RD Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles Ex 9.2 Solutions
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)2 sin x – (2sin xcos x)2 sin x
= 4(sin x cos2 x) ([1– sin2x]– sin2x) + ([1–sin2x] – sin2x)2 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)2 sin x – 4sin3 x [1 –sin2x]
= 4sin x (1 – sin2x) (1 – 2sin2 x)+ (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)2 sin x – (2sin xcos x)2 sin x
= 4(sin x cos2 x) ([1– sin2x]– sin2x) + ([1–sin2x] – sin2x)2 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)2 sin x – 4sin3 x [1 –sin2x]
= 4sin x (1 – sin2x) (1 – 2sin2 x)+ (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 : -