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• RD Sharma Class 11 Solutions for Maths

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

### 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 10o + sin3 20o) = 3 (cos 10o + sin 20o)

Solution:

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.

### 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 + 60o) + tan A tan (A – 60o) + tan (A + 60o) tan (A – 60o) = –3

Solution:

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.

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

Solution:

We have,

L.H.S. = tan A + tan (60o + A) – tan (60o – A)

= 3 tan 3A

= R.H.S.

Hence, proved.

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

Solution:

We have,

L.H.S. = cot A + cot (60o + A) – cot (60o – A)

= 3 cot 3A

= R.H.S.

Hence, proved.

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

Solution:

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.

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.

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