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RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles Ex-9.2 Interview Questions Answers

Question 1 : sin 5x = 5 sin x – 20 sin3 x + 16 sin5 x

Answer 1 :

Let us consider LHS:

sin 5x

Now,

sin 5x = sin (3x + 2x)

But we know,

Sin (x + y) = sin x cos y + cos x sin y…..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)

And

cos (x + y) = cos x cos y – sin x sin y……(iii)

Now substituting equation (i) and (iii) in equation(ii), we get

sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos2x cos x – sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos2 2x sin x+ (sin 2x cos 2x cos x – sin2 2x sin x)

= 2sin 2x cos 2x cos x + cos2 2x sin x– sin2 2x sin x …….(iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos2x – sin2x………(vi)

Substituting equation (v) and (vi) in equation (iv),we get

sin 5x = 2(2sin x cos x) (cos2x –sin2x)cos x + (cos2x – sin2x)sin x – (2sin xcos x)sin x

= 4(sin x cos2 x) ([1– sin2x]– sin2x) + ([1–sin2x] – sin2x)sinx – (4sin2 x cos2 x)sin x

(as cos2x + sin2x = 1  cos2x =1– sin2x)

sin 5x = 4(sin x [1 – sin2x]) (1 – 2sin2x)+ (1 – 2sin2x)sin x – 4sin3 x [1 –sin2x]

= 4sin x (1 – sin2x) (1 – 2sinx)+ (1 – 4sin2x + 4sin4x) sin x – 4sin3 x +4sin5x

= (4sin x – 4sin3x) (1 – 2sin2x)+ sin x – 4sin3x + 4sin5x – 4sin3 x +4sin5x

= 4sin x – 8sin3x – 4sin3x +8sin5x + sin x – 8sin3x + 8sin5x

= 5sin x – 20sin3x + 16sin5x

= RHS

Hence proved.

Question 2 : 4 (cos3 10o +sin3 20o) = 3 (cos 10o + sin 20o)

Answer 2 :

Let us consider LHS:

4 (cos3 10o + sin3 20o)

We know that, sin 60o = 3/2 = cos30o

Sin 30o = cos 60o =1/2

So,

Sin (3×20o) = cos (3×10o)

3sin 20°– 4sin320° = 4cos310° –3cos 10°

(we know, sin 3θ = 3sin θ – 4sin3 θand cos 3θ = 4cos3θ – 3cosθ)

So,

4(cos310°+sin320°) = 3(sin20°+cos 10°)

= RHS

Hence proved.

Question 3 :

cos3 xsin 3x + sin3 x cos 3x = 3/4 sin 4x

Answer 3 :

We know that,

cos 3θ = 4cos3θ – 3cosθ

So, 4 cos3θ = cos3θ + 3cosθ

cos3 θ = [cos3θ + 3cosθ]/4 …… (i)

Similarly,

sin 3θ = 3sin θ – 4sin3 θ

4 sin3θ = 3sinθ – sin 3θ

sin3θ = [3sinθ – sin 3θ]/4 …….. (ii)

Now,

Let us consider LHS:

cos3 x sin 3x + sin3 xcos 3x

Substituting the values from equation (i) and (ii), weget

cos3 x sin 3x + sin3 xcos 3x = (cos 3x + 3 cos x)/4 sin 3x + (3sin x – sin 3x)/4 cos 3x

= 1/4 (sin 3x cos 3x + 3 sin 3x cox x + 3sin x cos 3x– sin 3x cos 3x)

= 1/4 (3(sin 3x cos x + sin x cos 3x) + 0)

= 1/4 (3 sin (3x + x))

(We know, sin(x + y) = sin x cos y + cos x sin y)

= 3/4 sin 4x

= RHS

Hence proved.

Question 4 :

sin 5x =5 cos4 x sin x – 10 cos2 x sin3 x+ sin5 x

Answer 4 :

Let us consider LHS:

sin 5x

Now,

sin 5x = sin (3x + 2x)

But we know,

Sin (x + y) = sin x cos y + cos x sin y…..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)

And

cos (x + y) = cos x cos y – sin x sin y……(iii)

Now substituting equation (i) and (iii) in equation(ii), we get

sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos2x cos x – sin 2x sin x) sin 2x … (iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos2x – sin2x………(vi)

Substituting equation (v) and (vi) in equation (iv),we get

sin 5x = [(2 sin x cos x) cos x + (cos2x –sin2x) sin x] (cos2x – sin2x) + [(cos2x– sin2x) cos x – (2 sin x cos x) sin x)] (2 sin x cos x)

= [2 sin x cos2 x + sin x cos2x– sin3x] (cos2x – sin2x) + [cos3x –sin2x cos x – 2 sin2 x cos x] (2 sin x cos x)

= cos2x [3 sin x cos2 x –sin3x] – sin2x [3 sin x cos2 x – sin3x]+ 2 sin x cos4x – 2 sin3 x cos2 x –4 sin3 x cos2 x

= 3 sin x cos4 x – sin3xcos2x – 3 sin3 x cos2 x – sin5x+ 2 sin x cos4x – 2 sin3 x cos2 x –4 sin3 x cos2 x

= 5 sin x cos4 x –10sin3xcos2x+sin5x

= RHS

Hence proved.

Question 5 :

sin 5x =5 sin x – 20 sin3 x + 16 sin5 x

Answer 5 :

Let us consider LHS:

sin 5x

Now,

sin 5x = sin (3x + 2x)

But we know,

Sin (x + y) = sin x cos y + cos x sin y…..(i)

So,

sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin (2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)

And

cos (x + y) = cos x cos y – sin x sin y……(iii)

Now substituting equation (i) and (iii) in equation(ii), we get

sin 5x = (sin 2x cos x + cos 2x sin x ) cos 2x + ( cos2x cos x – sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos2 2x sin x+ (sin 2x cos 2x cos x – sin2 2x sin x)

= 2sin 2x cos 2x cos x + cos2 2x sin x– sin2 2x sin x …….(iv)

Now sin 2x = 2sin x cos x………(v)

And cos 2x = cos2x – sin2x………(vi)

Substituting equation (v) and (vi) in equation (iv),we get

sin 5x = 2(2sin x cos x) (cos2x –sin2x)cos x + (cos2x – sin2x)sin x – (2sin xcos x)sin x

= 4(sin x cos2 x) ([1– sin2x]– sin2x) + ([1–sin2x] – sin2x)sinx – (4sin2 x cos2 x)sin x

(as cos2x + sin2x = 1  cos2x =1– sin2x)

sin 5x = 4(sin x [1 – sin2x]) (1 – 2sin2x)+ (1 – 2sin2x)sin x – 4sin3 x [1 –sin2x]

= 4sin x (1 – sin2x) (1 – 2sinx)+ (1 – 4sin2x + 4sin4x) sin x – 4sin3 x +4sin5x

= (4sin x – 4sin3x) (1 – 2sin2x)+ sin x – 4sin3x + 4sin5x – 4sin3 x +4sin5x

= 4sin x – 8sin3x – 4sin3x +8sin5x + sin x – 8sin3x + 8sin5x

= 5sin x – 20sin3x + 16sin5x

= RHS

Hence proved.

Question 6 :

Answer 6 :



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RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles Ex-9.2 Contributors

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