Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.3

Prove that:

Question 1.  sin² 72^o – sin² 60^o = (√5 – 1)/8

Solution:

We have,

L.H.S. = sin² 72^o – sin² 60^o

= sin² (90^o–18^o) – sin² 60^o

= cos² 18^o – sin² 60^o

= 

= 

= 

= 

= 

= R.H.S.

Hence, proved.

Question 2. sin² 24^o – sin² 6^o = (√5 – 1)/8

Solution:

We have,

L.H.S. = sin² 24^o – sin² 6^o

= sin (24^o + 6^o) sin (24^o – 6^o)

= (sin 30^o) (sin 18^o)

= (1/2) × (√5 – 1)/4

= (√5 – 1)/8

= R.H.S.

Hence, proved.

Question 3. sin² 42^o – cos² 78^o = (√5 + 1)/8

Solution:

We have,

L.H.S. = sin² 42^o – cos² 78^o

= sin² (90^o–48^o) – cos² (90^o–12^o)

= cos² 48^o – sin² 12^o 

= cos (48^o + 12^o) cos (48^o – 12^o)

= cos 60^o cos 36^o

= (1/2) × (√5 + 1)/4

= (√5 + 1)/8

= R.H.S.

Hence, proved.

Question 4. cos 78^o cos 42^o cos 36^o = 1/8

Solution:

We have,

L.H.S. = cos 78^o cos 42^o cos 36^o

= (1/2) (2cos 78^o cos 42^o) (cos 36^o) 

= 1/2 [cos (78^o + 42^o) + cos (78^o – 42^o)] (cos 36^o) 

= 1/2 [(cos 120^o + cos 36^o)] (cos 36^o)

= 1/2 (cos (180^o – 60^o) + cos 36^o) (cos 36^o)

= 1/2 (–cos 60^o + cos 36^o) (cos 36^o)

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 5. 

Solution:

We have,

L.H.S. = 

= 

= 

= 

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 6. 

Solution:

We have,

L.H.S. = 

=  

= 

= 

= 

= 

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 7. cos 6^o cos 42^o cos 66^o cos 78^o = 1/16

Solution:

We have,

L.H.S. = cos 6^o cos 42^o cos 66^o cos 78^o

= (1/4) (2cos 6^o cos 66^o) (2cos 42^o cos 78^o)

= (1/4) (cos 72^o + cos 60^o) (cos 120^o + cos 36^o)

= (1/4) (sin 18^o + cos 60^o) (cos 36^o − cos 60^o)

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 8. sin 6^o sin 42^o sin 66^o sin 78^o = 1/16

Solution:

We have,

L.H.S. = sin 6^o sin 42^o sin 66^o sin 78^o

=  (1/4) (2sin 6^o sin 66^o) (2sin 42^o sin 78^o)

= (1/4) (cos 60^o − cos 72^o) (cos 36^o − cos 120^o)

= (1/4) (cos 60^o − sin 18^o) (cos 36^o + cos 60^o)

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 9. cos 36^o cos 42^o cos 60^o cos 78^o = 1/16

Solution:

We have,

L.H.S. = cos 36^o cos 42^o cos 60^o cos 78^o

= (1/2) cos 36^o cos 60^o (2cos 42^o cos 78^o)

= (1/2) cos 36^o cos 60^o (cos 120^o + cos 36^o)

= (1/2) cos 36^o cos 60^o (cos 36^o − cos 60^o)

= 

= 

= 

= 

= 

= 

= R.H.S.

Hence proved.

Question 10. sin 36^o sin 72^o sin 108^o sin 144^o = 5/16

Solution:

We have,

L.H.S. = sin 36^o sin 72^o sin 108^o sin 144^o 

= sin 36^o sin 72^o sin (180^o−72^o) sin (180^o−36^o) 

=  sin 36^o sin 72^o sin 72^o sin 36^o 

 = (1/4) (2sin 36^o sin 72^o)²

= (1/4) (2sin 36^o cos 18^o)²

= 

= 

= 

= 

= R.H.S.

Hence proved.