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

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)

= 

= 

= 

= 

= 

= –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)

= 

= 

= 

= 

= 

= 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)

= 

= 

= 

= 

= 

= 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)

= 

= 

= 

= 

= 

= 3 cot 3A

= R.H.S.

Hence, proved.

Question 9.

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= 

= 

= 

= 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.