GeeksforGeeks App
Open App
Browser
Continue

## Related Articles

• RD Sharma Class 11 Solutions for Maths

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

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

Solution:

We have,

L.H.S. = sin2 72o – sin2 60o

= sin2 (90o–18o) – sin2 60o

= cos2 18o – sin2 60o

= R.H.S.

Hence, proved.

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

Solution:

We have,

L.H.S. = sin2 24o – sin2 6o

= sin (24o + 6o) sin (24o – 6o)

= (sin 30o) (sin 18o)

= (1/2) × (√5 – 1)/4

= (√5 – 1)/8

= R.H.S.

Hence, proved.

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

Solution:

We have,

L.H.S. = sin2 42o – cos2 78o

= sin2 (90o–48o) – cos2 (90o–12o)

= cos2 48o – sin2 12o

= cos (48o + 12o) cos (48o – 12o)

= cos 60o cos 36o

= (1/2) × (√5 + 1)/4

= (√5 + 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 – 42o)] (cos 36o)

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

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

= 1/2 (–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 − 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 − cos 72o) (cos 36o − cos 120o)

= (1/4) (cos 60o − 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 − 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 (180o−72o) sin (180o−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.

My Personal Notes arrow_drop_up
Related Tutorials