Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.2

Prove that:

Question 1. sin 5θ = 5sin θ – 20 sin³θ + 16 sin⁵θ

Solution:

We have,
L.H.S. = sin 5θ
= sin (3θ + 2θ)
= sin 3θ cos 2θ + cos 3θ sin 2θ
= (3sin θ – 4sin³θ) (1 – 2sin²θ) + (4cos³θ – 3cos θ) (2sin θ cos θ)
= 3sin θ – 6sin³θ – 4sin³θ + 8sin⁵θ + 8sin θ cos⁴θ – 6sin θ cos²θ
= 3sin θ – 10sin³ θ + 8sin⁵ θ + 8sin θ (1 – sin²θ)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.

Question 2. 4 (cos³ 10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)

Solution:

Now we know,
sin θ = cos (90–θ)
For θ = 60^o, we have,
=> sin 60^o = cos 30^o
=> sin (3×20^o) = cos (3×10^o)
=> 3sin 20^o – 4sin³ 20^o = 4cos³ 10^o – 3cos 10^o
=> 4cos³ 10^o + 4sin³ 20^o = 3cos 10^o + 3sin 20^o
=> 4 (cos³10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)
Hence, proved.

Question 3. cos³ θ sin 3θ + sin³ θ cos 3θ = (3sin 4θ)/4

Solution:

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.

Question 4. sin 5A = 5 cos⁴ A sin A – 10 cos² A sin³ A + sin⁵ A

Solution:

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 cos²A + 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.

Question 5. tan A tan (A + 60^o) + tan A tan (A – 60^o) + tan (A + 60^o) tan (A – 60^o) = –3

Solution:

We have,
L.H.S. = tan A tan (A + 60^o) + tan A tan (A – 60^o) + tan (A + 60^o) tan (A – 60^o)
= $tanA\left[\frac{tanA+tan60^0}{1-tanAtan60^0}\right]+tanA\left[\frac{tanA-tan60^0}{1+tanAtan60^0}\right]+\left[\frac{tanA+tan60^0}{1-tanAtan60^0}\right]\left[\frac{tanA-tan60^0}{1+tanAtan60^0}\right]$
= $tanA\left[\frac{(tanA+tan60^0)(1+tanAtan60^0)}{1-tan^2Atan^260^0}\right]+tanA\left[\frac{(tanA-tan60^0)(1-tanAtan60^0)}{1-tan^2Atan^260^0}\right]+\left[\frac{tan^2A-tan^260^0}{1-tan^2tan^260^0}\right]$
= $\left[\frac{tanA(4tanA-\sqrt{3}-\sqrt{3}tan^2A+4tanA+\sqrt{3}+\sqrt{3}tan^2A)+tan^2A-3}{1-3tan^2A}\right]$
= $\frac{9tan^2A-3}{1-3tan^2A}$
= $\left[\frac{-3(1-3tan^2A)}{1-3tan^2A}\right]$
= –3
= R.H.S.
Hence, proved.

Question 6. tan A + tan (60^o+ A) – tan (60^o – A) = 3 tan 3A

Solution:

We have,
L.H.S. = tan A + tan (60^o + A) – tan (60^o – A)
= $tanA+\frac{tan60^o+tanA}{1-tan60^otanA}-\frac{tan60^o-tanA}{1+tan60^otanA}$
= $tanA+\frac{\sqrt{3}+tanA}{1-\sqrt{3}tanA}-\frac{\sqrt{3}-tanA}{1+\sqrt{3}tanA}$
= $tanA+\frac{8tanA}{1-3tan^2A}$
= $\frac{9tanA-3tan^3A}{1-3tan^2A}$
= $\frac{3(3tanA-tan^3A)}{1-3tan^2A}$
= 3 tan 3A
= R.H.S.
Hence, proved.

Question 7. cot A + cot (60^o + A) – cot (60^o – A) = 3 cot 3A

Solution:

We have,
L.H.S. = cot A + cot (60^o + A) – cot (60^o – A)
= $\frac{1}{tanA}+\frac{1-tan60^otanA}{tan60^o+tanA}-\frac{1+tan60^otanA}{tan60^o-tanA}$
= $\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}+tanA}-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}$
= $\frac{3-tan^2A-8tan^2A}{3tanA-tan^3A}$
= $\frac{3(1-3tan^2A)}{3tanA-tan^3A}$
= $\frac{3}{tan3A}$
= 3 cot 3A
= R.H.S.
Hence, proved.

Question 8. cot A + cot (60^o + A) + cot (120^o + A) = 3 cot 3A

Solution:

We have,
L.H.S. = cot A + cot (60^o + A) + cot (120^o + A)
= cot A + cot (60^o + A) – cot (180 – (120^o + A))
= cot A + cot (60^o + A) – cot (60^o – A)
= $\frac{1}{tanA}+\frac{1-tan60^otanA}{tan60^o+tanA}-\frac{1+tan60^otanA}{tan60^o-tanA}$
= $\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}+tanA}-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}$
= $\frac{3-tan^2A-8tan^2A}{3tanA-tan^3A}$
= $\frac{3(1-3tan^2A)}{3tanA-tan^3A}$
= $\frac{3}{tan3A}$
= 3 cot 3A
= R.H.S.
Hence, proved.

Question 9. $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)=\frac{-3}{4}sin3A$

Solution:

We have,
L.H.S. = $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)$
= $(\frac{3sinA-sin3A}{4})+\left[\frac{3sin(\frac{2π}{3}+A)-sin3(\frac{2π}{3}+A)}{4}\right]+\left[\frac{3sin(\frac{4π}{3}+A)-sin3(\frac{4π}{3}+A)}{4}\right]$
= $(\frac{3sinA-sin3A}{4})+\left[\frac{3sin(π-(\frac{π}{3}-A))-sin(2π+3A)}{4}\right]+\left[\frac{3sin(π+(\frac{π}{3}+A))-sin(4π+3A)}{4}\right]$
= $\frac{1}{4}\left[3sinA-sin3A+3sin(\frac{π}{3}-A)-sin3A-{3sin(\frac{π}{3}+A)-sin3A}\right]$
= $\frac{1}{4}\left[3sinA-3sin3A+3(sin(\frac{π}{3}-A)-sin(\frac{π}{3}+A))\right]$
= $\frac{1}{4}\left[3sinA-3sin3A+3(2cos\frac{\frac{π}{3}-A+\frac{π}{3}+A}{2}sin\frac{\frac{π}{3}-A-\frac{π}{3}-A}{2})\right]$
= $\frac{1}{4}\left[3sinA-3sin3A+6cos\frac{π}{3}sin(-A)\right]$
= $\frac{1}{4}\left[3sinA-3sin3A-3sinA\right]$
= $\frac{-3}{4}sin3A$
= R.H.S.
Hence, proved.

Question 10. |sin θ sin (60–θ) sin (60+θ)| ≤ 1/4 for all values of θ.

Solution:

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.

Question 11. |cos θ cos (60–θ) cos (60+θ)| ≤ 1/4 for all values of θ.

Solution:

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 – cos²θ))|
= |cos θ/4 (–3 + 4cos² θ)|
= |1/4 (4cos³ θ – 3cos θ)|
= |1/4 (cos 3θ)| ≤ 1/4
Hence, proved.

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Last Updated : 28 Apr, 2021

Related Articles

Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.2

Prove that:

Question 1. sin 5θ = 5sin θ – 20 sin³θ + 16 sin⁵θ

Question 2. 4 (cos³ 10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)

Question 3. cos³ θ sin 3θ + sin³ θ cos 3θ = (3sin 4θ)/4

Question 4. sin 5A = 5 cos⁴ A sin A – 10 cos² A sin³ A + sin⁵ A

Question 5. tan A tan (A + 60^o) + tan A tan (A – 60^o) + tan (A + 60^o) tan (A – 60^o) = –3

Question 6. tan A + tan (60^o+ A) – tan (60^o – A) = 3 tan 3A

Question 7. cot A + cot (60^o + A) – cot (60^o – A) = 3 cot 3A

Question 8. cot A + cot (60^o + A) + cot (120^o + A) = 3 cot 3A

Question 9. $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)=\frac{-3}{4}sin3A$

Question 10. |sin θ sin (60–θ) sin (60+θ)| ≤ 1/4 for all values of θ.

Question 11. |cos θ cos (60–θ) cos (60+θ)| ≤ 1/4 for all values of θ.

CBSE Class 12 Computer Science

GATE CS & IT 2024

Data Structures & Algorithms in JavaScript - Self Paced

Related Articles

Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.2

Prove that:

Question 1. sin 5θ = 5sin θ – 20 sin3 θ + 16 sin5 θ

Question 2. 4 (cos3 10o + sin3 20o) = 3 (cos 10o + sin 20o)

Question 3. cos3 θ sin 3θ + sin3 θ cos 3θ = (3sin 4θ)/4

Question 4. sin 5A = 5 cos4 A sin A – 10 cos2 A sin3 A + sin5 A

Question 5. tan A tan (A + 60o) + tan A tan (A – 60o) + tan (A + 60o) tan (A – 60o) = –3

Question 6. tan A + tan (60o + A) – tan (60o – A) = 3 tan 3A

Question 7. cot A + cot (60o + A) – cot (60o – A) = 3 cot 3A

Question 8. cot A + cot (60o + A) + cot (120o + A) = 3 cot 3A

Question 9.

Question 10. |sin θ sin (60–θ) sin (60+θ)| ≤ 1/4 for all values of θ.

Question 11. |cos θ cos (60–θ) cos (60+θ)| ≤ 1/4 for all values of θ.

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CBSE Class 12 Computer Science

GATE CS & IT 2024

Data Structures & Algorithms in JavaScript - Self Paced

Question 1. sin 5θ = 5sin θ – 20 sin³θ + 16 sin⁵θ

Question 2. 4 (cos³ 10^o + sin³ 20^o) = 3 (cos 10^o + sin 20^o)

Question 3. cos³ θ sin 3θ + sin³ θ cos 3θ = (3sin 4θ)/4

Question 4. sin 5A = 5 cos⁴ A sin A – 10 cos² A sin³ A + sin⁵ A

Question 5. tan A tan (A + 60^o) + tan A tan (A – 60^o) + tan (A + 60^o) tan (A – 60^o) = –3

Question 6. tan A + tan (60^o+ A) – tan (60^o – A) = 3 tan 3A

Question 7. cot A + cot (60^o + A) – cot (60^o – A) = 3 cot 3A

Question 8. cot A + cot (60^o + A) + cot (120^o + A) = 3 cot 3A

Question 9. $sin^3A+sin^3(\frac{2π}{3}+A)+sin^3(\frac{4π}{3}+A)=\frac{-3}{4}sin3A$