Класс 11. Решения RD Sharma. Глава 9. Тригонометрические соотношения кратных и дольных углов. Упражнение 9.2.
Докажи это:
Вопрос 1. sin 5 θ = 5sin θ – 20 sin 3 θ + 16 sin 5 θ
Решение:
We have,
L.H.S. = sin 5θ
= sin (3θ + 2θ)
= sin 3θ cos 2θ + cos 3θ sin 2θ
= (3sin θ – 4sin3 θ) (1 – 2sin2 θ) + (4cos3 θ – 3cos θ) (2sin θ cos θ)
= 3sin θ – 6sin3 θ – 4sin3 θ + 8sin5 θ + 8sin θ cos4 θ – 6sin θ cos2 θ
= 3sin θ – 10sin3 θ + 8sin5 θ + 8sin θ (1 – sin2 θ)2 – 6sin θ (1–sin2 θ)
= 3sin θ – 10sin3 θ + 8sin5 θ + 8sin θ (1 + sin4 θ – 2sin2θ) – 6sin θ + 6sin3 θ
= 3sin θ – 4sin3 θ + 8sin5 θ + 8sin θ + 8sin5 θ – 16sin3 θ – 6sin θ
= 5sin θ – 20 sin3 θ + 16 sin5 θ
= R.H.S.
Hence, proved.
Вопрос 2. 4 (cos 3 10 o + sin 3 20 o ) = 3 (cos 10 o + sin 20 o )
Решение:
Now we know,
sin θ = cos (90–θ)
For θ = 60o, we have,
=> sin 60o = cos 30o
=> sin (3×20o) = cos (3×10o)
=> 3sin 20o – 4sin3 20o = 4cos3 10o – 3cos 10o
=> 4cos3 10o + 4sin3 20o = 3cos 10o + 3sin 20o
=> 4 (cos3 10o + sin3 20o) = 3 (cos 10o + sin 20o)
Hence, proved.
Вопрос 3. cos 3 θ sin 3θ + sin 3 θ cos 3θ = (3sin 4θ)/4
Решение:
We have,
L.H.S. = cos3 θ sin 3θ + sin3 θ cos 3θ
= sin 3θ (cos 3θ + 3cos θ)/4 + cos 3θ (3sin θ – sin 3θ)/4
= (sin 3θ cos 3θ + 3sin 3θ cos θ + 3cos 3θ sin θ – cos 3θ sin 3θ)/4
= [3(sin 3θ cos θ + cos 3θ sin θ) ]/4
= [3sin (3θ+θ)]/4
= (3sin 4θ)/4
= R.H.S.
Hence, proved.
Вопрос 4. sin 5A = 5 cos 4 A sin A – 10 cos 2 A sin 3 A + sin 5 A
Решение:
We have,
L.H.S. = sin 5A
= sin (3A + 2A)
= sin 3A cos 2A + cos 3A sin 2A
= (3sin A – 4sin3 A) (2cos2 A – 1) + (4cos3 A – 3cos A) (2sin A cos A)
= 6sin A cos2 A – 3sin A – 8sin3 A cos2 A + 4sin3 A + 8sin A cos4 A – 6sin A cos2 A
= – 3sin A – 8sin3 A cos2 A + 4sin3 A + 8sin A cos4 A
= – 3sin A – 10 sin3 A cos2 A + 2sin3 A cos2 A + 4sin3 A + 5sin A cos4 A + 3sin A cos4 A
= 5sin A cos4 A – 10 sin3 A cos2 A – 3sin A (1 – cos4 A) + 2sin3 A (2 + cos2 A)
= 5sin A cos4 A – 10 sin3 A cos2 A – 3sin A (1 – cos2 A) (1 + cos2 A) + 2sin3 A (2 + cos2 A)
= 5sin A cos4 A – 10 sin3 A cos2 A – 3sin3 A (1 + cos2 A) + 2sin3 A (2 + cos2 A)
= 5sin A cos4 A – 10 sin3 A cos2 A – sin3 A (3 + 3cos2 A – 4 – 2cos2 A)
= 5sin A cos4 A – 10 sin3 A cos2 A – sin3 A (cos2 A – 1)
= 5sin A cos4 A – 10 sin3 A cos2 A + sin5 A
= R.H.S.
Hence, proved.
Вопрос 5. tan A tan (A + 60 o ) + tan A tan (A – 60 o ) + tan (A + 60 o ) tan (A – 60 o ) = –3
Решение:
We have,
L.H.S. = tan A tan (A + 60o) + tan A tan (A – 60o) + tan (A + 60o) tan (A – 60o)
=
=
=
=
=
= –3
= R.H.S.
Hence, proved.
Вопрос 6. загар А + загар (60 ° + А) – загар (60 ° – А) = 3 загар 3А
Решение:
We have,
L.H.S. = tan A + tan (60o + A) – tan (60o – A)
=
=
=
=
=
= 3 tan 3A
= R.H.S.
Hence, proved.
Вопрос 7. раскладушка А + раскладушка (60 ° + А) – раскладушка (60 ° – А) = 3 раскладушка 3А
Решение:
We have,
L.H.S. = cot A + cot (60o + A) – cot (60o – A)
=
=
=
=
=
= 3 cot 3A
= R.H.S.
Hence, proved.
Вопрос 8. раскладушка А + раскладушка (60 ° + А) + раскладушка (120 ° + А) = 3 раскладушка 3А
Решение:
We have,
L.H.S. = cot A + cot (60o + A) + cot (120o + A)
= cot A + cot (60o + A) – cot (180 – (120o + A))
= cot A + cot (60o + A) – cot (60o – A)
=
=
=
=
=
= 3 cot 3A
= R.H.S.
Hence, proved.
Вопрос 9. 
Решение:
We have,
L.H.S. =
=
=
=
=
=
=
=
=
= R.H.S.
Hence, proved.
Вопрос 10. |sin θ sin (60–θ) sin (60+θ)| ≤ 1/4 для всех значений θ.
Решение:
We have,
= |sin θ sin (60–θ) sin (60+θ)|
= |sin θ (sin2 60 – sin2 θ)|
= |sin θ (3/4 – sin2 θ)|
= |sin θ/4 (3 – 4sin2 θ)|
= |1/4 (3sin θ – 4sin3 θ)|
= |1/4 (sin 3θ)| ≤ 1/4
Hence, proved.
Вопрос 11. |cos θ cos (60–θ) cos (60+θ)| ≤ 1/4 для всех значений θ .
Решение:
We have,
= |cos θ cos (60–θ) cos (60+θ)|
= |cos θ (cos2 60 – sin2 θ)|
= |cos θ (1/4 – sin2 θ)|
= |cos θ/4 (1 – 4sin2 θ)|
= |cos θ/4 (1 – 4 (1 – cos2 θ))|
= |cos θ/4 (–3 + 4cos2 θ)|
= |1/4 (4cos3 θ – 3cos θ)|
= |1/4 (cos 3θ)| ≤ 1/4
Hence, proved.