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

• Last Updated : 28 Apr, 2021

### Question 1.  sin2 72o â€“ sin2 60o = (âˆš5 â€“ 1)/8

Solution:

We have,

L.H.S. = sin2 72o â€“ sin2 60o

= sin2 (90oâ€“18o) â€“ sin2 60o

= cos2 18o â€“ sin2 60o

= R.H.S.

Hence, proved.

### Question 2. sin2 24o â€“ sin2 6o = (âˆš5 â€“ 1)/8

Solution:

We have,

L.H.S. = sin2 24o â€“ sin2 6o

= sin (24o + 6o) sin (24o â€“ 6o)

= (sin 30o) (sin 18o)

= (1/2) Ã— (âˆš5 â€“ 1)/4

= (âˆš5 â€“ 1)/8

= R.H.S.

Hence, proved.

### Question 3. sin2 42o â€“ cos2 78o = (âˆš5 + 1)/8

Solution:

We have,

L.H.S. = sin2 42o â€“ cos2 78o

= sin2 (90oâ€“48o) â€“ cos2 (90oâ€“12o)

= cos2 48o â€“ sin2 12o

= cos (48o + 12o) cos (48o â€“ 12o)

= cos 60o cos 36o

= (1/2) Ã— (âˆš5 + 1)/4

= (âˆš5 + 1)/8

= R.H.S.

Hence, proved.

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

Solution:

We have,

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

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

= 1/2 [cos (78o + 42o) + cos (78o â€“ 42o)] (cos 36o)

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

= 1/2 (cos (180o â€“ 60o) + cos 36o) (cos 36o)

= 1/2 (â€“cos 60o + cos 36o) (cos 36o)

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

= R.H.S.

Hence proved.

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

Solution:

We have,

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

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

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

= (1/4) (sin 18o + cos 60o) (cos 36o âˆ’ cos 60o)

= R.H.S.

Hence proved.

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

Solution:

We have,

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

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

= (1/4) (cos 60o âˆ’ cos 72o) (cos 36o âˆ’ cos 120o)

= (1/4) (cos 60o âˆ’ sin 18o) (cos 36o + cos 60o)

= R.H.S.

Hence proved.

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

Solution:

We have,

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

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

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

= (1/2) cos 36o cos 60o (cos 36o âˆ’ cos 60o)

= R.H.S.

Hence proved.

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

Solution:

We have,

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

= sin 36o sin 72o sin (180oâˆ’72o) sin (180oâˆ’36o

=  sin 36o sin 72o sin 72o sin 36o

= (1/4) (2sin 36o sin 72o)2

= (1/4) (2sin 36o cos 18o)2

= R.H.S.

Hence proved.

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