This Exercise deals with problems based on the values of trigonometric functions at ‘x’ in terms of values at ‘x/3’. The solutions here are explained in the most simple and understandable language, which helps students to improve their problem-solving skills. The presentation of each solution in this exercise is done in a unique way by the expert tutors at BYJU’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.
RD Sharma Solutions for Class 11 Maths Exercise 9.3 Chapter 9 – Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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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 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 24o – sin2 6o = (√5 – 1)/8
Solution:
Let us consider LHS:
sin2 24o – sin2 6o
we know, sin (A + B) sin (A – B) = sin2A – sin2B
Then the above equation becomes,
sin2 24o – sin2 6o = sin (24o + 6o) – sin (24o – 6o)
= sin 30o – sin 18o
= sin 30o – (√5 – 1)/4 [since sin 18o = (√5 – 1)/4]
= 1/2 × (√5 – 1)/4
= (√5 – 1)/8
= RHS
Hence proved.
3. sin2 42o – cos2 78o = (√5 + 1)/8
Solution:
Let us consider LHS:
sin2 42o – cos2 78o = sin2 (90o – 48o) – cos2 (90o – 12o)
= cos2 48o – sin2 12o [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 (48o + 12o) cos (48o – 12o)
= cos 60o cos 36o [since, cos 36o = (√5 + 1)/4]
= 1/2 × (√5 + 1)/4
= (√5 + 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 – B)
Then the above equation becomes,
= 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 [since, cos(180° – A) = – A]
= 1/2 (-1/2 + (√5 + 1)/4) ((√5 + 1)/4) [since, cos 36o = (√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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