]

= 2 cos 4A [2 cos 6A cos 5A]

= 4 cos 4A cos 5A cos 6A

= R.H.S.

Hence proved.

(ii)cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A

解决方案：

Given L.H.S. = cos A + cos 3A + cos 5A + cos 7A

= (cos 3A + cos A) + (cos 7A + cos 5A)

We know, cos A + cos B = 2 cos (A+B)/2 cos (A–B)/2

= (cos 3A + cos A) + (cos 7A + cos 5A)

= [2 cos (3A+A)/2 cos (3A–A)/2] + [2 cos (7A+5A)/2 cos (7A–5A)/2]

= [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2]

= [2 cos 2A cos A] + [2 cos 6A cos A]

= 2 cos A (cos 6A + cos 2A)

= 2 cos A [2 cos (6A+2A)/2 cos (6A–2A)/2]

= 2 cos A [2 cos 8A/2 cos 4A/2]

= 2 cos A [2 cos 4A cos 2A]

= 4 cos A cos 2A cos 4A

= R.H.S.

Hence proved.

(iii)sin A + sin 2A + sin 4A + sin 5A = 4 cos A / 2 cos 3A / 2 sin 3A

解决方案：

Given, L.H.S. = sin A + sin 2A + sin 4A + sin 5A

= (sin 2A + sin A) + (sin 5A + sin 4A)

We know, sin A + sin B = 2 sin (A+B)/2 cos (A–B)/2

= (sin 2A + sin A) + (sin 5A + sin 4A)

= [2 sin (2A+A)/2 cos (2A–A)/2] + [2 sin (5A+4A)/2 cos (5A–4A)/2]

= [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2]

= 2 cos A/2 (sin 9A/2 + sin 3A/2)

= 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2]

= 2 cos A/2 [2 sin ((9A+3A)/2)/2 cos ((9A–3A)/2)/2]

= 2 cos A/2 [2 sin 12A/4 cos 6A/4]

= 2 cos A/2 [2 sin 3A cos 3A/2]

= 4 cos A/2 cos 3A/2 sin 3A

= R.H.S.

Hence proved.

(iv)sin 3A + sin 2A – sin A = 4 sin A cos A / 2 cos 3A / 2

解决方案：

Given, L.H.S. = sin 3A + sin 2A – sin A

= (sin 3A – sin A) + sin 2A

We know, sin A – sin B = 2 cos (A+B)/2 sin (A–B)/2

= (sin 3A – sin A) + sin 2A

= 2 cos (3A + A)/2 sin (3A – A)/2 + sin 2A

= 2 cos 4A/2 sin 2A/2 + sin 2A

= 2 cos 2A sin A + 2 sin A cos A

= 2 sin A (cos 2A + cos A)

= 2 sin A [2 cos (2A+A)/2 cos (2A-A)/2]

= 2 sin A [2 cos 3A/2 cos A/2]

= 4 sin A cos A/2 cos 3A/2

= R.H.S.

Hence proved.

(v)cos 20 ^o cos 100 ^o + cos 100 ^o cos 140 ^o – cos 140 ^o cos 200 ^o = – 3/4

解决方案：

Given L.H.S. = cos 20^o cos 100^o + cos 100^o cos 140^o – cos 140^o cos 200^o

= 1/2 [2 cos 100^o cos 20^o + 2 cos 140^o cos 100^o – 2 cos 200^o cos 140^o]

We know that, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (100^o + 20^o) + cos (100^o – 20^o) + cos (140^o + 100^o) + cos (140^o – 100^o) – cos (200^o + 140^o) – cos (200^o – 140^o)]]

= 1/2 [cos 120^o + cos 80^o + cos 240^o + cos 40^o – cos 340^o – cos 60^o]

= 1/2 [cos (90^o + 30^o) + cos 80^o + cos (180^o + 60^o) + cos 40^o – cos (360^o – 20^o) – cos 60^o]

= 1/2 [–sin 30^o + cos 80^o – cos 60^o + cos 40^o – cos 20^o – cos 60^o]

= 1/2 [–sin 30^o + cos 80^o + cos 40^o – cos 20^o – 2 cos 60^o]

= 1/2 [–sin 30^o + 2 cos (80^o+40^o)/2 cos (80^o–40^o)/2 – cos 20^o – 2 × 1/2]

= 1/2 [–sin 30^o + 2 cos 120^o/2 cos 40^o/2 – cos 20^o – 1]

= 1/2 [–sin 30^o + 2 cos 60^o cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 + 2×(1/2)×cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 + cos 20^o – cos 20^o – 1]

= 1/2 [–1/2 –1]

= 1/2 [–3/2]

= –3/4

= R.H.S.

Hence proved.

(vi)sin x / 2 sin 7x / 2 + sin 3x / 2 sin 11x / 2 = sin 2x sin 5x

解决方案：

Given L.H.S. = sin x/2 sin 7x/2 + sin 3x/2 sin 11x/2

= 1/2 [2 sin 7x/2 sin x/2 + 2 sin 11x/2 sin 3x/2]

We know, 2 sin A sin B = cos (A–B) – cos (A+B)

= 1/2 [cos (7x/2 – x/2) – cos (7x/2 + x/2) + cos (11x/2 – 3x/2) – cos (11x/2 + 3x/2)]

= 1/2 [cos (7x–x)/2 – cos (7x+x)/2 + cos (11x–3x)/2 – cos (11x+3x)/2]

= 1/2 [cos 6x/2 – cos 8x/2 + cos 8x/2 – cos 14x/2]

= 1/2 [cos 3x – cos 7x]

= –1/2 [cos 7x – cos 3x]

= –1/2 [–2 sin (7x+3x)/2 sin (7x–3x)/2]

= –1/2 [–2 sin 10x/2 sin 4x/2]

= –1/2 [–2 sin 5x sin 2x]

= –2/–2 sin 5x sin 2x

= sin 2x sin 5x

= R.H.S.

Hence proved.

(vii)cos x cos x / 2 – cos 3x cos 9x / 2 = sin 4x sin 7x / 2

解决方案：

Given L.H.S. = cos x cos x/2 – cos 3x cos 9x/2

= 1/2 [2 cos x cos x/2 – 2 cos 9x/2 cos 3x]

We know, 2 cos A cos B = cos (A+B) + cos (A–B)

= 1/2 [cos (x + x/2) + cos (x – x/2) – cos (9x/2 + 3x) – cos (9x/2 – 3x)]

= 1/2 [cos (2x+x)/2 + cos (2x–x)/2 – cos (9x+6x)/2 – cos (9x–6x)/2]

= 1/2 [cos 3x/2 + cos x/2 – cos 15x/2 – cos 3x/2]

= 1/2 [cos x/2 – cos 15x/2]

= – 1/2 [cos 15x/2 – cos x/2]

= – 1/2 [–2 sin (15x/2 + x/2)/2 sin (15x/2 – x/2)/2]

= -1/2 [–2 sin (16x/2)/2 sin (14x/2)/2]

= -1/2 [–2 sin 16x/4 sin 7x/2]

= – 1/2 [–2 sin 4x sin 7x/2]

= –2/–2 [sin 4x sin 7x/2]

= sin 4x sin 7x/2

= R.H.S.

Hence proved.

问题7.证明：

(一世) $\frac{sinA-sinB}{cosA+cosB}=tan\frac{A-B}{2}$

解决方案：

We have,

L.H.S. =$\frac{sinA-sinB}{cosA+cosB}$

=$\frac{2cos\frac{A+B}{2}sin\frac{A-B}{2}}{2cos\frac{A+B}{2}cos\frac{A-B}{2}}$

=$\frac{sin\frac{A-B}{2}}{cos\frac{A-B}{2}}$

=$tan\frac{A-B}{2}$

= R.H.S.

Hence proved.

(ii) $\frac{sinA+sinB}{sinA-sinB}=tan(\frac{A+B}{2})cot(\frac{A-B}{2})$

解决方案：

We have,

L.H.S. =$\frac{sinA+sinB}{sinA-sinB}$

=$\frac{2sin\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}sin\frac{A-B}{2}}$

=$\frac{sin\frac{A+B}{2}cos\frac{A-B}{2}}{cos\frac{A+B}{2}sin\frac{A-B}{2}}$

=$tan(\frac{A+B}{2})cot(\frac{A-B}{2})$

= R.H.S.

Hence proved.

(iii) $\frac{cosA+cosB}{cosB-cosA}=cot(\frac{A+B}{2})cot(\frac{A-B}{2})$

解决方案：

We have,

L.H.S. =$\frac{cosA+cosB}{cosB-cosA}$

=$\frac{2cos\frac{A+B}{2}cos\frac{A-B}{2}}{-2sin\frac{A+B}{2}sin\frac{B-A}{2}}$

=$\frac{cos\frac{A+B}{2}cos\frac{A-B}{2}}{sin\frac{A+B}{2}sin\frac{A-B}{2}}$

=$cot(\frac{A+B}{2})cot(\frac{A-B}{2})$

= R.H.S.

Hence proved.

问题8.证明：

(一世) $\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}=\frac{cos5A}{cos3A}$

解决方案：

We have,

L.H.S. =$\frac{cos3A+2cos5A+cos7A}{cosA+2cos3A+cos5A}$

=$\frac{2cos\frac{10A}{2}cos\frac{4A}{2}+2cos5A}{cos\frac{6A}{2}cos\frac{4A}{2}+2cos3A}$

=$\frac{2cos5Acos2A+2cos5A}{2cos3Acos2A+2cos3A}$

=$\frac{2cos5A(cos2A+1)}{2cos3A(cos2A+1)}$

=$\frac{cos5A}{cos3A}$

= R.H.S.

Hence proved.

(ii) $\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}=tan8A$

解决方案：

We have,

L.H.S. =$\frac{sin11AsinA+sin7Asin3A}{cos11AsinA+cos7Asin3A}$

=$\frac{cos10A-cos12A+cos4A-cos10A}{sin12A-sin10A+sin10A-sin4A}$

=$\frac{cos4A-cos12A}{sin12A-sin4A}$

=$\frac{2sin\frac{16A}{2}sin\frac{8A}{2}}{2cos\frac{16A}{2}sin\frac{8A}{2}}$

=$\frac{2sin8Asin4A}{2cos8Asin4A}$

= tan 8A

= R.H.S.

Hence proved.

(iii) $\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}=tanθ$

解决方案：

We have,

L.H.S. =$\frac{sin(θ+Ø)-2sinθ+sin(θ-Ø)}{cos(θ+Ø)-2cosθ+cos(θ-Ø)}$

=$\frac{sin(θ+Ø)+sin(θ-Ø)-2sinθ}{cos(θ+Ø)+cos(θ-Ø)-2cosθ}$

=$\frac{2sin\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2sinθ}{2cos\frac{(θ+Ø+θ-Ø)}{2}cos\frac{(θ+Ø-θ+Ø)}{2}-2cosθ}$

=$\frac{2sin\frac{2θ}{2}cos\frac{2Ø}{2}-2sinθ}{2cos\frac{2θ}{2}cos\frac{2Ø}{2}-2cosθ}$

=$\frac{2sinθcosØ-2sinθ}{2cosθcosØ-2cosθ}$

=$\frac{2sinθ(cosØ-1)}{2cosθ(cosØ-1)}$

=$\frac{sinθ}{cosθ}$

= tan θ

= R.H.S.

Hence proved.

问题9.证明：

(i)sinα+ sinβ+ sinγ– sin(α+β+γ)= 4 sin(α+β)/ 2 sin(β+γ)/ 2 sin(α+γ)/ 2

解决方案：

We have,

L.H.S. = sin α + sin β + sin γ – sin (α + β + γ)

=$2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{γ+α+β+γ}{2})sin(\frac{γ-α-β-γ}{2})$

=$2sin\frac{α+β}{2}cos\frac{α-β}{2}+2cos(\frac{α+β+2γ}{2})sin(\frac{-(α+β)}{2})$

=$2sin\frac{α+β}{2}cos\frac{α-β}{2}-2cos(\frac{α+β+2γ}{2})sin(\frac{α+β}{2})$

=$2sin\frac{α+β}{2}(cos\frac{α-β}{2}-cos\frac{α+β+2γ}{2})$

=$2sin\frac{α+β}{2}(-2sin\frac{\frac{α-β+α+β+2γ}{2}}{2}sin\frac{\frac{α-β-α-β-2γ}{2}}{2})$

=$2sin\frac{α+β}{2}(2sin\frac{α+γ}{2}sin\frac{β+γ}{2})$

= 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

= R.H.S.

Hence proved.

(ii)cos(A + B + C)+ cos(A – B + C)+ cos(A + B – C)+ cos(–A + B + C)= 4 cos A cos B cos C

解决方案：

We have,

L.H.S. = cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (–A + B + C)

=$2cos\frac{A+B+C+A-B+C}{2}cos\frac{A+B+C-A+B-C}{2}+2cos\frac{A+B-C-A+B+C}{2}cos\frac{A+B-C+A-B-C}{2}$

=$2cos\frac{2A+2C}{2}cos\frac{2B}{2}+2cos\frac{2B}{2}cos\frac{2A-2C}{2}$

= 2 cos (A + C) cos B + 2 cos B cos (A − C)

= 2 cos B [cos (A + C) + cos (A − C)]

= 2 cos B$\left[2cos\frac{A+C+A-C}{2}cos\frac{A+C-A+C}{2}\right]$

= 2 cos B [2 cos A cos C]

= 4 cos A cos B cos C

= R.H.S.

Hence proved.

问题10.如果cos A + cos B = 1/2且sin A + sin B = 1/4，则证明$tan\frac{A+B}{2}=\frac{1}{2}$ 。

解决方案：

We have,

cos A + cos B = 1/2

sin A + sin B = 1/4

=>$\frac{sinA+sinB}{cosA+cosB}=\frac{\frac{1}{4}}{\frac{1}{2}}$

=>$\frac{sinA+sinB}{cosA+cosB}=\frac{1}{2}$

=>$\frac{2sin\frac{A+B}{2}cos\frac{A-B}{2}}{2cos\frac{A+B}{2}cos\frac{A-B}{2}}=\frac{1}{2}$

=>$\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}=\frac{1}{2}$

=>$tan\frac{A+B}{2}=\frac{1}{2}$

Hence proved.