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

Chapter 9 of RD Sharma's Class 11 Mathematics textbook delves into the advanced concepts of the trigonometry specifically focusing on the trigonometric ratios of the multiple and submultiple angles. This chapter is crucial for students as it builds on the foundational knowledge of the trigonometric identities and extends it to the more complex scenarios which are essential for the solving higher-level problems in the trigonometry.

Trigonometric Ratios of Multiple and Submultiple Angles

The trigonometric ratios of multiple angles refer to the trigonometric functions of the angles that are multiples of a given angle such as the 2θ, 3θ and so on. The Submultiple angles involve dividing the angle such as the 𝜃\2. These concepts are vital in solving complex trigonometric equations and in the analysis of waveforms, oscillations and other periodic phenomena. Understanding these ratios allows for the simplification and solution of the trigonometric expressions in the various mathematical and physical applications.

Prove that:

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

Solution:

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.

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 (cos³ 10^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 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.

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 – cos² θ))|

= |cos θ/4 (–3 + 4cos² θ)|

= |1/4 (4cos³ θ – 3cos θ)|

= |1/4 (cos 3θ)| ≤ 1/4

Hence, proved.

Conclusion

In this exercise, students practice applying the formulas for the trigonometric ratios of the multiple and submultiple angles to solve problems. The Mastering these concepts is essential for the success in advanced trigonometry and related fields. By working through these problems students will gain a deeper understanding of how these ratios are derived and applied in the different contexts.