This Exercise deals with problems based on the values of trigonometric functions at â€˜xâ€™ in terms of values at â€˜x/3â€™. The solutions here are explained in the most simple and understandable language which helps students to improve their grasping abilities. The presentation of each solution in this exercise is described in a unique way by the expert tutors at BYJUâ€™S. To score good marks, practising RD Sharma Class 11 Maths Solutions can help to a great extent. The pdf of RD Sharma Solutions for Class 11 Maths Chapter 9 exercise 9.3 is provided here, which students can download for free easily.

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Exercise 9.1 Solutions

Exercise 9.2 solutions

### Access answers to RD Sharma Solutions for Class 11 Maths Exercise 9.3 Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle

Prove that:

1. sin2 2Ï€/5 â€“ sin2 Ï€/3 = (âˆš5 – 1)/8

Solution:

Let us consider LHS:

sin2 2Ï€/5 â€“ sin2 Ï€/3 = sin2 (Ï€/2 – Ï€/10) â€“ sin2 Ï€/3

we know, sin (90Â°â€“ A) = cos A

So, sin2 (Ï€/2 – Ï€/10) = cos2 Ï€/10

Sin Ï€/3 = âˆš3/2

Then the above equation becomes,

= Cos2 Ï€/10 â€“ (âˆš3/2)2

We know, cos Ï€/10 = âˆš(10+2âˆš5)/4

the above equation becomes,

= [âˆš(10+2âˆš5)/4]2 â€“ 3/4

= [10 + 2âˆš5]/16 â€“ 3/4

= [10 + 2âˆš5 – 12]/16

= [2âˆš5 â€“ 2]/16

= [âˆš5 â€“ 1]/8

= RHS

Hence proved.

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

Solution:

Let us consider LHS:

sin2 24o â€“ sin2 6o

we know, sin (A + B) sin (A â€“ B) = sin2A â€“ sin2B

Then the above equation becomes,

sin2 24o â€“ sin2 6o = sin (24o + 6o) â€“ sin (24o â€“ 6o)

= sin 30o â€“ sin 18o

= sin 30o â€“ (âˆš5 – 1)/4 [since, sin 18o = (âˆš5 – 1)/4]

= 1/2 Ã— (âˆš5 – 1)/4

= (âˆš5 – 1)/8

= RHS

Hence proved.

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

Solution:

Let us consider LHS:

sin2 42o â€“ cos2 78o = sin2 (90o â€“ 48o) â€“ cos2 (90o â€“ 12o)

= cos2 48o â€“ sin2 12o [since, sin (90 – A) = cos A and cos (90 – A) = sin A]

We know, cos (A + B) cos (A â€“ B) = cos2A â€“ sin2B

Then the above equation becomes,

= cos2 (48o + 12o) cos (48o â€“ 12o)

= cos 60o cos 36o [since, cos 36o = (âˆš5 + 1)/4]

= 1/2 Ã— (âˆš5 + 1)/4

= (âˆš5 + 1)/8

= RHS

Hence proved.

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

Solution:

Let us consider LHS:

cos 78o cos 42o cos 36o

Let us multiply and divide by 2 we get,

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

We know, 2 cos A cos B = cos (A + B) + cos (A â€“ B)

Then the above equation becomes,

= 1/2 (cos (78o + 42o) + cos (78o â€“ 42o)) Ã— cos 36o

= 1/2 (cos 120o + cos 36o) Ã— cos 36o

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

= 1/2 (-cos (60o) + cos 36o) Ã— cos 36o [since, cos(180Â° â€“ A) = â€“ A]

= 1/2 (-1/2 + (âˆš5 + 1)/4) ((âˆš5 + 1)/4) [since, cos 36o = (âˆš5 + 1)/4]

= 1/2 (âˆš5 + 1 – 2)/4 ((âˆš5 + 1)/4)

= 1/2 (âˆš5 – 1)/4) ((âˆš5 + 1)/4)

= 1/2 ((âˆš5)2 – 12)/16

= 1/2 (5-1)/16

= 1/2 (4/16)

= 1/8

= RHS

Hence proved.

5. cos Ï€/15 cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15 = 1/16

Solution:

Let us consider LHS:

cos Ï€/15 cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15

Let us multiply and divide by 2 sin Ï€/15, we get,

= [2 sin Ï€/15 cos Ï€/15] cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15] / 2 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 2Ï€/15) cos 2Ï€/15 cos 4Ï€/15 cos 7Ï€/15] / 2 sin Ï€/15

Now, multiply and divide by 2 we get,

= [(2 sin 2Ï€/15 cos 2Ï€/15) cos 4Ï€/15 cos 7Ï€/15] / 2 Ã— 2 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 4Ï€/15) cos 4Ï€/15 cos 7Ï€/15] / 4 sin Ï€/15

Now, multiply and divide by 2 we get,

= [(2 sin 4Ï€/15 cos 4Ï€/15) cos 7Ï€/15] / 2 Ã— 4 sin Ï€/15

We know, 2sin A cos A = sin 2A

Then the above equation becomes,

= [(sin 8Ï€/15) cos 7Ï€/15] / 8 sin Ï€/15

Now, multiply and divide by 2 we get,

= [2 sin 8Ï€/15 cos 7Ï€/15] / 2 Ã— 8 sin Ï€/15

We know, 2sin A cos B = sin (A+B) + sin (Aâ€“B)

Then the above equation becomes,

= [sin (8Ï€/15 + 7Ï€/15) + sin (8Ï€/15 – 7Ï€/15)] / 16 sin Ï€/15

= [sin (Ï€) + sin (Ï€/15)] / 16 sin Ï€/15

= [0 + sin (Ï€/15)] / 16 sin Ï€/15

= sin (Ï€/15) / 16 sin Ï€/15

= 1/16

= RHS

Hence proved.