RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3 (Updated For 2024)

RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3: Featuring a large number of worked examples and examples, RD Sharma Solutions Class 11 Maths provides step-by-step explanations of many difficult concepts and includes a wide variety of questions to practice. Prepare for your Class 11 Maths exams by downloading the Free PDF of RD Sharma Solutions Class 11 Maths Chapter 9 Exercise 9.3.

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RD Sharma Class 11 Solutions Chapter 9 Exercise 9.3

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Prove that:

1. sin² 2π/5 – sin² π/3 = (√5 – 1)/8

Solution:

Let us consider LHS:

sin² 2π/5 – sin² π/3 = sin² (π/2 – π/10) – sin² π/3

we know, sin (90°– A) = cos A

So, sin² (π/2 – π/10) = cos² π/10

Sin π/3 = √3/2

Then the above equation becomes,

= Cos² π/10 – (√3/2)²

We know, cos π/10 = √(10+2√5)/4

the above equation becomes,

= [√(10+2√5)/4]² – 3/4

= [10 + 2√5]/16 – 3/4

= [10 + 2√5 – 12]/16

= [2√5 – 2]/16

= [√5 – 1]/8

= RHS

Hence proved.

2. sin² 24^o – sin² 6^o = (√5 – 1)/8

Solution:

Let us consider LHS:

sin² 24^o – sin² 6^o

we know, sin (A + B) sin (A – B) = sin²A – sin²B

Then the above equation becomes,

sin² 24^o – sin² 6^o = sin (24^o + 6^o) – sin (24^o – 6^o)

= sin 30^o – sin 18^o

= sin 30^o – (√5 – 1)/4 [since sin 18^o = (√5 – 1)/4]

= 1/2 × (√5 – 1)/4

= (√5 – 1)/8

= RHS

Hence proved.

3. sin² 42^o – cos² 78^o = (√5 + 1)/8

Solution:

Let us consider LHS:

sin² 42^o – cos² 78^o = sin² (90^o – 48^o) – cos² (90^o – 12^o)

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

We know, cos (A + B) cos (A – B) = cos²A – sin²B

Then the above equation becomes,

= cos² (48^o + 12^o) cos (48^o – 12^o)

= cos 60^o cos 36^o [since, cos 36^o = (√5 + 1)/4]

= 1/2 × (√5 + 1)/4

= (√5 + 1)/8

= RHS

Hence proved.

4. cos 78^o cos 42^o cos 36^o = 1/8

Solution:

Let us consider LHS:

cos 78^o cos 42^o cos 36^o

Let us multiply and divide by 2 we get,

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

We know, 2 cos A cos B = cos (A + B) + cos (A – B)

Then the above equation becomes,

= 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 [since, cos(180° – A) = – A]

= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36^o = (√5 + 1)/4]

= 1/2 (√5 + 1 – 2)/4 ((√5 + 1)/4)

= 1/2 (√5 – 1)/4) ((√5 + 1)/4)

= 1/2 ((√5)² – 1²)/16

= 1/2 (5-1)/16

= 1/2 (4/16)

= 1/8

= RHS

Hence proved.

5. cos π/15 cos 2π/15 cos 4π/15 cos 7π/15 = 1/16

Solution:

Let us consider LHS:

cos π/15 cos 2π/15 cos 4π/15 cos 7π/15

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

= [2 sin π/15 cos π/15] cos 2π/15 cos 4π/15 cos 7π/15] / 2 sin π/15

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

Then the above equation becomes,

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

Now, multiply and divide by 2 we get,

= [(2 sin 2π/15 cos 2π/15) cos 4π/15 cos 7π/15] / 2 × 2 sin π/15

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

Then the above equation becomes,

= [(sin 4π/15) cos 4π/15 cos 7π/15] / 4 sin π/15

Now, multiply and divide by 2, and we get,

= [(2 sin 4π/15 cos 4π/15) cos 7π/15] / 2 × 4 sin π/15

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

Then the above equation becomes,

= [(sin 8π/15) cos 7π/15] / 8 sin π/15

Now, multiply and divide by 2, and we get,

= [2 sin 8π/15 cos 7π/15] / 2 × 8 sin π/15

We know, 2sin A cos B = sin (A+B) + sin (A–B)

Then the above equation becomes,

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

= [sin (π) + sin (π/15)] / 16 sin π/15

= [0 + sin (π/15)] / 16 sin π/15

= sin (π/15) / 16 sin π/15

= 1/16

= RHS

Hence proved.

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