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# RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles Ex-9.3 Interview Questions Answers

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Question 1 :

Provethat:

sin2 2π/5– sin2 π/3 = (√5 – 1)/8

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.

Question 2 :

sin2 24o –sin2 6o = (√5 – 1)/8

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.

Question 3 :

sin2 42o –cos2 78o = (√5 + 1)/8

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.

Question 4 :

cos 78o cos42o cos 36o = 1/8

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.

Question 5 : cos π/15 cos 2π/15 cos 4π/15 cos 7π/15 = 1/16

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, 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 sinπ/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 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 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 sinπ/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.

Todays Deals  ### RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles Ex-9.3 Contributors krishan

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